YES

We show the termination of the TRS R:

  active(fst(|0|(),Z)) -> mark(nil())
  active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
  active(from(X)) -> mark(cons(X,from(s(X))))
  active(add(|0|(),X)) -> mark(X)
  active(add(s(X),Y)) -> mark(s(add(X,Y)))
  active(len(nil())) -> mark(|0|())
  active(len(cons(X,Z))) -> mark(s(len(Z)))
  mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
  mark(|0|()) -> active(|0|())
  mark(nil()) -> active(nil())
  mark(s(X)) -> active(s(X))
  mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
  mark(from(X)) -> active(from(mark(X)))
  mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
  mark(len(X)) -> active(len(mark(X)))
  fst(mark(X1),X2) -> fst(X1,X2)
  fst(X1,mark(X2)) -> fst(X1,X2)
  fst(active(X1),X2) -> fst(X1,X2)
  fst(X1,active(X2)) -> fst(X1,X2)
  s(mark(X)) -> s(X)
  s(active(X)) -> s(X)
  cons(mark(X1),X2) -> cons(X1,X2)
  cons(X1,mark(X2)) -> cons(X1,X2)
  cons(active(X1),X2) -> cons(X1,X2)
  cons(X1,active(X2)) -> cons(X1,X2)
  from(mark(X)) -> from(X)
  from(active(X)) -> from(X)
  add(mark(X1),X2) -> add(X1,X2)
  add(X1,mark(X2)) -> add(X1,X2)
  add(active(X1),X2) -> add(X1,X2)
  add(X1,active(X2)) -> add(X1,X2)
  len(mark(X)) -> len(X)
  len(active(X)) -> len(X)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(fst(|0|(),Z)) -> mark#(nil())
p2: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p3: active#(fst(s(X),cons(Y,Z))) -> cons#(Y,fst(X,Z))
p4: active#(fst(s(X),cons(Y,Z))) -> fst#(X,Z)
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: active#(from(X)) -> cons#(X,from(s(X)))
p7: active#(from(X)) -> from#(s(X))
p8: active#(from(X)) -> s#(X)
p9: active#(add(|0|(),X)) -> mark#(X)
p10: active#(add(s(X),Y)) -> mark#(s(add(X,Y)))
p11: active#(add(s(X),Y)) -> s#(add(X,Y))
p12: active#(add(s(X),Y)) -> add#(X,Y)
p13: active#(len(nil())) -> mark#(|0|())
p14: active#(len(cons(X,Z))) -> mark#(s(len(Z)))
p15: active#(len(cons(X,Z))) -> s#(len(Z))
p16: active#(len(cons(X,Z))) -> len#(Z)
p17: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))
p18: mark#(fst(X1,X2)) -> fst#(mark(X1),mark(X2))
p19: mark#(fst(X1,X2)) -> mark#(X1)
p20: mark#(fst(X1,X2)) -> mark#(X2)
p21: mark#(|0|()) -> active#(|0|())
p22: mark#(nil()) -> active#(nil())
p23: mark#(s(X)) -> active#(s(X))
p24: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p25: mark#(cons(X1,X2)) -> cons#(mark(X1),X2)
p26: mark#(cons(X1,X2)) -> mark#(X1)
p27: mark#(from(X)) -> active#(from(mark(X)))
p28: mark#(from(X)) -> from#(mark(X))
p29: mark#(from(X)) -> mark#(X)
p30: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p31: mark#(add(X1,X2)) -> add#(mark(X1),mark(X2))
p32: mark#(add(X1,X2)) -> mark#(X1)
p33: mark#(add(X1,X2)) -> mark#(X2)
p34: mark#(len(X)) -> active#(len(mark(X)))
p35: mark#(len(X)) -> len#(mark(X))
p36: mark#(len(X)) -> mark#(X)
p37: fst#(mark(X1),X2) -> fst#(X1,X2)
p38: fst#(X1,mark(X2)) -> fst#(X1,X2)
p39: fst#(active(X1),X2) -> fst#(X1,X2)
p40: fst#(X1,active(X2)) -> fst#(X1,X2)
p41: s#(mark(X)) -> s#(X)
p42: s#(active(X)) -> s#(X)
p43: cons#(mark(X1),X2) -> cons#(X1,X2)
p44: cons#(X1,mark(X2)) -> cons#(X1,X2)
p45: cons#(active(X1),X2) -> cons#(X1,X2)
p46: cons#(X1,active(X2)) -> cons#(X1,X2)
p47: from#(mark(X)) -> from#(X)
p48: from#(active(X)) -> from#(X)
p49: add#(mark(X1),X2) -> add#(X1,X2)
p50: add#(X1,mark(X2)) -> add#(X1,X2)
p51: add#(active(X1),X2) -> add#(X1,X2)
p52: add#(X1,active(X2)) -> add#(X1,X2)
p53: len#(mark(X)) -> len#(X)
p54: len#(active(X)) -> len#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p2, p5, p9, p10, p14, p17, p19, p20, p23, p24, p26, p27, p29, p30, p32, p33, p34, p36}
  {p43, p44, p45, p46}
  {p37, p38, p39, p40}
  {p47, p48}
  {p41, p42}
  {p49, p50, p51, p52}
  {p53, p54}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(len(X)) -> active#(len(mark(X)))
p2: active#(len(cons(X,Z))) -> mark#(s(len(Z)))
p3: mark#(len(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(add(X1,X2)) -> mark#(X1)
p6: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p7: active#(add(s(X),Y)) -> mark#(s(add(X,Y)))
p8: mark#(from(X)) -> mark#(X)
p9: mark#(from(X)) -> active#(from(mark(X)))
p10: active#(add(|0|(),X)) -> mark#(X)
p11: mark#(cons(X1,X2)) -> mark#(X1)
p12: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p13: active#(from(X)) -> mark#(cons(X,from(s(X))))
p14: mark#(s(X)) -> active#(s(X))
p15: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p16: mark#(fst(X1,X2)) -> mark#(X2)
p17: mark#(fst(X1,X2)) -> mark#(X1)
p18: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{23, x1 - 56}
        len_A(x1) = max{57, x1 + 24}
        active#_A(x1) = max{4, x1 - 56}
        mark_A(x1) = max{16, x1}
        cons_A(x1,x2) = max{80, x1 + 62}
        s_A(x1) = 28
        add_A(x1,x2) = max{80, x1 + 14, x2 + 29}
        from_A(x1) = max{137, x1 + 62}
        |0|_A = 66
        fst_A(x1,x2) = max{x1, x2}
        active_A(x1) = max{1, x1}
        nil_A = 43
  
    precedence:
    
      s = add > mark# = active# > len = cons > mark = |0| = active > from > fst > nil
    
    partial status:
  
      pi(mark#) = []
      pi(len) = []
      pi(active#) = []
      pi(mark) = []
      pi(cons) = []
      pi(s) = []
      pi(add) = []
      pi(from) = []
      pi(|0|) = []
      pi(fst) = []
      pi(active) = []
      pi(nil) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(len(X)) -> active#(len(mark(X)))
p2: mark#(len(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> mark#(X2)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p6: active#(add(s(X),Y)) -> mark#(s(add(X,Y)))
p7: mark#(from(X)) -> mark#(X)
p8: mark#(from(X)) -> active#(from(mark(X)))
p9: active#(add(|0|(),X)) -> mark#(X)
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p12: active#(from(X)) -> mark#(cons(X,from(s(X))))
p13: mark#(s(X)) -> active#(s(X))
p14: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p15: mark#(fst(X1,X2)) -> mark#(X2)
p16: mark#(fst(X1,X2)) -> mark#(X1)
p17: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(len(X)) -> active#(len(mark(X)))
p2: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p3: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(fst(X1,X2)) -> mark#(X1)
p6: mark#(fst(X1,X2)) -> mark#(X2)
p7: mark#(s(X)) -> active#(s(X))
p8: active#(add(|0|(),X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: active#(add(s(X),Y)) -> mark#(s(add(X,Y)))
p11: mark#(cons(X1,X2)) -> mark#(X1)
p12: mark#(from(X)) -> active#(from(mark(X)))
p13: mark#(from(X)) -> mark#(X)
p14: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p15: mark#(add(X1,X2)) -> mark#(X1)
p16: mark#(add(X1,X2)) -> mark#(X2)
p17: mark#(len(X)) -> mark#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 114
        len_A(x1) = max{107, x1}
        active#_A(x1) = max{112, x1 + 108}
        mark_A(x1) = x1
        fst_A(x1,x2) = max{x1 + 47, x2 + 110}
        s_A(x1) = 70
        cons_A(x1,x2) = x1 + 71
        from_A(x1) = x1 + 185
        add_A(x1,x2) = max{x1 + 114, x2 + 114}
        |0|_A = 50
        active_A(x1) = max{48, x1}
        nil_A = 49
  
    precedence:
    
      len > mark# = active# = mark = fst = s = cons = from = add = |0| = active = nil
    
    partial status:
  
      pi(mark#) = []
      pi(len) = []
      pi(active#) = []
      pi(mark) = [1]
      pi(fst) = []
      pi(s) = []
      pi(cons) = []
      pi(from) = []
      pi(add) = []
      pi(|0|) = []
      pi(active) = [1]
      pi(nil) = []

The next rules are strictly ordered:

  p10

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(len(X)) -> active#(len(mark(X)))
p2: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p3: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(fst(X1,X2)) -> mark#(X1)
p6: mark#(fst(X1,X2)) -> mark#(X2)
p7: mark#(s(X)) -> active#(s(X))
p8: active#(add(|0|(),X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(from(X)) -> active#(from(mark(X)))
p12: mark#(from(X)) -> mark#(X)
p13: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p14: mark#(add(X1,X2)) -> mark#(X1)
p15: mark#(add(X1,X2)) -> mark#(X2)
p16: mark#(len(X)) -> mark#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(len(X)) -> active#(len(mark(X)))
p2: active#(add(|0|(),X)) -> mark#(X)
p3: mark#(len(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(add(X1,X2)) -> mark#(X1)
p6: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p7: active#(from(X)) -> mark#(cons(X,from(s(X))))
p8: mark#(from(X)) -> mark#(X)
p9: mark#(from(X)) -> active#(from(mark(X)))
p10: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p11: mark#(cons(X1,X2)) -> mark#(X1)
p12: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p13: mark#(s(X)) -> active#(s(X))
p14: mark#(fst(X1,X2)) -> mark#(X2)
p15: mark#(fst(X1,X2)) -> mark#(X1)
p16: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{43, x1 - 4}
        len_A(x1) = max{190, x1 + 148}
        active#_A(x1) = max{23, x1 - 128}
        mark_A(x1) = max{59, x1 + 17}
        add_A(x1,x2) = max{228, x1 + 113, x2 + 177}
        |0|_A = 191
        from_A(x1) = max{182, x1 + 124}
        cons_A(x1,x2) = max{44, x1}
        s_A(x1) = 28
        fst_A(x1,x2) = max{169, x1 + 111, x2 + 127}
        active_A(x1) = max{51, x1}
        nil_A = 109
  
    precedence:
    
      active# > mark > mark# = len > add = |0| = from = cons = s = fst = active = nil
    
    partial status:
  
      pi(mark#) = []
      pi(len) = [1]
      pi(active#) = []
      pi(mark) = []
      pi(add) = []
      pi(|0|) = []
      pi(from) = [1]
      pi(cons) = []
      pi(s) = []
      pi(fst) = []
      pi(active) = [1]
      pi(nil) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(add(|0|(),X)) -> mark#(X)
p2: mark#(len(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> mark#(X2)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p6: active#(from(X)) -> mark#(cons(X,from(s(X))))
p7: mark#(from(X)) -> mark#(X)
p8: mark#(from(X)) -> active#(from(mark(X)))
p9: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p12: mark#(s(X)) -> active#(s(X))
p13: mark#(fst(X1,X2)) -> mark#(X2)
p14: mark#(fst(X1,X2)) -> mark#(X1)
p15: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(add(|0|(),X)) -> mark#(X)
p2: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))
p3: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p4: mark#(fst(X1,X2)) -> mark#(X1)
p5: mark#(fst(X1,X2)) -> mark#(X2)
p6: mark#(s(X)) -> active#(s(X))
p7: active#(from(X)) -> mark#(cons(X,from(s(X))))
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(from(X)) -> active#(from(mark(X)))
p11: mark#(from(X)) -> mark#(X)
p12: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p13: mark#(add(X1,X2)) -> mark#(X1)
p14: mark#(add(X1,X2)) -> mark#(X2)
p15: mark#(len(X)) -> mark#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = x1 + 8
        add_A(x1,x2) = max{x1 + 3, x2 + 7}
        |0|_A = 24
        mark#_A(x1) = max{12, x1 + 8}
        fst_A(x1,x2) = max{x1 + 30, x2 + 31}
        mark_A(x1) = x1
        s_A(x1) = x1
        cons_A(x1,x2) = max{13, x1 + 7, x2}
        from_A(x1) = x1 + 14
        len_A(x1) = x1 + 15
        active_A(x1) = x1
        nil_A = 23
  
    precedence:
    
      active# = add = |0| = mark# = fst = mark = s = cons = from = len = active = nil
    
    partial status:
  
      pi(active#) = [1]
      pi(add) = []
      pi(|0|) = []
      pi(mark#) = [1]
      pi(fst) = []
      pi(mark) = [1]
      pi(s) = []
      pi(cons) = []
      pi(from) = []
      pi(len) = []
      pi(active) = [1]
      pi(nil) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))
p2: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p3: mark#(fst(X1,X2)) -> mark#(X1)
p4: mark#(fst(X1,X2)) -> mark#(X2)
p5: mark#(s(X)) -> active#(s(X))
p6: active#(from(X)) -> mark#(cons(X,from(s(X))))
p7: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(from(X)) -> active#(from(mark(X)))
p10: mark#(from(X)) -> mark#(X)
p11: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p12: mark#(add(X1,X2)) -> mark#(X1)
p13: mark#(add(X1,X2)) -> mark#(X2)
p14: mark#(len(X)) -> mark#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))
p2: active#(from(X)) -> mark#(cons(X,from(s(X))))
p3: mark#(len(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(add(X1,X2)) -> mark#(X1)
p6: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p7: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p8: mark#(from(X)) -> mark#(X)
p9: mark#(from(X)) -> active#(from(mark(X)))
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p12: mark#(s(X)) -> active#(s(X))
p13: mark#(fst(X1,X2)) -> mark#(X2)
p14: mark#(fst(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 25
        fst_A(x1,x2) = max{x1 + 13, x2 + 12}
        active#_A(x1) = max{36, x1 + 25}
        mark_A(x1) = x1
        from_A(x1) = max{35, x1 + 18}
        cons_A(x1,x2) = max{x1 + 15, x2}
        s_A(x1) = max{17, x1}
        len_A(x1) = x1 + 26
        add_A(x1,x2) = max{x1 + 39, x2 + 38}
        active_A(x1) = max{13, x1}
        |0|_A = 39
        nil_A = 14
  
    precedence:
    
      mark# = fst = active# = mark = from = cons = s = len = add = active = |0| = nil
    
    partial status:
  
      pi(mark#) = [1]
      pi(fst) = []
      pi(active#) = [1]
      pi(mark) = [1]
      pi(from) = []
      pi(cons) = []
      pi(s) = []
      pi(len) = []
      pi(add) = []
      pi(active) = [1]
      pi(|0|) = []
      pi(nil) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))
p2: active#(from(X)) -> mark#(cons(X,from(s(X))))
p3: mark#(add(X1,X2)) -> mark#(X2)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p6: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p7: mark#(from(X)) -> mark#(X)
p8: mark#(from(X)) -> active#(from(mark(X)))
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p11: mark#(s(X)) -> active#(s(X))
p12: mark#(fst(X1,X2)) -> mark#(X2)
p13: mark#(fst(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))
p2: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p3: mark#(fst(X1,X2)) -> mark#(X1)
p4: mark#(fst(X1,X2)) -> mark#(X2)
p5: mark#(s(X)) -> active#(s(X))
p6: active#(from(X)) -> mark#(cons(X,from(s(X))))
p7: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(from(X)) -> active#(from(mark(X)))
p10: mark#(from(X)) -> mark#(X)
p11: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p12: mark#(add(X1,X2)) -> mark#(X1)
p13: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{105, x1 + 48}
        fst_A(x1,x2) = max{x1 + 208, x2 + 167}
        active#_A(x1) = max{104, x1 + 40}
        mark_A(x1) = x1
        s_A(x1) = 66
        cons_A(x1,x2) = max{54, x1 + 13}
        from_A(x1) = x1 + 65
        add_A(x1,x2) = max{x1 + 88, x2 + 48}
        active_A(x1) = max{54, x1}
        |0|_A = 105
        nil_A = 104
        len_A(x1) = x1 + 54
  
    precedence:
    
      mark# = fst = active# = mark = s = cons = from = add = active = |0| = nil = len
    
    partial status:
  
      pi(mark#) = [1]
      pi(fst) = []
      pi(active#) = [1]
      pi(mark) = [1]
      pi(s) = []
      pi(cons) = []
      pi(from) = []
      pi(add) = []
      pi(active) = [1]
      pi(|0|) = []
      pi(nil) = []
      pi(len) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p2: mark#(fst(X1,X2)) -> mark#(X1)
p3: mark#(fst(X1,X2)) -> mark#(X2)
p4: mark#(s(X)) -> active#(s(X))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: mark#(from(X)) -> active#(from(mark(X)))
p9: mark#(from(X)) -> mark#(X)
p10: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p11: mark#(add(X1,X2)) -> mark#(X1)
p12: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p2: mark#(add(X1,X2)) -> mark#(X2)
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> active#(from(mark(X)))
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(s(X)) -> active#(s(X))
p11: mark#(fst(X1,X2)) -> mark#(X2)
p12: mark#(fst(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{49, x1 + 3}
        fst_A(x1,x2) = max{x1 + 101, x2 + 235}
        s_A(x1) = 50
        cons_A(x1,x2) = x1 + 60
        mark#_A(x1) = x1 + 60
        add_A(x1,x2) = max{x1 + 74, x2 + 100}
        mark_A(x1) = x1 + 50
        from_A(x1) = x1 + 120
        active_A(x1) = max{29, x1}
        |0|_A = 50
        nil_A = 152
        len_A(x1) = 100
  
    precedence:
    
      fst = s = cons = mark = from = active = |0| = nil > mark# > active# > add = len
    
    partial status:
  
      pi(active#) = [1]
      pi(fst) = []
      pi(s) = []
      pi(cons) = []
      pi(mark#) = [1]
      pi(add) = [2]
      pi(mark) = []
      pi(from) = []
      pi(active) = []
      pi(|0|) = []
      pi(nil) = []
      pi(len) = []

The next rules are strictly ordered:

  p10

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p2: mark#(add(X1,X2)) -> mark#(X2)
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> active#(from(mark(X)))
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(fst(X1,X2)) -> mark#(X2)
p11: mark#(fst(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p2: mark#(fst(X1,X2)) -> mark#(X1)
p3: mark#(fst(X1,X2)) -> mark#(X2)
p4: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(from(X)) -> active#(from(mark(X)))
p8: mark#(from(X)) -> mark#(X)
p9: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p10: mark#(add(X1,X2)) -> mark#(X1)
p11: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{49, x1 - 14}
        fst_A(x1,x2) = max{x1 + 108, x2 + 93}
        s_A(x1) = 0
        cons_A(x1,x2) = max{31, x1}
        mark#_A(x1) = max{47, x1 + 18}
        mark_A(x1) = x1 + 27
        from_A(x1) = max{58, x1 + 32}
        add_A(x1,x2) = max{x1 + 48, x2 + 48}
        active_A(x1) = max{24, x1}
        |0|_A = 0
        nil_A = 67
        len_A(x1) = 27
  
    precedence:
    
      active# = cons = mark# = mark = add = active = nil > |0| > fst = s = from > len
    
    partial status:
  
      pi(active#) = []
      pi(fst) = [1, 2]
      pi(s) = []
      pi(cons) = []
      pi(mark#) = []
      pi(mark) = []
      pi(from) = []
      pi(add) = []
      pi(active) = []
      pi(|0|) = []
      pi(nil) = []
      pi(len) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> mark#(X1)
p2: mark#(fst(X1,X2)) -> mark#(X2)
p3: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(from(X)) -> active#(from(mark(X)))
p7: mark#(from(X)) -> mark#(X)
p8: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p9: mark#(add(X1,X2)) -> mark#(X1)
p10: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> mark#(X1)
p2: mark#(add(X1,X2)) -> mark#(X2)
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> active#(from(mark(X)))
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(fst(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = 21
        fst_A(x1,x2) = 21
        add_A(x1,x2) = 5
        active#_A(x1) = max{19, x1 + 15}
        mark_A(x1) = 21
        from_A(x1) = 6
        cons_A(x1,x2) = 3
        s_A(x1) = 4
        active_A(x1) = max{21, x1}
        |0|_A = 20
        nil_A = 19
        len_A(x1) = 21
  
    precedence:
    
      fst = mark = active > mark# = add = active# = s = |0| = nil = len > from = cons
    
    partial status:
  
      pi(mark#) = []
      pi(fst) = []
      pi(add) = []
      pi(active#) = []
      pi(mark) = []
      pi(from) = []
      pi(cons) = []
      pi(s) = []
      pi(active) = []
      pi(|0|) = []
      pi(nil) = []
      pi(len) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> mark#(X1)
p2: mark#(add(X1,X2)) -> mark#(X2)
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(from(X)) -> mark#(X)
p6: mark#(from(X)) -> active#(from(mark(X)))
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(fst(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> mark#(X1)
p2: mark#(fst(X1,X2)) -> mark#(X2)
p3: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(from(X)) -> active#(from(mark(X)))
p7: mark#(from(X)) -> mark#(X)
p8: mark#(add(X1,X2)) -> mark#(X1)
p9: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{18, x1 + 14}
        fst_A(x1,x2) = max{x1 + 246, x2 + 253}
        cons_A(x1,x2) = x1 + 18
        active#_A(x1) = max{16, x1 - 43}
        mark_A(x1) = x1 + 14
        from_A(x1) = x1 + 117
        s_A(x1) = 15
        add_A(x1,x2) = max{182, x1 + 174, x2 + 181}
        active_A(x1) = max{29, x1}
        |0|_A = 16
        nil_A = 31
        len_A(x1) = max{28, x1 + 19}
  
    precedence:
    
      nil > fst = cons = active# = mark = from = s = add = active = |0| > len > mark#
    
    partial status:
  
      pi(mark#) = [1]
      pi(fst) = []
      pi(cons) = []
      pi(active#) = []
      pi(mark) = [1]
      pi(from) = []
      pi(s) = []
      pi(add) = []
      pi(active) = []
      pi(|0|) = []
      pi(nil) = []
      pi(len) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> mark#(X1)
p2: mark#(fst(X1,X2)) -> mark#(X2)
p3: active#(from(X)) -> mark#(cons(X,from(s(X))))
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(from(X)) -> active#(from(mark(X)))
p6: mark#(from(X)) -> mark#(X)
p7: mark#(add(X1,X2)) -> mark#(X1)
p8: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> mark#(X1)
p2: mark#(add(X1,X2)) -> mark#(X2)
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(from(X)) -> active#(from(mark(X)))
p6: active#(from(X)) -> mark#(cons(X,from(s(X))))
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: mark#(fst(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{32, x1 - 18}
        fst_A(x1,x2) = max{x1 + 70, x2 + 21}
        add_A(x1,x2) = max{x1 + 24, x2 + 21}
        from_A(x1) = x1 + 72
        active#_A(x1) = max{1, x1 - 40}
        mark_A(x1) = x1 + 21
        cons_A(x1,x2) = x1 + 50
        s_A(x1) = 24
        active_A(x1) = max{18, x1}
        |0|_A = 22
        nil_A = 48
        len_A(x1) = max{17, x1 - 5}
  
    precedence:
    
      fst > mark# = add = from = active# = mark = cons = s = active = |0| = nil = len
    
    partial status:
  
      pi(mark#) = []
      pi(fst) = []
      pi(add) = []
      pi(from) = []
      pi(active#) = []
      pi(mark) = []
      pi(cons) = []
      pi(s) = []
      pi(active) = []
      pi(|0|) = []
      pi(nil) = []
      pi(len) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(add(X1,X2)) -> mark#(X2)
p2: mark#(add(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> mark#(X)
p4: mark#(from(X)) -> active#(from(mark(X)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(fst(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(add(X1,X2)) -> mark#(X2)
p2: mark#(fst(X1,X2)) -> mark#(X2)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> active#(from(mark(X)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(from(X)) -> mark#(X)
p7: mark#(add(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{2, x1}
        add_A(x1,x2) = max{x1 + 4, x2 + 4}
        fst_A(x1,x2) = x2 + 13
        cons_A(x1,x2) = x1 + 3
        from_A(x1) = x1 + 17
        active#_A(x1) = max{17, x1 - 1}
        mark_A(x1) = x1
        s_A(x1) = 0
        active_A(x1) = x1
        |0|_A = 10
        nil_A = 1
        len_A(x1) = x1 + 10
  
    precedence:
    
      mark# = add = fst = cons = from = active# = mark = s = active = |0| = nil = len
    
    partial status:
  
      pi(mark#) = [1]
      pi(add) = []
      pi(fst) = []
      pi(cons) = []
      pi(from) = []
      pi(active#) = []
      pi(mark) = [1]
      pi(s) = []
      pi(active) = [1]
      pi(|0|) = []
      pi(nil) = []
      pi(len) = []

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(add(X1,X2)) -> mark#(X2)
p2: mark#(fst(X1,X2)) -> mark#(X2)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> active#(from(mark(X)))
p5: mark#(from(X)) -> mark#(X)
p6: mark#(add(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p5, p6}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(add(X1,X2)) -> mark#(X2)
p2: mark#(add(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(fst(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{3, x1 + 2}
        add_A(x1,x2) = max{x1, x2}
        from_A(x1) = max{1, x1}
        cons_A(x1,x2) = max{x1 + 3, x2 + 3}
        fst_A(x1,x2) = x2
  
    precedence:
    
      mark# = add = from = cons = fst
    
    partial status:
  
      pi(mark#) = [1]
      pi(add) = [1, 2]
      pi(from) = [1]
      pi(cons) = [1, 2]
      pi(fst) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(add(X1,X2)) -> mark#(X1)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(fst(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(add(X1,X2)) -> mark#(X1)
p2: mark#(fst(X1,X2)) -> mark#(X2)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> mark#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{5, x1 + 2}
        add_A(x1,x2) = max{x1, x2}
        fst_A(x1,x2) = x2 + 3
        cons_A(x1,x2) = max{x1, x2}
        from_A(x1) = max{3, x1}
  
    precedence:
    
      mark# = add = fst = cons = from
    
    partial status:
  
      pi(mark#) = [1]
      pi(add) = [1, 2]
      pi(fst) = [2]
      pi(cons) = [1, 2]
      pi(from) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> mark#(X2)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> mark#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(fst(X1,X2)) -> mark#(X2)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{3, x1 + 2}
        fst_A(x1,x2) = x2
        from_A(x1) = x1 + 3
        cons_A(x1,x2) = max{1, x1, x2}
  
    precedence:
    
      mark# = fst = from = cons
    
    partial status:
  
      pi(mark#) = [1]
      pi(fst) = [2]
      pi(from) = [1]
      pi(cons) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(from(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(from(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 1
        from_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      mark# = from = cons
    
    partial status:
  
      pi(mark#) = [1]
      pi(from) = [1]
      pi(cons) = [1, 2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 2
        cons_A(x1,x2) = max{x1 + 1, x2 + 1}
  
    precedence:
    
      mark# = cons
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(mark(X1),X2) -> cons#(X1,X2)
p2: cons#(X1,active(X2)) -> cons#(X1,X2)
p3: cons#(active(X1),X2) -> cons#(X1,X2)
p4: cons#(X1,mark(X2)) -> cons#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = max{0, x2 - 2}
        mark_A(x1) = x1
        active_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      cons# = mark = active
    
    partial status:
  
      pi(cons#) = []
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(mark(X1),X2) -> cons#(X1,X2)
p2: cons#(active(X1),X2) -> cons#(X1,X2)
p3: cons#(X1,mark(X2)) -> cons#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(mark(X1),X2) -> cons#(X1,X2)
p2: cons#(X1,mark(X2)) -> cons#(X1,X2)
p3: cons#(active(X1),X2) -> cons#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = max{0, x1 - 2}
        mark_A(x1) = max{3, x1 + 1}
        active_A(x1) = x1
  
    precedence:
    
      cons# = mark = active
    
    partial status:
  
      pi(cons#) = []
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(X1,mark(X2)) -> cons#(X1,X2)
p2: cons#(active(X1),X2) -> cons#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(X1,mark(X2)) -> cons#(X1,X2)
p2: cons#(active(X1),X2) -> cons#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = x2 + 1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
  
    precedence:
    
      cons# = mark = active
    
    partial status:
  
      pi(cons#) = []
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(active(X1),X2) -> cons#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: cons#(active(X1),X2) -> cons#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = max{1, x1, x2}
        active_A(x1) = x1 + 1
  
    precedence:
    
      cons# = active
    
    partial status:
  
      pi(cons#) = [1, 2]
      pi(active) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: fst#(mark(X1),X2) -> fst#(X1,X2)
p2: fst#(X1,active(X2)) -> fst#(X1,X2)
p3: fst#(active(X1),X2) -> fst#(X1,X2)
p4: fst#(X1,mark(X2)) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fst#_A(x1,x2) = max{0, x2 - 2}
        mark_A(x1) = x1
        active_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      fst# = mark = active
    
    partial status:
  
      pi(fst#) = []
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: fst#(mark(X1),X2) -> fst#(X1,X2)
p2: fst#(active(X1),X2) -> fst#(X1,X2)
p3: fst#(X1,mark(X2)) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: fst#(mark(X1),X2) -> fst#(X1,X2)
p2: fst#(X1,mark(X2)) -> fst#(X1,X2)
p3: fst#(active(X1),X2) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fst#_A(x1,x2) = max{0, x1 - 2}
        mark_A(x1) = max{3, x1 + 1}
        active_A(x1) = x1
  
    precedence:
    
      fst# = mark = active
    
    partial status:
  
      pi(fst#) = []
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: fst#(X1,mark(X2)) -> fst#(X1,X2)
p2: fst#(active(X1),X2) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: fst#(X1,mark(X2)) -> fst#(X1,X2)
p2: fst#(active(X1),X2) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fst#_A(x1,x2) = x2 + 1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
  
    precedence:
    
      fst# = mark = active
    
    partial status:
  
      pi(fst#) = []
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: fst#(active(X1),X2) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: fst#(active(X1),X2) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fst#_A(x1,x2) = max{1, x1, x2}
        active_A(x1) = x1 + 1
  
    precedence:
    
      fst# = active
    
    partial status:
  
      pi(fst#) = [1, 2]
      pi(active) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: from#(mark(X)) -> from#(X)
p2: from#(active(X)) -> from#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        from#_A(x1) = x1 + 1
        mark_A(x1) = x1
        active_A(x1) = x1
  
    precedence:
    
      from# = mark = active
    
    partial status:
  
      pi(from#) = [1]
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: from#(active(X)) -> from#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: from#(active(X)) -> from#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        from#_A(x1) = x1 + 1
        active_A(x1) = x1
  
    precedence:
    
      from# = active
    
    partial status:
  
      pi(from#) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: s#(mark(X)) -> s#(X)
p2: s#(active(X)) -> s#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        s#_A(x1) = max{3, x1 + 2}
        mark_A(x1) = x1
        active_A(x1) = x1 + 1
  
    precedence:
    
      s# = mark = active
    
    partial status:
  
      pi(s#) = [1]
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: s#(active(X)) -> s#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: s#(active(X)) -> s#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        s#_A(x1) = x1 + 1
        active_A(x1) = x1
  
    precedence:
    
      s# = active
    
    partial status:
  
      pi(s#) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: add#(mark(X1),X2) -> add#(X1,X2)
p2: add#(X1,active(X2)) -> add#(X1,X2)
p3: add#(active(X1),X2) -> add#(X1,X2)
p4: add#(X1,mark(X2)) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        add#_A(x1,x2) = max{0, x2 - 2}
        mark_A(x1) = x1
        active_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      add# = mark = active
    
    partial status:
  
      pi(add#) = []
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: add#(mark(X1),X2) -> add#(X1,X2)
p2: add#(active(X1),X2) -> add#(X1,X2)
p3: add#(X1,mark(X2)) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: add#(mark(X1),X2) -> add#(X1,X2)
p2: add#(X1,mark(X2)) -> add#(X1,X2)
p3: add#(active(X1),X2) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        add#_A(x1,x2) = max{0, x1 - 2}
        mark_A(x1) = max{3, x1 + 1}
        active_A(x1) = x1
  
    precedence:
    
      add# = mark = active
    
    partial status:
  
      pi(add#) = []
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: add#(X1,mark(X2)) -> add#(X1,X2)
p2: add#(active(X1),X2) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: add#(X1,mark(X2)) -> add#(X1,X2)
p2: add#(active(X1),X2) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        add#_A(x1,x2) = x2 + 1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
  
    precedence:
    
      add# = mark = active
    
    partial status:
  
      pi(add#) = []
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: add#(active(X1),X2) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: add#(active(X1),X2) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        add#_A(x1,x2) = max{1, x1, x2}
        active_A(x1) = x1 + 1
  
    precedence:
    
      add# = active
    
    partial status:
  
      pi(add#) = [1, 2]
      pi(active) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: len#(mark(X)) -> len#(X)
p2: len#(active(X)) -> len#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        len#_A(x1) = x1 + 1
        mark_A(x1) = x1
        active_A(x1) = x1
  
    precedence:
    
      len# = mark = active
    
    partial status:
  
      pi(len#) = [1]
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: len#(active(X)) -> len#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: len#(active(X)) -> len#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        len#_A(x1) = x1 + 1
        active_A(x1) = x1
  
    precedence:
    
      len# = active
    
    partial status:
  
      pi(len#) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.