YES

We show the termination of the TRS R:

  active(eq(|0|(),|0|())) -> mark(true())
  active(eq(s(X),s(Y))) -> mark(eq(X,Y))
  active(eq(X,Y)) -> mark(false())
  active(inf(X)) -> mark(cons(X,inf(s(X))))
  active(take(|0|(),X)) -> mark(nil())
  active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
  active(length(nil())) -> mark(|0|())
  active(length(cons(X,L))) -> mark(s(length(L)))
  mark(eq(X1,X2)) -> active(eq(X1,X2))
  mark(|0|()) -> active(|0|())
  mark(true()) -> active(true())
  mark(s(X)) -> active(s(X))
  mark(false()) -> active(false())
  mark(inf(X)) -> active(inf(mark(X)))
  mark(cons(X1,X2)) -> active(cons(X1,X2))
  mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
  mark(nil()) -> active(nil())
  mark(length(X)) -> active(length(mark(X)))
  eq(mark(X1),X2) -> eq(X1,X2)
  eq(X1,mark(X2)) -> eq(X1,X2)
  eq(active(X1),X2) -> eq(X1,X2)
  eq(X1,active(X2)) -> eq(X1,X2)
  s(mark(X)) -> s(X)
  s(active(X)) -> s(X)
  inf(mark(X)) -> inf(X)
  inf(active(X)) -> inf(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)
  take(mark(X1),X2) -> take(X1,X2)
  take(X1,mark(X2)) -> take(X1,X2)
  take(active(X1),X2) -> take(X1,X2)
  take(X1,active(X2)) -> take(X1,X2)
  length(mark(X)) -> length(X)
  length(active(X)) -> length(X)

-- SCC decomposition.

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

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

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(X)

The estimated dependency graph contains the following SCCs:

  {p2, p5, p10, p14, p17, p20, p22, p24, p25, p26, p28, p29, p31, p33}
  {p34, p35, p36, p37}
  {p42, p43, p44, p45}
  {p40, p41}
  {p38, p39}
  {p46, p47, p48, p49}
  {p50, p51}


-- Reduction pair.

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

p1: mark#(length(X)) -> active#(length(mark(X)))
p2: active#(length(cons(X,L))) -> mark#(s(length(L)))
p3: mark#(length(X)) -> mark#(X)
p4: mark#(take(X1,X2)) -> mark#(X2)
p5: mark#(take(X1,X2)) -> mark#(X1)
p6: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p7: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p8: mark#(cons(X1,X2)) -> active#(cons(X1,X2))
p9: active#(inf(X)) -> mark#(cons(X,inf(s(X))))
p10: mark#(inf(X)) -> mark#(X)
p11: mark#(inf(X)) -> active#(inf(mark(X)))
p12: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p13: mark#(s(X)) -> active#(s(X))
p14: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{5, x1}
        length_A(x1) = x1 + 17
        active#_A(x1) = max{1, x1}
        mark_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2 - 10}
        s_A(x1) = max{2, x1 - 10}
        take_A(x1,x2) = max{x1 + 16, x2 + 16}
        inf_A(x1) = x1 + 7
        eq_A(x1,x2) = 17
        active_A(x1) = x1
        |0|_A = 7
        true_A = 10
        false_A = 17
        nil_A = 1
  
    precedence:
    
      take > mark# = active# = eq = true > length > |0| > mark = cons = s = inf = active = false = nil
    
    partial status:
  
      pi(mark#) = [1]
      pi(length) = []
      pi(active#) = [1]
      pi(mark) = []
      pi(cons) = []
      pi(s) = []
      pi(take) = []
      pi(inf) = []
      pi(eq) = []
      pi(active) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(nil) = []

The next rules are strictly ordered:

  p9

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(length(X)) -> active#(length(mark(X)))
p2: active#(length(cons(X,L))) -> mark#(s(length(L)))
p3: mark#(length(X)) -> mark#(X)
p4: mark#(take(X1,X2)) -> mark#(X2)
p5: mark#(take(X1,X2)) -> mark#(X1)
p6: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p7: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p8: mark#(cons(X1,X2)) -> active#(cons(X1,X2))
p9: mark#(inf(X)) -> mark#(X)
p10: mark#(inf(X)) -> active#(inf(mark(X)))
p11: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p12: mark#(s(X)) -> active#(s(X))
p13: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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#(length(X)) -> active#(length(mark(X)))
p2: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p3: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p4: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p5: mark#(s(X)) -> active#(s(X))
p6: active#(length(cons(X,L))) -> mark#(s(length(L)))
p7: mark#(inf(X)) -> active#(inf(mark(X)))
p8: mark#(inf(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> active#(cons(X1,X2))
p10: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p11: mark#(take(X1,X2)) -> mark#(X1)
p12: mark#(take(X1,X2)) -> mark#(X2)
p13: mark#(length(X)) -> mark#(X)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{1, x1 - 16}
        length_A(x1) = max{136, x1 + 72}
        active#_A(x1) = max{11, x1 - 16}
        mark_A(x1) = max{30, x1}
        eq_A(x1,x2) = 29
        s_A(x1) = max{27, x1 - 54}
        take_A(x1,x2) = max{143, x1 + 98, x2 + 113}
        cons_A(x1,x2) = max{27, x1 - 7, x2 - 54}
        inf_A(x1) = max{83, x1 + 28}
        active_A(x1) = max{30, x1}
        |0|_A = 137
        true_A = 28
        false_A = 28
        nil_A = 113
  
    precedence:
    
      length > mark# = active# = mark = eq = s = take = cons = inf = active = |0| = true = false = nil
    
    partial status:
  
      pi(mark#) = []
      pi(length) = []
      pi(active#) = []
      pi(mark) = [1]
      pi(eq) = []
      pi(s) = []
      pi(take) = []
      pi(cons) = []
      pi(inf) = []
      pi(active) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(nil) = []

The next rules are strictly ordered:

  p13

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(length(X)) -> active#(length(mark(X)))
p2: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p3: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p4: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p5: mark#(s(X)) -> active#(s(X))
p6: active#(length(cons(X,L))) -> mark#(s(length(L)))
p7: mark#(inf(X)) -> active#(inf(mark(X)))
p8: mark#(inf(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> active#(cons(X1,X2))
p10: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p11: mark#(take(X1,X2)) -> mark#(X1)
p12: mark#(take(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: mark#(length(X)) -> active#(length(mark(X)))
p2: active#(length(cons(X,L))) -> mark#(s(length(L)))
p3: mark#(take(X1,X2)) -> mark#(X2)
p4: mark#(take(X1,X2)) -> mark#(X1)
p5: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p6: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p7: mark#(cons(X1,X2)) -> active#(cons(X1,X2))
p8: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p9: mark#(inf(X)) -> mark#(X)
p10: mark#(inf(X)) -> active#(inf(mark(X)))
p11: mark#(s(X)) -> active#(s(X))
p12: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{1, x1 - 5}
        length_A(x1) = max{59, x1 + 33}
        active#_A(x1) = max{12, x1 - 15}
        mark_A(x1) = max{23, x1 + 7}
        cons_A(x1,x2) = x1 + 60
        s_A(x1) = 18
        take_A(x1,x2) = max{66, x1 + 44, x2 + 29}
        eq_A(x1,x2) = 17
        inf_A(x1) = max{92, x1 + 68}
        active_A(x1) = max{24, x1}
        |0|_A = 61
        true_A = 17
        false_A = 17
        nil_A = 67
  
    precedence:
    
      mark# = active# = mark = eq = active = true > length = nil > false > cons = take > s > inf = |0|
    
    partial status:
  
      pi(mark#) = []
      pi(length) = [1]
      pi(active#) = []
      pi(mark) = []
      pi(cons) = []
      pi(s) = []
      pi(take) = [1]
      pi(eq) = []
      pi(inf) = []
      pi(active) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(nil) = []

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(length(X)) -> active#(length(mark(X)))
p2: active#(length(cons(X,L))) -> mark#(s(length(L)))
p3: mark#(take(X1,X2)) -> mark#(X2)
p4: mark#(take(X1,X2)) -> mark#(X1)
p5: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p6: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p7: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p8: mark#(inf(X)) -> mark#(X)
p9: mark#(inf(X)) -> active#(inf(mark(X)))
p10: mark#(s(X)) -> active#(s(X))
p11: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: mark#(length(X)) -> active#(length(mark(X)))
p2: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p3: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p4: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p5: mark#(s(X)) -> active#(s(X))
p6: active#(length(cons(X,L))) -> mark#(s(length(L)))
p7: mark#(inf(X)) -> active#(inf(mark(X)))
p8: mark#(inf(X)) -> mark#(X)
p9: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p10: mark#(take(X1,X2)) -> mark#(X1)
p11: mark#(take(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{0, x1 - 19}
        length_A(x1) = max{120, x1 + 71}
        active#_A(x1) = max{17, x1 - 35}
        mark_A(x1) = max{50, x1 + 14}
        eq_A(x1,x2) = 36
        s_A(x1) = 98
        take_A(x1,x2) = max{122, x1 + 86, x2 + 70}
        cons_A(x1,x2) = x1 + 111
        inf_A(x1) = max{161, x1 + 125}
        active_A(x1) = max{50, x1}
        |0|_A = 121
        true_A = 21
        false_A = 5
        nil_A = 123
  
    precedence:
    
      length > mark# = active# > mark = eq = s = take = cons = inf = active = |0| = true = false = nil
    
    partial status:
  
      pi(mark#) = []
      pi(length) = []
      pi(active#) = []
      pi(mark) = []
      pi(eq) = []
      pi(s) = []
      pi(take) = []
      pi(cons) = []
      pi(inf) = []
      pi(active) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(nil) = []

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(length(X)) -> active#(length(mark(X)))
p2: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p3: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p4: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p5: mark#(s(X)) -> active#(s(X))
p6: active#(length(cons(X,L))) -> mark#(s(length(L)))
p7: mark#(inf(X)) -> active#(inf(mark(X)))
p8: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p9: mark#(take(X1,X2)) -> mark#(X1)
p10: mark#(take(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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#(length(X)) -> active#(length(mark(X)))
p2: active#(length(cons(X,L))) -> mark#(s(length(L)))
p3: mark#(take(X1,X2)) -> mark#(X2)
p4: mark#(take(X1,X2)) -> mark#(X1)
p5: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p6: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p7: mark#(inf(X)) -> active#(inf(mark(X)))
p8: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p9: mark#(s(X)) -> active#(s(X))
p10: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{18, x1}
        length_A(x1) = max{16, x1 - 9}
        active#_A(x1) = max{17, x1}
        mark_A(x1) = x1
        cons_A(x1,x2) = max{62, x1 + 51}
        s_A(x1) = 15
        take_A(x1,x2) = max{x1 + 16, x2 + 19}
        inf_A(x1) = max{63, x1 + 61}
        eq_A(x1,x2) = 19
        active_A(x1) = max{1, x1}
        |0|_A = 11
        true_A = 18
        false_A = 2
        nil_A = 10
  
    precedence:
    
      mark# = length = active# = mark = cons = s = take = inf = eq = active = |0| = true = false = nil
    
    partial status:
  
      pi(mark#) = [1]
      pi(length) = []
      pi(active#) = [1]
      pi(mark) = [1]
      pi(cons) = []
      pi(s) = []
      pi(take) = []
      pi(inf) = []
      pi(eq) = []
      pi(active) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      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#(length(X)) -> active#(length(mark(X)))
p2: mark#(take(X1,X2)) -> mark#(X2)
p3: mark#(take(X1,X2)) -> mark#(X1)
p4: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p5: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p6: mark#(inf(X)) -> active#(inf(mark(X)))
p7: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p8: mark#(s(X)) -> active#(s(X))
p9: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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#(length(X)) -> active#(length(mark(X)))
p2: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p3: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p4: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p5: mark#(s(X)) -> active#(s(X))
p6: mark#(inf(X)) -> active#(inf(mark(X)))
p7: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p8: mark#(take(X1,X2)) -> mark#(X1)
p9: mark#(take(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 22
        length_A(x1) = 31
        active#_A(x1) = max{22, x1 + 7}
        mark_A(x1) = x1 + 2
        eq_A(x1,x2) = 0
        s_A(x1) = max{20, x1 - 2}
        take_A(x1,x2) = max{x1, x2 + 22}
        cons_A(x1,x2) = x1 + 20
        inf_A(x1) = x1 + 22
        active_A(x1) = max{2, x1}
        |0|_A = 28
        true_A = 0
        false_A = 0
        nil_A = 1
  
    precedence:
    
      mark# = length = active# = mark = s = inf = active = |0| = true = false = nil > eq = take = cons
    
    partial status:
  
      pi(mark#) = []
      pi(length) = []
      pi(active#) = []
      pi(mark) = []
      pi(eq) = []
      pi(s) = []
      pi(take) = [2]
      pi(cons) = []
      pi(inf) = []
      pi(active) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(nil) = []

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(length(X)) -> active#(length(mark(X)))
p2: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p3: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p4: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p5: mark#(s(X)) -> active#(s(X))
p6: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p7: mark#(take(X1,X2)) -> mark#(X1)
p8: mark#(take(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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#(length(X)) -> active#(length(mark(X)))
p2: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p3: mark#(take(X1,X2)) -> mark#(X2)
p4: mark#(take(X1,X2)) -> mark#(X1)
p5: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p6: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p7: mark#(s(X)) -> active#(s(X))
p8: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{38, x1 + 24}
        length_A(x1) = x1 + 3
        active#_A(x1) = x1 + 24
        mark_A(x1) = x1
        take_A(x1,x2) = max{x1 + 19, x2 + 19}
        s_A(x1) = 13
        cons_A(x1,x2) = x1 + 10
        eq_A(x1,x2) = 19
        active_A(x1) = x1
        |0|_A = 0
        true_A = 18
        false_A = 0
        inf_A(x1) = max{28, x1 + 14}
        nil_A = 18
  
    precedence:
    
      length = mark = active = true = false = inf = nil > mark# = active# > take = s = cons = eq = |0|
    
    partial status:
  
      pi(mark#) = []
      pi(length) = []
      pi(active#) = []
      pi(mark) = []
      pi(take) = [2]
      pi(s) = []
      pi(cons) = [1]
      pi(eq) = []
      pi(active) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(inf) = []
      pi(nil) = []

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(length(X)) -> active#(length(mark(X)))
p2: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p3: mark#(take(X1,X2)) -> mark#(X2)
p4: mark#(take(X1,X2)) -> mark#(X1)
p5: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p6: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p7: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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#(length(X)) -> active#(length(mark(X)))
p2: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p3: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p4: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p5: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p6: mark#(take(X1,X2)) -> mark#(X1)
p7: mark#(take(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{46, x1 - 70}
        length_A(x1) = x1 + 124
        active#_A(x1) = 46
        mark_A(x1) = x1 + 7
        eq_A(x1,x2) = 0
        s_A(x1) = 0
        take_A(x1,x2) = max{x1 + 71, x2 + 77}
        cons_A(x1,x2) = 47
        active_A(x1) = max{7, x1}
        |0|_A = 0
        true_A = 0
        false_A = 0
        inf_A(x1) = 54
        nil_A = 0
  
    precedence:
    
      mark# = length = active# = take = cons = |0| > mark = active = nil > eq = true > s = false = inf
    
    partial status:
  
      pi(mark#) = []
      pi(length) = []
      pi(active#) = []
      pi(mark) = []
      pi(eq) = []
      pi(s) = []
      pi(take) = []
      pi(cons) = []
      pi(active) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(inf) = []
      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#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p3: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p4: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p5: mark#(take(X1,X2)) -> mark#(X1)
p6: mark#(take(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(take(X1,X2)) -> mark#(X2)
p3: mark#(take(X1,X2)) -> mark#(X1)
p4: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p5: active#(take(s(X),cons(Y,L))) -> mark#(cons(Y,take(X,L)))
p6: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{4, x1 - 49}
        eq_A(x1,x2) = 31
        s_A(x1) = 20
        mark#_A(x1) = max{4, x1 - 27}
        take_A(x1,x2) = max{81, x1 + 69, x2 + 75}
        mark_A(x1) = max{5, x1}
        cons_A(x1,x2) = x1 + 19
        active_A(x1) = max{3, x1}
        |0|_A = 9
        true_A = 4
        false_A = 6
        inf_A(x1) = max{42, x1 + 21}
        nil_A = 4
        length_A(x1) = max{21, x1 + 4}
  
    precedence:
    
      active# = eq = s = mark# = take = mark = cons = active = |0| = true = false = inf = nil = length
    
    partial status:
  
      pi(active#) = []
      pi(eq) = []
      pi(s) = []
      pi(mark#) = []
      pi(take) = []
      pi(mark) = [1]
      pi(cons) = [1]
      pi(active) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(inf) = []
      pi(nil) = []
      pi(length) = []

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: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(take(X1,X2)) -> mark#(X2)
p3: mark#(take(X1,X2)) -> mark#(X1)
p4: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p5: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p3: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p4: mark#(take(X1,X2)) -> mark#(X1)
p5: mark#(take(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{21, x1 - 19}
        eq_A(x1,x2) = 21
        s_A(x1) = 5
        mark#_A(x1) = max{1, x1}
        take_A(x1,x2) = max{78, x1 + 31, x2 + 54}
        mark_A(x1) = max{30, x1 + 6}
        active_A(x1) = max{30, x1}
        |0|_A = 25
        true_A = 20
        false_A = 14
        inf_A(x1) = max{28, x1 - 1}
        cons_A(x1,x2) = 24
        nil_A = 31
        length_A(x1) = max{18, x1}
  
    precedence:
    
      eq = mark = active > s > active# = mark# = take = true > inf = cons > |0| = nil > false = length
    
    partial status:
  
      pi(active#) = []
      pi(eq) = []
      pi(s) = []
      pi(mark#) = []
      pi(take) = [2]
      pi(mark) = []
      pi(active) = []
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(inf) = []
      pi(cons) = []
      pi(nil) = []
      pi(length) = []

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: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p3: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p4: mark#(take(X1,X2)) -> mark#(X1)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(take(X1,X2)) -> mark#(X1)
p3: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p4: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{91, x1 + 27}
        eq_A(x1,x2) = 5
        s_A(x1) = 14
        mark#_A(x1) = max{91, x1 + 59}
        take_A(x1,x2) = max{x1 + 6, x2 + 6}
        mark_A(x1) = x1
        active_A(x1) = max{3, x1}
        |0|_A = 35
        true_A = 4
        false_A = 3
        inf_A(x1) = x1 + 9
        cons_A(x1,x2) = 8
        nil_A = 5
        length_A(x1) = x1 + 35
  
    precedence:
    
      active# = eq = s = mark# = take = mark = active = |0| = true = false = inf = cons = nil = length
    
    partial status:
  
      pi(active#) = [1]
      pi(eq) = []
      pi(s) = []
      pi(mark#) = [1]
      pi(take) = []
      pi(mark) = [1]
      pi(active) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(inf) = []
      pi(cons) = []
      pi(nil) = []
      pi(length) = []

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: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))
p3: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(eq(X1,X2)) -> active#(eq(X1,X2))
p3: mark#(take(X1,X2)) -> active#(take(mark(X1),mark(X2)))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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, r34, r35, r36

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{66, x1 + 31}
        eq_A(x1,x2) = 16
        s_A(x1) = max{9, x1 - 10}
        mark#_A(x1) = x1 + 50
        take_A(x1,x2) = max{x1 + 20, x2 + 19}
        mark_A(x1) = x1
        active_A(x1) = max{3, x1}
        |0|_A = 14
        true_A = 15
        false_A = 4
        inf_A(x1) = max{4, x1 + 2}
        cons_A(x1,x2) = max{4, x1 + 2, x2 - 10}
        nil_A = 4
        length_A(x1) = 15
  
    precedence:
    
      s > active# = eq = mark# = take = mark = active = |0| = true = false = inf = cons = nil = length
    
    partial status:
  
      pi(active#) = []
      pi(eq) = []
      pi(s) = []
      pi(mark#) = []
      pi(take) = []
      pi(mark) = [1]
      pi(active) = [1]
      pi(|0|) = []
      pi(true) = []
      pi(false) = []
      pi(inf) = []
      pi(cons) = []
      pi(nil) = []
      pi(length) = []

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: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: active#(eq(s(X),s(Y))) -> mark#(eq(X,Y))
p2: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(X)

The set of usable rules consists of

  r19, r20, r21, r22

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{1, x1}
        eq_A(x1,x2) = max{3, x1 - 3, x2 - 6}
        s_A(x1) = max{12, x1 + 10}
        mark#_A(x1) = x1 + 5
        mark_A(x1) = x1
        active_A(x1) = x1
  
    precedence:
    
      active# = eq = s = mark# = mark = active
    
    partial status:
  
      pi(active#) = [1]
      pi(eq) = []
      pi(s) = [1]
      pi(mark#) = [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: mark#(eq(X1,X2)) -> active#(eq(X1,X2))

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(X)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

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

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        eq#_A(x1,x2) = max{0, x2 - 2}
        mark_A(x1) = x1
        active_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      eq# = mark = active
    
    partial status:
  
      pi(eq#) = []
      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: eq#(mark(X1),X2) -> eq#(X1,X2)
p2: eq#(active(X1),X2) -> eq#(X1,X2)
p3: eq#(X1,mark(X2)) -> eq#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: eq#(mark(X1),X2) -> eq#(X1,X2)
p2: eq#(X1,mark(X2)) -> eq#(X1,X2)
p3: eq#(active(X1),X2) -> eq#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        eq#_A(x1,x2) = max{0, x1 - 2}
        mark_A(x1) = max{3, x1 + 1}
        active_A(x1) = x1
  
    precedence:
    
      eq# = mark = active
    
    partial status:
  
      pi(eq#) = []
      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: eq#(X1,mark(X2)) -> eq#(X1,X2)
p2: eq#(active(X1),X2) -> eq#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: eq#(X1,mark(X2)) -> eq#(X1,X2)
p2: eq#(active(X1),X2) -> eq#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        eq#_A(x1,x2) = x2 + 1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
  
    precedence:
    
      eq# = mark = active
    
    partial status:
  
      pi(eq#) = []
      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: eq#(active(X1),X2) -> eq#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        eq#_A(x1,x2) = max{1, x1, x2}
        active_A(x1) = x1 + 1
  
    precedence:
    
      eq# = active
    
    partial status:
  
      pi(eq#) = [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: 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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: inf#(mark(X)) -> inf#(X)
p2: inf#(active(X)) -> inf#(X)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        inf#_A(x1) = x1 + 1
        mark_A(x1) = x1
        active_A(x1) = x1
  
    precedence:
    
      inf# = mark = active
    
    partial status:
  
      pi(inf#) = [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: inf#(active(X)) -> inf#(X)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        inf#_A(x1) = x1 + 1
        active_A(x1) = x1
  
    precedence:
    
      inf# = active
    
    partial status:
  
      pi(inf#) = [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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: take#(mark(X1),X2) -> take#(X1,X2)
p2: take#(X1,active(X2)) -> take#(X1,X2)
p3: take#(active(X1),X2) -> take#(X1,X2)
p4: take#(X1,mark(X2)) -> take#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        take#_A(x1,x2) = max{0, x2 - 1}
        mark_A(x1) = x1
        active_A(x1) = x1 + 2
  
    precedence:
    
      take# = mark = active
    
    partial status:
  
      pi(take#) = []
      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: take#(mark(X1),X2) -> take#(X1,X2)
p2: take#(active(X1),X2) -> take#(X1,X2)
p3: take#(X1,mark(X2)) -> take#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: take#(mark(X1),X2) -> take#(X1,X2)
p2: take#(X1,mark(X2)) -> take#(X1,X2)
p3: take#(active(X1),X2) -> take#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        take#_A(x1,x2) = max{0, x1 - 2}
        mark_A(x1) = max{3, x1 + 1}
        active_A(x1) = x1
  
    precedence:
    
      take# = mark = active
    
    partial status:
  
      pi(take#) = []
      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: take#(X1,mark(X2)) -> take#(X1,X2)
p2: take#(active(X1),X2) -> take#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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: take#(X1,mark(X2)) -> take#(X1,X2)
p2: take#(active(X1),X2) -> take#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        take#_A(x1,x2) = max{0, x2 - 2}
        mark_A(x1) = max{3, x1 + 1}
        active_A(x1) = x1 + 1
  
    precedence:
    
      take# = mark = active
    
    partial status:
  
      pi(take#) = []
      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: take#(active(X1),X2) -> take#(X1,X2)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        take#_A(x1,x2) = max{1, x1, x2}
        active_A(x1) = x1 + 1
  
    precedence:
    
      take# = active
    
    partial status:
  
      pi(take#) = [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: length#(mark(X)) -> length#(X)
p2: length#(active(X)) -> length#(X)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        length#_A(x1) = x1 + 1
        mark_A(x1) = x1
        active_A(x1) = x1
  
    precedence:
    
      length# = mark = active
    
    partial status:
  
      pi(length#) = [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: length#(active(X)) -> length#(X)

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: active(eq(|0|(),|0|())) -> mark(true())
r2: active(eq(s(X),s(Y))) -> mark(eq(X,Y))
r3: active(eq(X,Y)) -> mark(false())
r4: active(inf(X)) -> mark(cons(X,inf(s(X))))
r5: active(take(|0|(),X)) -> mark(nil())
r6: active(take(s(X),cons(Y,L))) -> mark(cons(Y,take(X,L)))
r7: active(length(nil())) -> mark(|0|())
r8: active(length(cons(X,L))) -> mark(s(length(L)))
r9: mark(eq(X1,X2)) -> active(eq(X1,X2))
r10: mark(|0|()) -> active(|0|())
r11: mark(true()) -> active(true())
r12: mark(s(X)) -> active(s(X))
r13: mark(false()) -> active(false())
r14: mark(inf(X)) -> active(inf(mark(X)))
r15: mark(cons(X1,X2)) -> active(cons(X1,X2))
r16: mark(take(X1,X2)) -> active(take(mark(X1),mark(X2)))
r17: mark(nil()) -> active(nil())
r18: mark(length(X)) -> active(length(mark(X)))
r19: eq(mark(X1),X2) -> eq(X1,X2)
r20: eq(X1,mark(X2)) -> eq(X1,X2)
r21: eq(active(X1),X2) -> eq(X1,X2)
r22: eq(X1,active(X2)) -> eq(X1,X2)
r23: s(mark(X)) -> s(X)
r24: s(active(X)) -> s(X)
r25: inf(mark(X)) -> inf(X)
r26: inf(active(X)) -> inf(X)
r27: cons(mark(X1),X2) -> cons(X1,X2)
r28: cons(X1,mark(X2)) -> cons(X1,X2)
r29: cons(active(X1),X2) -> cons(X1,X2)
r30: cons(X1,active(X2)) -> cons(X1,X2)
r31: take(mark(X1),X2) -> take(X1,X2)
r32: take(X1,mark(X2)) -> take(X1,X2)
r33: take(active(X1),X2) -> take(X1,X2)
r34: take(X1,active(X2)) -> take(X1,X2)
r35: length(mark(X)) -> length(X)
r36: length(active(X)) -> length(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:
      
        length#_A(x1) = x1 + 1
        active_A(x1) = x1
  
    precedence:
    
      length# = active
    
    partial status:
  
      pi(length#) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

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