YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__fst#_A(x1,x2) = max{x1 + 8, x2 + 6}
        s_A(x1) = x1
        cons_A(x1,x2) = max{3, x1 + 1, x2}
        mark#_A(x1) = x1
        len_A(x1) = x1
        add_A(x1,x2) = max{x1 + 4, x2 + 5}
        a__add#_A(x1,x2) = max{x1 + 3, x2 + 3}
        mark_A(x1) = x1
        |0|_A = 1
        from_A(x1) = x1 + 3
        a__from#_A(x1) = x1 + 3
        fst_A(x1,x2) = max{x1 + 9, x2 + 7}
        a__fst_A(x1,x2) = max{x1 + 9, x2 + 7}
        nil_A = 5
        a__from_A(x1) = x1 + 3
        a__add_A(x1,x2) = max{x1 + 4, x2 + 5}
        a__len_A(x1) = x1
  
    precedence:
    
      from = a__from > a__from# > cons = mark# = len = mark = |0| = fst = a__fst = nil = a__len > a__fst# = a__add# > s = add = a__add
    
    partial status:
  
      pi(a__fst#) = [2]
      pi(s) = []
      pi(cons) = [1]
      pi(mark#) = [1]
      pi(len) = [1]
      pi(add) = []
      pi(a__add#) = [1, 2]
      pi(mark) = [1]
      pi(|0|) = []
      pi(from) = []
      pi(a__from#) = []
      pi(fst) = []
      pi(a__fst) = [2]
      pi(nil) = []
      pi(a__from) = []
      pi(a__add) = []
      pi(a__len) = [1]

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__fst#_A(x1,x2) = max{59, x1 + 27, x2 + 27}
        s_A(x1) = 17
        cons_A(x1,x2) = max{83, x1 + 44}
        mark#_A(x1) = max{43, x1 + 14}
        fst_A(x1,x2) = max{45, x1 + 13, x2 + 15}
        mark_A(x1) = max{19, x1}
        from_A(x1) = max{83, x1 + 64}
        a__from#_A(x1) = max{63, x1 + 44}
        add_A(x1,x2) = max{37, x1 + 13, x2 + 18}
        len_A(x1) = max{84, x1 + 42}
        a__fst_A(x1,x2) = max{45, x1 + 13, x2 + 15}
        |0|_A = 85
        nil_A = 44
        a__from_A(x1) = max{83, x1 + 64}
        a__add_A(x1,x2) = max{37, x1 + 13, x2 + 18}
        a__len_A(x1) = max{84, x1 + 42}
  
    precedence:
    
      a__fst# = s = cons = mark# = fst = mark = from = a__from# = add = len = a__fst = |0| = nil = a__from = a__add = a__len
    
    partial status:
  
      pi(a__fst#) = [1, 2]
      pi(s) = []
      pi(cons) = []
      pi(mark#) = [1]
      pi(fst) = []
      pi(mark) = [1]
      pi(from) = []
      pi(a__from#) = [1]
      pi(add) = []
      pi(len) = [1]
      pi(a__fst) = [2]
      pi(|0|) = []
      pi(nil) = []
      pi(a__from) = [1]
      pi(a__add) = []
      pi(a__len) = [1]

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__fst#_A(x1,x2) = max{2, x1 - 1, x2}
        s_A(x1) = max{4, x1}
        cons_A(x1,x2) = max{x1 + 3, x2}
        mark#_A(x1) = x1
        len_A(x1) = x1 + 5
        add_A(x1,x2) = max{3, x1, x2}
        from_A(x1) = max{10, x1 + 5}
        a__from#_A(x1) = x1 + 4
        mark_A(x1) = x1
        fst_A(x1,x2) = max{4, x1, x2 + 3}
        a__fst_A(x1,x2) = max{4, x1, x2 + 3}
        |0|_A = 2
        nil_A = 1
        a__from_A(x1) = max{10, x1 + 5}
        a__add_A(x1,x2) = max{3, x1, x2}
        a__len_A(x1) = x1 + 5
  
    precedence:
    
      a__fst# = s = cons = mark# = len = add = from = a__from# = mark = fst = a__fst = |0| = nil = a__from = a__add = a__len
    
    partial status:
  
      pi(a__fst#) = [2]
      pi(s) = []
      pi(cons) = []
      pi(mark#) = []
      pi(len) = []
      pi(add) = []
      pi(from) = []
      pi(a__from#) = [1]
      pi(mark) = []
      pi(fst) = []
      pi(a__fst) = []
      pi(|0|) = []
      pi(nil) = []
      pi(a__from) = []
      pi(a__add) = []
      pi(a__len) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 12
        cons_A(x1,x2) = max{20, x1 + 8}
        fst_A(x1,x2) = max{x1 + 17, x2 + 18}
        from_A(x1) = x1 + 31
        a__from#_A(x1) = x1 + 12
        mark_A(x1) = x1 + 18
        add_A(x1,x2) = max{x1 + 15, x2 + 19}
        len_A(x1) = x1 + 20
        a__fst_A(x1,x2) = max{19, x1 + 17, x2 + 18}
        |0|_A = 22
        nil_A = 17
        s_A(x1) = 9
        a__from_A(x1) = x1 + 31
        a__add_A(x1,x2) = max{28, x1 + 15, x2 + 19}
        a__len_A(x1) = max{21, x1 + 20}
  
    precedence:
    
      |0| > a__from# > nil > mark# = mark = len = a__len > a__from > a__add > fst = a__fst > s > cons > add > from
    
    partial status:
  
      pi(mark#) = [1]
      pi(cons) = []
      pi(fst) = []
      pi(from) = []
      pi(a__from#) = []
      pi(mark) = [1]
      pi(add) = [2]
      pi(len) = [1]
      pi(a__fst) = []
      pi(|0|) = []
      pi(nil) = []
      pi(s) = []
      pi(a__from) = []
      pi(a__add) = []
      pi(a__len) = [1]

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{4, x1 + 3}
        cons_A(x1,x2) = max{x1, x2 + 1}
        len_A(x1) = x1 + 2
        add_A(x1,x2) = max{x1 + 1, x2}
        fst_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      mark# = cons = len = add = fst
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]
      pi(len) = [1]
      pi(add) = [2]
      pi(fst) = [2]

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(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: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(fst(X1,X2)) -> mark#(X1)
p3: mark#(fst(X1,X2)) -> mark#(X2)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: mark#(add(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 4
        cons_A(x1,x2) = max{x1 + 3, x2 + 3}
        fst_A(x1,x2) = max{x1, x2}
        add_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      mark# = cons = fst = add
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]
      pi(fst) = [2]
      pi(add) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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