YES

We show the termination of the TRS R:

  fst(|0|(),Z) -> nil()
  fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
  from(X) -> cons(X,n__from(n__s(X)))
  add(|0|(),X) -> X
  add(s(X),Y) -> s(n__add(activate(X),Y))
  len(nil()) -> |0|()
  len(cons(X,Z)) -> s(n__len(activate(Z)))
  fst(X1,X2) -> n__fst(X1,X2)
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  add(X1,X2) -> n__add(X1,X2)
  len(X) -> n__len(X)
  activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(X)
  activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
  activate(n__len(X)) -> len(activate(X))
  activate(X) -> X

-- SCC decomposition.

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

p1: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: fst#(s(X),cons(Y,Z)) -> activate#(Z)
p3: add#(s(X),Y) -> s#(n__add(activate(X),Y))
p4: add#(s(X),Y) -> activate#(X)
p5: len#(cons(X,Z)) -> s#(n__len(activate(Z)))
p6: len#(cons(X,Z)) -> activate#(Z)
p7: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))
p8: activate#(n__fst(X1,X2)) -> activate#(X1)
p9: activate#(n__fst(X1,X2)) -> activate#(X2)
p10: activate#(n__from(X)) -> from#(activate(X))
p11: activate#(n__from(X)) -> activate#(X)
p12: activate#(n__s(X)) -> s#(X)
p13: activate#(n__add(X1,X2)) -> add#(activate(X1),activate(X2))
p14: activate#(n__add(X1,X2)) -> activate#(X1)
p15: activate#(n__add(X1,X2)) -> activate#(X2)
p16: activate#(n__len(X)) -> len#(activate(X))
p17: activate#(n__len(X)) -> activate#(X)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__len(X)) -> activate#(X)
p3: activate#(n__len(X)) -> len#(activate(X))
p4: len#(cons(X,Z)) -> activate#(Z)
p5: activate#(n__add(X1,X2)) -> activate#(X2)
p6: activate#(n__add(X1,X2)) -> activate#(X1)
p7: activate#(n__add(X1,X2)) -> add#(activate(X1),activate(X2))
p8: add#(s(X),Y) -> activate#(X)
p9: activate#(n__from(X)) -> activate#(X)
p10: activate#(n__fst(X1,X2)) -> activate#(X2)
p11: activate#(n__fst(X1,X2)) -> activate#(X1)
p12: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))
p13: fst#(s(X),cons(Y,Z)) -> activate#(Z)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fst#_A(x1,x2) = max{x1 + 4, x2 + 9}
        s_A(x1) = max{0, x1 - 3}
        cons_A(x1,x2) = max{14, x1 - 3, x2 - 3}
        activate#_A(x1) = max{5, x1}
        n__len_A(x1) = max{27, x1 + 18}
        len#_A(x1) = x1 + 9
        activate_A(x1) = x1
        n__add_A(x1,x2) = max{x1 + 16, x2 + 10}
        add#_A(x1,x2) = max{x1 + 9, x2 + 9}
        n__from_A(x1) = max{15, x1}
        n__fst_A(x1,x2) = max{x1 + 27, x2 + 12}
        fst_A(x1,x2) = max{x1 + 27, x2 + 12}
        |0|_A = 28
        nil_A = 11
        from_A(x1) = max{15, x1}
        n__s_A(x1) = max{0, x1 - 3}
        add_A(x1,x2) = max{x1 + 16, x2 + 10}
        len_A(x1) = max{27, x1 + 18}
  
    precedence:
    
      n__len = |0| = len > n__fst = fst = nil > s = activate = n__add = n__from = from = n__s = add > fst# = activate# = add# > len# > cons
    
    partial status:
  
      pi(fst#) = [1]
      pi(s) = []
      pi(cons) = []
      pi(activate#) = [1]
      pi(n__len) = []
      pi(len#) = []
      pi(activate) = [1]
      pi(n__add) = [2]
      pi(add#) = [1]
      pi(n__from) = [1]
      pi(n__fst) = []
      pi(fst) = []
      pi(|0|) = []
      pi(nil) = []
      pi(from) = [1]
      pi(n__s) = []
      pi(add) = [2]
      pi(len) = []

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: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__len(X)) -> activate#(X)
p3: activate#(n__len(X)) -> len#(activate(X))
p4: len#(cons(X,Z)) -> activate#(Z)
p5: activate#(n__add(X1,X2)) -> activate#(X2)
p6: activate#(n__add(X1,X2)) -> activate#(X1)
p7: activate#(n__add(X1,X2)) -> add#(activate(X1),activate(X2))
p8: add#(s(X),Y) -> activate#(X)
p9: activate#(n__from(X)) -> activate#(X)
p10: activate#(n__fst(X1,X2)) -> activate#(X2)
p11: activate#(n__fst(X1,X2)) -> activate#(X1)
p12: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))
p3: activate#(n__fst(X1,X2)) -> activate#(X1)
p4: activate#(n__fst(X1,X2)) -> activate#(X2)
p5: activate#(n__from(X)) -> activate#(X)
p6: activate#(n__add(X1,X2)) -> add#(activate(X1),activate(X2))
p7: add#(s(X),Y) -> activate#(X)
p8: activate#(n__add(X1,X2)) -> activate#(X1)
p9: activate#(n__add(X1,X2)) -> activate#(X2)
p10: activate#(n__len(X)) -> len#(activate(X))
p11: len#(cons(X,Z)) -> activate#(Z)
p12: activate#(n__len(X)) -> activate#(X)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fst#_A(x1,x2) = max{13, x1 + 8, x2 - 4}
        s_A(x1) = max{3, x1}
        cons_A(x1,x2) = max{17, x1 + 5, x2}
        activate#_A(x1) = max{12, x1 + 8}
        n__fst_A(x1,x2) = max{12, x1, x2}
        activate_A(x1) = max{3, x1}
        n__from_A(x1) = max{21, x1 + 18}
        n__add_A(x1,x2) = max{21, x1 + 17, x2 + 15}
        add#_A(x1,x2) = max{x1 + 8, x2 + 13}
        n__len_A(x1) = max{21, x1 + 15}
        len#_A(x1) = max{22, x1 + 8}
        fst_A(x1,x2) = max{12, x1, x2}
        |0|_A = 11
        nil_A = 11
        from_A(x1) = max{21, x1 + 18}
        n__s_A(x1) = x1
        add_A(x1,x2) = max{21, x1 + 17, x2 + 15}
        len_A(x1) = max{21, x1 + 15}
  
    precedence:
    
      n__fst = activate = add# = n__len = fst = |0| = nil = len > fst# = s = cons = activate# = n__from = n__add = len# = from = n__s = add
    
    partial status:
  
      pi(fst#) = []
      pi(s) = []
      pi(cons) = []
      pi(activate#) = []
      pi(n__fst) = []
      pi(activate) = []
      pi(n__from) = []
      pi(n__add) = []
      pi(add#) = []
      pi(n__len) = []
      pi(len#) = []
      pi(fst) = []
      pi(|0|) = []
      pi(nil) = []
      pi(from) = []
      pi(n__s) = []
      pi(add) = [2]
      pi(len) = []

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: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))
p3: activate#(n__fst(X1,X2)) -> activate#(X1)
p4: activate#(n__fst(X1,X2)) -> activate#(X2)
p5: activate#(n__from(X)) -> activate#(X)
p6: add#(s(X),Y) -> activate#(X)
p7: activate#(n__add(X1,X2)) -> activate#(X1)
p8: activate#(n__add(X1,X2)) -> activate#(X2)
p9: activate#(n__len(X)) -> len#(activate(X))
p10: len#(cons(X,Z)) -> activate#(Z)
p11: activate#(n__len(X)) -> activate#(X)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__len(X)) -> activate#(X)
p3: activate#(n__len(X)) -> len#(activate(X))
p4: len#(cons(X,Z)) -> activate#(Z)
p5: activate#(n__add(X1,X2)) -> activate#(X2)
p6: activate#(n__add(X1,X2)) -> activate#(X1)
p7: activate#(n__from(X)) -> activate#(X)
p8: activate#(n__fst(X1,X2)) -> activate#(X2)
p9: activate#(n__fst(X1,X2)) -> activate#(X1)
p10: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fst#_A(x1,x2) = max{x1 + 12, x2 + 12}
        s_A(x1) = max{4, x1}
        cons_A(x1,x2) = max{3, x1 + 2, x2}
        activate#_A(x1) = max{3, x1 - 6}
        n__len_A(x1) = x1 + 11
        len#_A(x1) = max{5, x1 + 4}
        activate_A(x1) = x1
        n__add_A(x1,x2) = max{x1 + 5, x2 + 7}
        n__from_A(x1) = max{9, x1 + 5}
        n__fst_A(x1,x2) = max{x1 + 20, x2 + 19}
        fst_A(x1,x2) = max{x1 + 20, x2 + 19}
        |0|_A = 12
        nil_A = 11
        from_A(x1) = max{9, x1 + 5}
        n__s_A(x1) = max{4, x1}
        add_A(x1,x2) = max{x1 + 5, x2 + 7}
        len_A(x1) = x1 + 11
  
    precedence:
    
      fst# = s = cons = activate# = n__len = len# = activate = n__add = n__from = n__fst = fst = |0| = nil = from = n__s = add = len
    
    partial status:
  
      pi(fst#) = []
      pi(s) = []
      pi(cons) = []
      pi(activate#) = []
      pi(n__len) = []
      pi(len#) = []
      pi(activate) = [1]
      pi(n__add) = [2]
      pi(n__from) = [1]
      pi(n__fst) = []
      pi(fst) = []
      pi(|0|) = []
      pi(nil) = []
      pi(from) = [1]
      pi(n__s) = []
      pi(add) = [2]
      pi(len) = [1]

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: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__len(X)) -> activate#(X)
p3: activate#(n__len(X)) -> len#(activate(X))
p4: len#(cons(X,Z)) -> activate#(Z)
p5: activate#(n__add(X1,X2)) -> activate#(X2)
p6: activate#(n__add(X1,X2)) -> activate#(X1)
p7: activate#(n__from(X)) -> activate#(X)
p8: activate#(n__fst(X1,X2)) -> activate#(X1)
p9: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))
p3: activate#(n__fst(X1,X2)) -> activate#(X1)
p4: activate#(n__from(X)) -> activate#(X)
p5: activate#(n__add(X1,X2)) -> activate#(X1)
p6: activate#(n__add(X1,X2)) -> activate#(X2)
p7: activate#(n__len(X)) -> len#(activate(X))
p8: len#(cons(X,Z)) -> activate#(Z)
p9: activate#(n__len(X)) -> activate#(X)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fst#_A(x1,x2) = max{x1 + 2, x2 + 2}
        s_A(x1) = max{2, x1}
        cons_A(x1,x2) = max{x1 - 1, x2}
        activate#_A(x1) = x1
        n__fst_A(x1,x2) = max{x1 + 3, x2 + 5}
        activate_A(x1) = x1
        n__from_A(x1) = max{5, x1 + 3}
        n__add_A(x1,x2) = max{x1 + 1, x2 + 4}
        n__len_A(x1) = x1 + 5
        len#_A(x1) = x1
        fst_A(x1,x2) = max{x1 + 3, x2 + 5}
        |0|_A = 4
        nil_A = 0
        from_A(x1) = max{5, x1 + 3}
        n__s_A(x1) = max{2, x1}
        add_A(x1,x2) = max{x1 + 1, x2 + 4}
        len_A(x1) = x1 + 5
  
    precedence:
    
      activate = nil = from = len > n__fst = n__from = n__add = n__len = fst = |0| = add > s = cons = activate# = n__s > fst# > len#
    
    partial status:
  
      pi(fst#) = [1, 2]
      pi(s) = [1]
      pi(cons) = [2]
      pi(activate#) = [1]
      pi(n__fst) = []
      pi(activate) = [1]
      pi(n__from) = []
      pi(n__add) = []
      pi(n__len) = []
      pi(len#) = [1]
      pi(fst) = [1]
      pi(|0|) = []
      pi(nil) = []
      pi(from) = [1]
      pi(n__s) = [1]
      pi(add) = [2]
      pi(len) = []

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: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))
p3: activate#(n__fst(X1,X2)) -> activate#(X1)
p4: activate#(n__from(X)) -> activate#(X)
p5: activate#(n__add(X1,X2)) -> activate#(X1)
p6: activate#(n__add(X1,X2)) -> activate#(X2)
p7: activate#(n__len(X)) -> len#(activate(X))
p8: activate#(n__len(X)) -> activate#(X)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__len(X)) -> activate#(X)
p3: activate#(n__add(X1,X2)) -> activate#(X2)
p4: activate#(n__add(X1,X2)) -> activate#(X1)
p5: activate#(n__from(X)) -> activate#(X)
p6: activate#(n__fst(X1,X2)) -> activate#(X1)
p7: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fst#_A(x1,x2) = max{21, x1 + 18, x2 + 2}
        s_A(x1) = max{4, x1 - 5}
        cons_A(x1,x2) = max{x1 + 20, x2 - 5}
        activate#_A(x1) = max{22, x1 + 12}
        n__len_A(x1) = x1
        n__add_A(x1,x2) = max{x1 + 2, x2 + 11}
        n__from_A(x1) = x1 + 23
        n__fst_A(x1,x2) = max{x1 + 23, x2 + 24}
        activate_A(x1) = x1
        fst_A(x1,x2) = max{x1 + 23, x2 + 24}
        |0|_A = 24
        nil_A = 25
        from_A(x1) = x1 + 23
        n__s_A(x1) = max{4, x1 - 5}
        add_A(x1,x2) = max{x1 + 2, x2 + 11}
        len_A(x1) = x1
  
    precedence:
    
      fst# > activate# = n__fst = fst > cons = activate = |0| = from = add > n__add > n__from = nil > s > n__s = len > n__len
    
    partial status:
  
      pi(fst#) = []
      pi(s) = []
      pi(cons) = [1]
      pi(activate#) = [1]
      pi(n__len) = [1]
      pi(n__add) = []
      pi(n__from) = [1]
      pi(n__fst) = []
      pi(activate) = [1]
      pi(fst) = []
      pi(|0|) = []
      pi(nil) = []
      pi(from) = [1]
      pi(n__s) = []
      pi(add) = []
      pi(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: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__len(X)) -> activate#(X)
p3: activate#(n__add(X1,X2)) -> activate#(X2)
p4: activate#(n__add(X1,X2)) -> activate#(X1)
p5: activate#(n__from(X)) -> activate#(X)
p6: activate#(n__fst(X1,X2)) -> activate#(X1)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: activate#(n__len(X)) -> activate#(X)
p2: activate#(n__fst(X1,X2)) -> activate#(X1)
p3: activate#(n__from(X)) -> activate#(X)
p4: activate#(n__add(X1,X2)) -> activate#(X1)
p5: activate#(n__add(X1,X2)) -> activate#(X2)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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:
      
        activate#_A(x1) = x1 + 2
        n__len_A(x1) = x1
        n__fst_A(x1,x2) = max{x1, x2}
        n__from_A(x1) = x1 + 1
        n__add_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      activate# = n__len = n__fst = n__from = n__add
    
    partial status:
  
      pi(activate#) = []
      pi(n__len) = [1]
      pi(n__fst) = [2]
      pi(n__from) = [1]
      pi(n__add) = [2]

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: activate#(n__len(X)) -> activate#(X)
p2: activate#(n__fst(X1,X2)) -> activate#(X1)
p3: activate#(n__add(X1,X2)) -> activate#(X1)
p4: activate#(n__add(X1,X2)) -> activate#(X2)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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: activate#(n__len(X)) -> activate#(X)
p2: activate#(n__add(X1,X2)) -> activate#(X2)
p3: activate#(n__add(X1,X2)) -> activate#(X1)
p4: activate#(n__fst(X1,X2)) -> activate#(X1)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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:
      
        activate#_A(x1) = x1 + 2
        n__len_A(x1) = x1 + 1
        n__add_A(x1,x2) = max{x1, x2}
        n__fst_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      activate# = n__len = n__add = n__fst
    
    partial status:
  
      pi(activate#) = []
      pi(n__len) = [1]
      pi(n__add) = [2]
      pi(n__fst) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__add(X1,X2)) -> activate#(X2)
p2: activate#(n__add(X1,X2)) -> activate#(X1)
p3: activate#(n__fst(X1,X2)) -> activate#(X1)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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: activate#(n__add(X1,X2)) -> activate#(X2)
p2: activate#(n__fst(X1,X2)) -> activate#(X1)
p3: activate#(n__add(X1,X2)) -> activate#(X1)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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:
      
        activate#_A(x1) = max{4, x1 + 3}
        n__add_A(x1,x2) = max{x1, x2 + 2}
        n__fst_A(x1,x2) = max{x1 + 1, x2 + 1}
  
    precedence:
    
      activate# = n__add = n__fst
    
    partial status:
  
      pi(activate#) = []
      pi(n__add) = [2]
      pi(n__fst) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__fst(X1,X2)) -> activate#(X1)
p2: activate#(n__add(X1,X2)) -> activate#(X1)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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: activate#(n__fst(X1,X2)) -> activate#(X1)
p2: activate#(n__add(X1,X2)) -> activate#(X1)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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:
      
        activate#_A(x1) = x1 + 2
        n__fst_A(x1,x2) = max{x1 + 2, x2 + 2}
        n__add_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      activate# = n__fst = n__add
    
    partial status:
  
      pi(activate#) = []
      pi(n__fst) = [2]
      pi(n__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: activate#(n__add(X1,X2)) -> activate#(X1)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: activate#(n__add(X1,X2)) -> activate#(X1)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> 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:
      
        activate#_A(x1) = x1 + 3
        n__add_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      activate# = n__add
    
    partial status:
  
      pi(activate#) = []
      pi(n__add) = [2]

The next rules are strictly ordered:

  p1

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