YES

We show the termination of the TRS R:

  and(true(),X) -> activate(X)
  and(false(),Y) -> false()
  if(true(),X,Y) -> activate(X)
  if(false(),X,Y) -> activate(Y)
  add(|0|(),X) -> activate(X)
  add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
  first(|0|(),X) -> nil()
  first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
  from(X) -> cons(activate(X),n__from(n__s(activate(X))))
  add(X1,X2) -> n__add(X1,X2)
  first(X1,X2) -> n__first(X1,X2)
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  activate(n__add(X1,X2)) -> add(activate(X1),X2)
  activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
  activate(n__from(X)) -> from(X)
  activate(n__s(X)) -> s(X)
  activate(X) -> X

-- SCC decomposition.

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

p1: and#(true(),X) -> activate#(X)
p2: if#(true(),X,Y) -> activate#(X)
p3: if#(false(),X,Y) -> activate#(Y)
p4: add#(|0|(),X) -> activate#(X)
p5: add#(s(X),Y) -> s#(n__add(activate(X),activate(Y)))
p6: add#(s(X),Y) -> activate#(X)
p7: add#(s(X),Y) -> activate#(Y)
p8: first#(s(X),cons(Y,Z)) -> activate#(Y)
p9: first#(s(X),cons(Y,Z)) -> activate#(X)
p10: first#(s(X),cons(Y,Z)) -> activate#(Z)
p11: from#(X) -> activate#(X)
p12: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p13: activate#(n__add(X1,X2)) -> activate#(X1)
p14: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p15: activate#(n__first(X1,X2)) -> activate#(X1)
p16: activate#(n__first(X1,X2)) -> activate#(X2)
p17: activate#(n__from(X)) -> from#(X)
p18: activate#(n__s(X)) -> s#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p2: add#(s(X),Y) -> activate#(Y)
p3: activate#(n__from(X)) -> from#(X)
p4: from#(X) -> activate#(X)
p5: activate#(n__first(X1,X2)) -> activate#(X2)
p6: activate#(n__first(X1,X2)) -> activate#(X1)
p7: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p8: first#(s(X),cons(Y,Z)) -> activate#(Z)
p9: activate#(n__add(X1,X2)) -> activate#(X1)
p10: first#(s(X),cons(Y,Z)) -> activate#(X)
p11: first#(s(X),cons(Y,Z)) -> activate#(Y)
p12: add#(s(X),Y) -> activate#(X)
p13: add#(|0|(),X) -> activate#(X)

and R consists of:

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

The set of usable rules consists of

  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:
      
        activate#_A(x1) = max{1, x1 - 3}
        n__add_A(x1,x2) = max{x1 + 17, x2 + 15}
        add#_A(x1,x2) = max{x1 + 11, x2 + 2}
        activate_A(x1) = x1
        s_A(x1) = max{7, x1 - 1}
        n__from_A(x1) = max{14, x1 + 6}
        from#_A(x1) = x1 + 2
        n__first_A(x1,x2) = max{x1 + 10, x2 + 13}
        first#_A(x1,x2) = max{x1 + 4, x2 + 4}
        cons_A(x1,x2) = max{7, x1 + 6, x2 - 2}
        |0|_A = 2
        add_A(x1,x2) = max{x1 + 17, x2 + 15}
        first_A(x1,x2) = max{x1 + 10, x2 + 13}
        nil_A = 1
        from_A(x1) = max{14, x1 + 6}
        n__s_A(x1) = max{7, x1 - 1}
  
    precedence:
    
      activate# > add# = activate = s = add > |0| > n__add = from# = from > first# > n__from > cons = first > n__first = nil = n__s
    
    partial status:
  
      pi(activate#) = []
      pi(n__add) = [1, 2]
      pi(add#) = [1, 2]
      pi(activate) = [1]
      pi(s) = []
      pi(n__from) = []
      pi(from#) = []
      pi(n__first) = [1, 2]
      pi(first#) = [1, 2]
      pi(cons) = []
      pi(|0|) = []
      pi(add) = [2]
      pi(first) = [1, 2]
      pi(nil) = []
      pi(from) = [1]
      pi(n__s) = []

The next rules are strictly ordered:

  p11

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)) -> add#(activate(X1),X2)
p2: add#(s(X),Y) -> activate#(Y)
p3: activate#(n__from(X)) -> from#(X)
p4: from#(X) -> activate#(X)
p5: activate#(n__first(X1,X2)) -> activate#(X2)
p6: activate#(n__first(X1,X2)) -> activate#(X1)
p7: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p8: first#(s(X),cons(Y,Z)) -> activate#(Z)
p9: activate#(n__add(X1,X2)) -> activate#(X1)
p10: first#(s(X),cons(Y,Z)) -> activate#(X)
p11: add#(s(X),Y) -> activate#(X)
p12: add#(|0|(),X) -> activate#(X)

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  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:
      
        activate#_A(x1) = x1 + 4
        n__add_A(x1,x2) = max{x1 + 13, x2 + 11}
        add#_A(x1,x2) = max{x1 + 8, x2 + 5}
        activate_A(x1) = x1
        |0|_A = 2
        n__first_A(x1,x2) = max{x1 + 30, x2 + 42}
        first#_A(x1,x2) = max{x1 + 29, x2 + 29}
        s_A(x1) = max{5, x1 - 3}
        cons_A(x1,x2) = max{3, x1 - 24, x2 - 11}
        n__from_A(x1) = x1 + 6
        from#_A(x1) = x1 + 4
        add_A(x1,x2) = max{x1 + 13, x2 + 11}
        first_A(x1,x2) = max{x1 + 30, x2 + 42}
        nil_A = 1
        from_A(x1) = x1 + 6
        n__s_A(x1) = max{5, x1 - 3}
  
    precedence:
    
      activate = from > n__from = from# > add# > activate# > |0| = first# = s = add > n__add = n__first = cons = first > nil = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__add) = [1, 2]
      pi(add#) = []
      pi(activate) = [1]
      pi(|0|) = []
      pi(n__first) = []
      pi(first#) = []
      pi(s) = []
      pi(cons) = []
      pi(n__from) = [1]
      pi(from#) = []
      pi(add) = [2]
      pi(first) = []
      pi(nil) = []
      pi(from) = [1]
      pi(n__s) = []

The next rules are strictly ordered:

  p12

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)) -> add#(activate(X1),X2)
p2: add#(|0|(),X) -> activate#(X)
p3: activate#(n__add(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p5: first#(s(X),cons(Y,Z)) -> activate#(X)
p6: activate#(n__first(X1,X2)) -> activate#(X1)
p7: activate#(n__first(X1,X2)) -> activate#(X2)
p8: activate#(n__from(X)) -> from#(X)
p9: from#(X) -> activate#(X)
p10: first#(s(X),cons(Y,Z)) -> activate#(Z)
p11: add#(s(X),Y) -> activate#(X)

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  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:
      
        activate#_A(x1) = max{62, x1 + 34}
        n__add_A(x1,x2) = max{x1 + 30, x2 + 38}
        add#_A(x1,x2) = max{x1 + 56, x2 + 62}
        activate_A(x1) = max{7, x1}
        s_A(x1) = max{7, x1 - 9}
        n__from_A(x1) = x1 + 29
        from#_A(x1) = x1 + 62
        n__first_A(x1,x2) = max{46, x1 + 25, x2 + 18}
        first#_A(x1,x2) = max{x1 + 55, x2 + 47}
        cons_A(x1,x2) = max{x1 + 15, x2 - 12}
        |0|_A = 6
        add_A(x1,x2) = max{x1 + 30, x2 + 38}
        first_A(x1,x2) = max{46, x1 + 25, x2 + 18}
        nil_A = 17
        from_A(x1) = x1 + 29
        n__s_A(x1) = max{1, x1 - 9}
  
    precedence:
    
      add# = from# > activate# = first# > n__add = activate = s = n__from = n__first = cons = |0| = add = first = nil = from = n__s
    
    partial status:
  
      pi(activate#) = []
      pi(n__add) = [1, 2]
      pi(add#) = [1]
      pi(activate) = [1]
      pi(s) = []
      pi(n__from) = [1]
      pi(from#) = []
      pi(n__first) = []
      pi(first#) = []
      pi(cons) = []
      pi(|0|) = []
      pi(add) = [1, 2]
      pi(first) = [2]
      pi(nil) = []
      pi(from) = [1]
      pi(n__s) = []

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__add(X1,X2)) -> add#(activate(X1),X2)
p2: add#(s(X),Y) -> activate#(X)
p3: from#(X) -> activate#(X)
p4: activate#(n__first(X1,X2)) -> activate#(X2)
p5: activate#(n__first(X1,X2)) -> activate#(X1)
p6: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p7: first#(s(X),cons(Y,Z)) -> activate#(Z)
p8: activate#(n__add(X1,X2)) -> activate#(X1)
p9: first#(s(X),cons(Y,Z)) -> activate#(X)
p10: add#(|0|(),X) -> activate#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p2: add#(|0|(),X) -> activate#(X)
p3: activate#(n__add(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p5: first#(s(X),cons(Y,Z)) -> activate#(X)
p6: activate#(n__first(X1,X2)) -> activate#(X1)
p7: activate#(n__first(X1,X2)) -> activate#(X2)
p8: first#(s(X),cons(Y,Z)) -> activate#(Z)
p9: add#(s(X),Y) -> activate#(X)

and R consists of:

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

The set of usable rules consists of

  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:
      
        activate#_A(x1) = max{25, x1 + 23}
        n__add_A(x1,x2) = max{x1 + 22, x2 + 16}
        add#_A(x1,x2) = max{x1 + 41, x2 + 38}
        activate_A(x1) = x1
        |0|_A = 1
        n__first_A(x1,x2) = max{x1 + 28, x2 + 24}
        first#_A(x1,x2) = max{x1 + 51, x2 + 46}
        s_A(x1) = max{10, x1 - 3}
        cons_A(x1,x2) = max{x1 + 10, x2 - 20}
        add_A(x1,x2) = max{x1 + 22, x2 + 16}
        first_A(x1,x2) = max{x1 + 28, x2 + 24}
        nil_A = 0
        from_A(x1) = x1 + 10
        n__from_A(x1) = x1 + 10
        n__s_A(x1) = max{10, x1 - 3}
  
    precedence:
    
      activate# > add# > n__add = activate = |0| = n__first = first# = s = cons = add = first = nil = from = n__from = n__s
    
    partial status:
  
      pi(activate#) = []
      pi(n__add) = []
      pi(add#) = [2]
      pi(activate) = [1]
      pi(|0|) = []
      pi(n__first) = []
      pi(first#) = []
      pi(s) = []
      pi(cons) = []
      pi(add) = []
      pi(first) = []
      pi(nil) = []
      pi(from) = [1]
      pi(n__from) = [1]
      pi(n__s) = []

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: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p2: add#(|0|(),X) -> activate#(X)
p3: activate#(n__add(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p5: first#(s(X),cons(Y,Z)) -> activate#(X)
p6: activate#(n__first(X1,X2)) -> activate#(X1)
p7: activate#(n__first(X1,X2)) -> activate#(X2)
p8: add#(s(X),Y) -> activate#(X)

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  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:
      
        activate#_A(x1) = max{1, x1}
        n__add_A(x1,x2) = max{x1 + 4, x2 + 11}
        add#_A(x1,x2) = max{x1 + 2, x2 + 10}
        activate_A(x1) = max{6, x1}
        s_A(x1) = max{14, x1}
        n__first_A(x1,x2) = max{24, x1 + 9, x2 + 16}
        first#_A(x1,x2) = max{x1 + 2, x2 + 11}
        cons_A(x1,x2) = max{5, x1 - 5, x2}
        |0|_A = 3
        add_A(x1,x2) = max{x1 + 4, x2 + 11}
        first_A(x1,x2) = max{24, x1 + 9, x2 + 16}
        nil_A = 4
        from_A(x1) = max{10, x1 - 5}
        n__from_A(x1) = max{10, x1 - 5}
        n__s_A(x1) = max{14, x1}
  
    precedence:
    
      n__first = first = nil > activate = cons = |0| = add = from > n__add = add# > n__from > activate# = s = first# = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__add) = []
      pi(add#) = []
      pi(activate) = [1]
      pi(s) = [1]
      pi(n__first) = []
      pi(first#) = [1]
      pi(cons) = []
      pi(|0|) = []
      pi(add) = []
      pi(first) = []
      pi(nil) = []
      pi(from) = []
      pi(n__from) = []
      pi(n__s) = [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: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p2: add#(s(X),Y) -> activate#(X)
p3: activate#(n__first(X1,X2)) -> activate#(X2)
p4: activate#(n__first(X1,X2)) -> activate#(X1)
p5: first#(s(X),cons(Y,Z)) -> activate#(X)
p6: activate#(n__add(X1,X2)) -> activate#(X1)
p7: add#(|0|(),X) -> activate#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p2: add#(|0|(),X) -> activate#(X)
p3: activate#(n__add(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> activate#(X1)
p5: activate#(n__first(X1,X2)) -> activate#(X2)
p6: add#(s(X),Y) -> activate#(X)

and R consists of:

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

The set of usable rules consists of

  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:
      
        activate#_A(x1) = x1 + 3
        n__add_A(x1,x2) = max{10, x1 + 4, x2 + 9}
        add#_A(x1,x2) = max{8, x1 + 3, x2 + 4}
        activate_A(x1) = x1
        |0|_A = 8
        n__first_A(x1,x2) = max{x1 + 8, x2 + 8}
        s_A(x1) = x1
        add_A(x1,x2) = max{10, x1 + 4, x2 + 9}
        first_A(x1,x2) = max{x1 + 8, x2 + 8}
        nil_A = 7
        cons_A(x1,x2) = max{4, x1}
        from_A(x1) = max{4, x1}
        n__from_A(x1) = max{4, x1}
        n__s_A(x1) = x1
  
    precedence:
    
      n__add = activate = s = add = from > add# = n__first = first = cons = n__s > activate# > |0| = nil = n__from
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__add) = []
      pi(add#) = [1, 2]
      pi(activate) = []
      pi(|0|) = []
      pi(n__first) = []
      pi(s) = []
      pi(add) = []
      pi(first) = []
      pi(nil) = []
      pi(cons) = []
      pi(from) = []
      pi(n__from) = []
      pi(n__s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: add#(|0|(),X) -> activate#(X)
p2: activate#(n__add(X1,X2)) -> activate#(X1)
p3: activate#(n__first(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> activate#(X2)
p5: add#(s(X),Y) -> activate#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}


-- Reduction pair.

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

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

and R consists of:

r1: and(true(),X) -> activate(X)
r2: and(false(),Y) -> false()
r3: if(true(),X,Y) -> activate(X)
r4: if(false(),X,Y) -> activate(Y)
r5: add(|0|(),X) -> activate(X)
r6: add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
r7: first(|0|(),X) -> nil()
r8: first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: add(X1,X2) -> n__add(X1,X2)
r11: first(X1,X2) -> n__first(X1,X2)
r12: from(X) -> n__from(X)
r13: s(X) -> n__s(X)
r14: activate(n__add(X1,X2)) -> add(activate(X1),X2)
r15: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r16: activate(n__from(X)) -> from(X)
r17: activate(n__s(X)) -> s(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, x2 + 1}
        n__first_A(x1,x2) = max{x1, x2 + 1}
  
    precedence:
    
      activate# = n__add = n__first
    
    partial status:
  
      pi(activate#) = []
      pi(n__add) = [2]
      pi(n__first) = [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: activate#(n__add(X1,X2)) -> activate#(X1)
p2: activate#(n__first(X1,X2)) -> activate#(X1)

and R consists of:

r1: and(true(),X) -> activate(X)
r2: and(false(),Y) -> false()
r3: if(true(),X,Y) -> activate(X)
r4: if(false(),X,Y) -> activate(Y)
r5: add(|0|(),X) -> activate(X)
r6: add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
r7: first(|0|(),X) -> nil()
r8: first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: add(X1,X2) -> n__add(X1,X2)
r11: first(X1,X2) -> n__first(X1,X2)
r12: from(X) -> n__from(X)
r13: s(X) -> n__s(X)
r14: activate(n__add(X1,X2)) -> add(activate(X1),X2)
r15: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r16: activate(n__from(X)) -> from(X)
r17: activate(n__s(X)) -> s(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__add(X1,X2)) -> activate#(X1)
p2: activate#(n__first(X1,X2)) -> activate#(X1)

and R consists of:

r1: and(true(),X) -> activate(X)
r2: and(false(),Y) -> false()
r3: if(true(),X,Y) -> activate(X)
r4: if(false(),X,Y) -> activate(Y)
r5: add(|0|(),X) -> activate(X)
r6: add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
r7: first(|0|(),X) -> nil()
r8: first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: add(X1,X2) -> n__add(X1,X2)
r11: first(X1,X2) -> n__first(X1,X2)
r12: from(X) -> n__from(X)
r13: s(X) -> n__s(X)
r14: activate(n__add(X1,X2)) -> add(activate(X1),X2)
r15: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r16: activate(n__from(X)) -> from(X)
r17: activate(n__s(X)) -> s(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__add_A(x1,x2) = max{x1 + 2, x2 + 2}
        n__first_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      activate# = n__add = n__first
    
    partial status:
  
      pi(activate#) = []
      pi(n__add) = [2]
      pi(n__first) = [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__first(X1,X2)) -> activate#(X1)

and R consists of:

r1: and(true(),X) -> activate(X)
r2: and(false(),Y) -> false()
r3: if(true(),X,Y) -> activate(X)
r4: if(false(),X,Y) -> activate(Y)
r5: add(|0|(),X) -> activate(X)
r6: add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
r7: first(|0|(),X) -> nil()
r8: first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: add(X1,X2) -> n__add(X1,X2)
r11: first(X1,X2) -> n__first(X1,X2)
r12: from(X) -> n__from(X)
r13: s(X) -> n__s(X)
r14: activate(n__add(X1,X2)) -> add(activate(X1),X2)
r15: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r16: activate(n__from(X)) -> from(X)
r17: activate(n__s(X)) -> s(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__first(X1,X2)) -> activate#(X1)

and R consists of:

r1: and(true(),X) -> activate(X)
r2: and(false(),Y) -> false()
r3: if(true(),X,Y) -> activate(X)
r4: if(false(),X,Y) -> activate(Y)
r5: add(|0|(),X) -> activate(X)
r6: add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
r7: first(|0|(),X) -> nil()
r8: first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: add(X1,X2) -> n__add(X1,X2)
r11: first(X1,X2) -> n__first(X1,X2)
r12: from(X) -> n__from(X)
r13: s(X) -> n__s(X)
r14: activate(n__add(X1,X2)) -> add(activate(X1),X2)
r15: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r16: activate(n__from(X)) -> from(X)
r17: activate(n__s(X)) -> s(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__first_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      activate# = n__first
    
    partial status:
  
      pi(activate#) = []
      pi(n__first) = [2]

The next rules are strictly ordered:

  p1

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