YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(n__s(X)))
  |2ndspos|(|0|(),Z) -> rnil()
  |2ndspos|(s(N),cons(X,Z)) -> |2ndspos|(s(N),cons2(X,activate(Z)))
  |2ndspos|(s(N),cons2(X,cons(Y,Z))) -> rcons(posrecip(Y),|2ndsneg|(N,activate(Z)))
  |2ndsneg|(|0|(),Z) -> rnil()
  |2ndsneg|(s(N),cons(X,Z)) -> |2ndsneg|(s(N),cons2(X,activate(Z)))
  |2ndsneg|(s(N),cons2(X,cons(Y,Z))) -> rcons(negrecip(Y),|2ndspos|(N,activate(Z)))
  pi(X) -> |2ndspos|(X,from(|0|()))
  plus(|0|(),Y) -> Y
  plus(s(X),Y) -> s(plus(X,Y))
  times(|0|(),Y) -> |0|()
  times(s(X),Y) -> plus(Y,times(X,Y))
  square(X) -> times(X,X)
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(activate(X))
  activate(X) -> X

-- SCC decomposition.

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

p1: |2ndspos|#(s(N),cons(X,Z)) -> |2ndspos|#(s(N),cons2(X,activate(Z)))
p2: |2ndspos|#(s(N),cons(X,Z)) -> activate#(Z)
p3: |2ndspos|#(s(N),cons2(X,cons(Y,Z))) -> |2ndsneg|#(N,activate(Z))
p4: |2ndspos|#(s(N),cons2(X,cons(Y,Z))) -> activate#(Z)
p5: |2ndsneg|#(s(N),cons(X,Z)) -> |2ndsneg|#(s(N),cons2(X,activate(Z)))
p6: |2ndsneg|#(s(N),cons(X,Z)) -> activate#(Z)
p7: |2ndsneg|#(s(N),cons2(X,cons(Y,Z))) -> |2ndspos|#(N,activate(Z))
p8: |2ndsneg|#(s(N),cons2(X,cons(Y,Z))) -> activate#(Z)
p9: pi#(X) -> |2ndspos|#(X,from(|0|()))
p10: pi#(X) -> from#(|0|())
p11: plus#(s(X),Y) -> s#(plus(X,Y))
p12: plus#(s(X),Y) -> plus#(X,Y)
p13: times#(s(X),Y) -> plus#(Y,times(X,Y))
p14: times#(s(X),Y) -> times#(X,Y)
p15: square#(X) -> times#(X,X)
p16: activate#(n__from(X)) -> from#(activate(X))
p17: activate#(n__from(X)) -> activate#(X)
p18: activate#(n__s(X)) -> s#(activate(X))
p19: activate#(n__s(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: |2ndspos|(|0|(),Z) -> rnil()
r3: |2ndspos|(s(N),cons(X,Z)) -> |2ndspos|(s(N),cons2(X,activate(Z)))
r4: |2ndspos|(s(N),cons2(X,cons(Y,Z))) -> rcons(posrecip(Y),|2ndsneg|(N,activate(Z)))
r5: |2ndsneg|(|0|(),Z) -> rnil()
r6: |2ndsneg|(s(N),cons(X,Z)) -> |2ndsneg|(s(N),cons2(X,activate(Z)))
r7: |2ndsneg|(s(N),cons2(X,cons(Y,Z))) -> rcons(negrecip(Y),|2ndspos|(N,activate(Z)))
r8: pi(X) -> |2ndspos|(X,from(|0|()))
r9: plus(|0|(),Y) -> Y
r10: plus(s(X),Y) -> s(plus(X,Y))
r11: times(|0|(),Y) -> |0|()
r12: times(s(X),Y) -> plus(Y,times(X,Y))
r13: square(X) -> times(X,X)
r14: from(X) -> n__from(X)
r15: s(X) -> n__s(X)
r16: activate(n__from(X)) -> from(activate(X))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p3, p5, p7}
  {p17, p19}
  {p14}
  {p12}


-- Reduction pair.

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

p1: |2ndspos|#(s(N),cons(X,Z)) -> |2ndspos|#(s(N),cons2(X,activate(Z)))
p2: |2ndspos|#(s(N),cons2(X,cons(Y,Z))) -> |2ndsneg|#(N,activate(Z))
p3: |2ndsneg|#(s(N),cons2(X,cons(Y,Z))) -> |2ndspos|#(N,activate(Z))
p4: |2ndsneg|#(s(N),cons(X,Z)) -> |2ndsneg|#(s(N),cons2(X,activate(Z)))

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: |2ndspos|(|0|(),Z) -> rnil()
r3: |2ndspos|(s(N),cons(X,Z)) -> |2ndspos|(s(N),cons2(X,activate(Z)))
r4: |2ndspos|(s(N),cons2(X,cons(Y,Z))) -> rcons(posrecip(Y),|2ndsneg|(N,activate(Z)))
r5: |2ndsneg|(|0|(),Z) -> rnil()
r6: |2ndsneg|(s(N),cons(X,Z)) -> |2ndsneg|(s(N),cons2(X,activate(Z)))
r7: |2ndsneg|(s(N),cons2(X,cons(Y,Z))) -> rcons(negrecip(Y),|2ndspos|(N,activate(Z)))
r8: pi(X) -> |2ndspos|(X,from(|0|()))
r9: plus(|0|(),Y) -> Y
r10: plus(s(X),Y) -> s(plus(X,Y))
r11: times(|0|(),Y) -> |0|()
r12: times(s(X),Y) -> plus(Y,times(X,Y))
r13: square(X) -> times(X,X)
r14: from(X) -> n__from(X)
r15: s(X) -> n__s(X)
r16: activate(n__from(X)) -> from(activate(X))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The set of usable rules consists of

  r1, r14, r15, r16, r17, r18

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        activate > from > cons2 > s > n__s > n__from > cons > |2ndsneg|# > |2ndspos|#
      
      argument filter:
    
        pi(|2ndspos|#) = 1
        pi(s) = [1]
        pi(cons) = 2
        pi(cons2) = []
        pi(activate) = [1]
        pi(|2ndsneg|#) = [1]
        pi(from) = [1]
        pi(n__from) = [1]
        pi(n__s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        n__s > from > cons > |2ndsneg|# > |2ndspos|# > activate > n__from > cons2 > s
      
      argument filter:
    
        pi(|2ndspos|#) = []
        pi(s) = [1]
        pi(cons) = 2
        pi(cons2) = []
        pi(activate) = [1]
        pi(|2ndsneg|#) = []
        pi(from) = 1
        pi(n__from) = 1
        pi(n__s) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        n__from > activate > n__s > from > cons > |2ndspos|# > cons2 > |2ndsneg|# > s
      
      argument filter:
    
        pi(|2ndspos|#) = []
        pi(s) = [1]
        pi(cons) = 2
        pi(cons2) = []
        pi(activate) = 1
        pi(|2ndsneg|#) = []
        pi(from) = 1
        pi(n__from) = 1
        pi(n__s) = 1
    

The next rules are strictly ordered:

  p2, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: |2ndspos|#(s(N),cons(X,Z)) -> |2ndspos|#(s(N),cons2(X,activate(Z)))
p2: |2ndsneg|#(s(N),cons(X,Z)) -> |2ndsneg|#(s(N),cons2(X,activate(Z)))

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: |2ndspos|(|0|(),Z) -> rnil()
r3: |2ndspos|(s(N),cons(X,Z)) -> |2ndspos|(s(N),cons2(X,activate(Z)))
r4: |2ndspos|(s(N),cons2(X,cons(Y,Z))) -> rcons(posrecip(Y),|2ndsneg|(N,activate(Z)))
r5: |2ndsneg|(|0|(),Z) -> rnil()
r6: |2ndsneg|(s(N),cons(X,Z)) -> |2ndsneg|(s(N),cons2(X,activate(Z)))
r7: |2ndsneg|(s(N),cons2(X,cons(Y,Z))) -> rcons(negrecip(Y),|2ndspos|(N,activate(Z)))
r8: pi(X) -> |2ndspos|(X,from(|0|()))
r9: plus(|0|(),Y) -> Y
r10: plus(s(X),Y) -> s(plus(X,Y))
r11: times(|0|(),Y) -> |0|()
r12: times(s(X),Y) -> plus(Y,times(X,Y))
r13: square(X) -> times(X,X)
r14: from(X) -> n__from(X)
r15: s(X) -> n__s(X)
r16: activate(n__from(X)) -> from(activate(X))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: activate#(n__s(X)) -> activate#(X)
p2: activate#(n__from(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: |2ndspos|(|0|(),Z) -> rnil()
r3: |2ndspos|(s(N),cons(X,Z)) -> |2ndspos|(s(N),cons2(X,activate(Z)))
r4: |2ndspos|(s(N),cons2(X,cons(Y,Z))) -> rcons(posrecip(Y),|2ndsneg|(N,activate(Z)))
r5: |2ndsneg|(|0|(),Z) -> rnil()
r6: |2ndsneg|(s(N),cons(X,Z)) -> |2ndsneg|(s(N),cons2(X,activate(Z)))
r7: |2ndsneg|(s(N),cons2(X,cons(Y,Z))) -> rcons(negrecip(Y),|2ndspos|(N,activate(Z)))
r8: pi(X) -> |2ndspos|(X,from(|0|()))
r9: plus(|0|(),Y) -> Y
r10: plus(s(X),Y) -> s(plus(X,Y))
r11: times(|0|(),Y) -> |0|()
r12: times(s(X),Y) -> plus(Y,times(X,Y))
r13: square(X) -> times(X,X)
r14: from(X) -> n__from(X)
r15: s(X) -> n__s(X)
r16: activate(n__from(X)) -> from(activate(X))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        activate# > n__from > n__s
      
      argument filter:
    
        pi(activate#) = 1
        pi(n__s) = [1]
        pi(n__from) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        activate# > n__from > n__s
      
      argument filter:
    
        pi(activate#) = 1
        pi(n__s) = [1]
        pi(n__from) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        activate# > n__from > n__s
      
      argument filter:
    
        pi(activate#) = 1
        pi(n__s) = [1]
        pi(n__from) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

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

-- Reduction pair.

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

p1: times#(s(X),Y) -> times#(X,Y)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: |2ndspos|(|0|(),Z) -> rnil()
r3: |2ndspos|(s(N),cons(X,Z)) -> |2ndspos|(s(N),cons2(X,activate(Z)))
r4: |2ndspos|(s(N),cons2(X,cons(Y,Z))) -> rcons(posrecip(Y),|2ndsneg|(N,activate(Z)))
r5: |2ndsneg|(|0|(),Z) -> rnil()
r6: |2ndsneg|(s(N),cons(X,Z)) -> |2ndsneg|(s(N),cons2(X,activate(Z)))
r7: |2ndsneg|(s(N),cons2(X,cons(Y,Z))) -> rcons(negrecip(Y),|2ndspos|(N,activate(Z)))
r8: pi(X) -> |2ndspos|(X,from(|0|()))
r9: plus(|0|(),Y) -> Y
r10: plus(s(X),Y) -> s(plus(X,Y))
r11: times(|0|(),Y) -> |0|()
r12: times(s(X),Y) -> plus(Y,times(X,Y))
r13: square(X) -> times(X,X)
r14: from(X) -> n__from(X)
r15: s(X) -> n__s(X)
r16: activate(n__from(X)) -> from(activate(X))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        times# > s
      
      argument filter:
    
        pi(times#) = 1
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > times#
      
      argument filter:
    
        pi(times#) = [1]
        pi(s) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        times# > s
      
      argument filter:
    
        pi(times#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: plus#(s(X),Y) -> plus#(X,Y)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: |2ndspos|(|0|(),Z) -> rnil()
r3: |2ndspos|(s(N),cons(X,Z)) -> |2ndspos|(s(N),cons2(X,activate(Z)))
r4: |2ndspos|(s(N),cons2(X,cons(Y,Z))) -> rcons(posrecip(Y),|2ndsneg|(N,activate(Z)))
r5: |2ndsneg|(|0|(),Z) -> rnil()
r6: |2ndsneg|(s(N),cons(X,Z)) -> |2ndsneg|(s(N),cons2(X,activate(Z)))
r7: |2ndsneg|(s(N),cons2(X,cons(Y,Z))) -> rcons(negrecip(Y),|2ndspos|(N,activate(Z)))
r8: pi(X) -> |2ndspos|(X,from(|0|()))
r9: plus(|0|(),Y) -> Y
r10: plus(s(X),Y) -> s(plus(X,Y))
r11: times(|0|(),Y) -> |0|()
r12: times(s(X),Y) -> plus(Y,times(X,Y))
r13: square(X) -> times(X,X)
r14: from(X) -> n__from(X)
r15: s(X) -> n__s(X)
r16: activate(n__from(X)) -> from(activate(X))
r17: activate(n__s(X)) -> s(activate(X))
r18: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        plus# > s
      
      argument filter:
    
        pi(plus#) = [1, 2]
        pi(s) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > plus#
      
      argument filter:
    
        pi(plus#) = 1
        pi(s) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > plus#
      
      argument filter:
    
        pi(plus#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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