YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(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)
  activate(n__from(X)) -> from(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) -> plus#(X,Y)
p12: times#(s(X),Y) -> plus#(Y,times(X,Y))
p13: times#(s(X),Y) -> times#(X,Y)
p14: square#(X) -> times#(X,X)
p15: activate#(n__from(X)) -> from#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(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: activate(n__from(X)) -> from(X)
r16: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p3, p5, p7}
  {p13}
  {p11}


-- 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(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: activate(n__from(X)) -> from(X)
r16: activate(X) -> X

The set of usable rules consists of

  r1, r14, r15, r16

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      |2ndspos|#_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1
      s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (0,1,1)
      cons_A(x1,x2) = (1,3,1)
      cons2_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1 + ((0,0,0),(1,0,0),(0,0,0)) x2 + (2,0,1)
      activate_A(x1) = x1 + (2,1,1)
      |2ndsneg|#_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1
      from_A(x1) = (2,2,2)
      n__from_A(x1) = ((0,0,0),(0,0,0),(1,0,0)) x1 + (1,1,3)

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(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: activate(n__from(X)) -> from(X)
r16: 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: times#(s(X),Y) -> times#(X,Y)

and R consists of:

r1: from(X) -> cons(X,n__from(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: activate(n__from(X)) -> from(X)
r16: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      times#_A(x1,x2) = x1 + ((1,0,0),(0,1,0),(0,1,1)) x2
      s_A(x1) = ((1,0,0),(0,0,0),(1,1,0)) x1 + (1,1,1)

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(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: activate(n__from(X)) -> from(X)
r16: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      plus#_A(x1,x2) = x1 + ((0,0,0),(1,0,0),(0,1,0)) x2
      s_A(x1) = ((1,0,0),(0,0,0),(1,1,0)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1

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