YES

We show the termination of the TRS R:

  f(|0|()) -> cons(|0|(),n__f(n__s(n__0())))
  f(s(|0|())) -> f(p(s(|0|())))
  p(s(X)) -> X
  f(X) -> n__f(X)
  s(X) -> n__s(X)
  |0|() -> n__0()
  activate(n__f(X)) -> f(activate(X))
  activate(n__s(X)) -> s(activate(X))
  activate(n__0()) -> |0|()
  activate(X) -> X

-- SCC decomposition.

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

p1: f#(s(|0|())) -> f#(p(s(|0|())))
p2: f#(s(|0|())) -> p#(s(|0|()))
p3: activate#(n__f(X)) -> f#(activate(X))
p4: activate#(n__f(X)) -> activate#(X)
p5: activate#(n__s(X)) -> s#(activate(X))
p6: activate#(n__s(X)) -> activate#(X)
p7: activate#(n__0()) -> |0|#()

and R consists of:

r1: f(|0|()) -> cons(|0|(),n__f(n__s(n__0())))
r2: f(s(|0|())) -> f(p(s(|0|())))
r3: p(s(X)) -> X
r4: f(X) -> n__f(X)
r5: s(X) -> n__s(X)
r6: |0|() -> n__0()
r7: activate(n__f(X)) -> f(activate(X))
r8: activate(n__s(X)) -> s(activate(X))
r9: activate(n__0()) -> |0|()
r10: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p4, p6}
  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: f(|0|()) -> cons(|0|(),n__f(n__s(n__0())))
r2: f(s(|0|())) -> f(p(s(|0|())))
r3: p(s(X)) -> X
r4: f(X) -> n__f(X)
r5: s(X) -> n__s(X)
r6: |0|() -> n__0()
r7: activate(n__f(X)) -> f(activate(X))
r8: activate(n__s(X)) -> s(activate(X))
r9: activate(n__0()) -> |0|()
r10: 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__f > n__s
      
      argument filter:
    
        pi(activate#) = 1
        pi(n__s) = [1]
        pi(n__f) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        activate# > n__f > n__s
      
      argument filter:
    
        pi(activate#) = 1
        pi(n__s) = [1]
        pi(n__f) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        activate# > n__f > n__s
      
      argument filter:
    
        pi(activate#) = 1
        pi(n__s) = [1]
        pi(n__f) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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: f#(s(|0|())) -> f#(p(s(|0|())))

and R consists of:

r1: f(|0|()) -> cons(|0|(),n__f(n__s(n__0())))
r2: f(s(|0|())) -> f(p(s(|0|())))
r3: p(s(X)) -> X
r4: f(X) -> n__f(X)
r5: s(X) -> n__s(X)
r6: |0|() -> n__0()
r7: activate(n__f(X)) -> f(activate(X))
r8: activate(n__s(X)) -> s(activate(X))
r9: activate(n__0()) -> |0|()
r10: activate(X) -> X

The set of usable rules consists of

  r3, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        |0| > n__0 > n__s > p > f# > s
      
      argument filter:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(|0|) = []
        pi(p) = 1
        pi(n__s) = 1
        pi(n__0) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        n__0 > n__s > |0| > p > s > f#
      
      argument filter:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(|0|) = []
        pi(p) = []
        pi(n__s) = 1
        pi(n__0) = []
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        n__0 > n__s > p > |0| > s > f#
      
      argument filter:
    
        pi(f#) = [1]
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(n__s) = 1
        pi(n__0) = []
    

The next rules are strictly ordered:

  p1

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