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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = x1 + 1
          n__s_A(x1) = x1
          n__f_A(x1) = x1
    
      precedence:
      
        activate# = n__s = n__f
      
      partial status:
    
        pi(activate#) = []
        pi(n__s) = []
        pi(n__f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = x1 + 1
          n__s_A(x1) = x1
          n__f_A(x1) = x1
    
      precedence:
      
        activate# = n__s = n__f
      
      partial status:
    
        pi(activate#) = [1]
        pi(n__s) = [1]
        pi(n__f) = [1]
    

The next rules are strictly ordered:

  p1, p2

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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{8, x1 + 7}
          s_A(x1) = x1 + 3
          |0|_A = 4
          p_A(x1) = max{3, x1}
          n__s_A(x1) = max{3, x1 + 2}
          n__0_A = 0
    
      precedence:
      
        s = |0| > f# = p = n__s = n__0
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(|0|) = []
        pi(p) = []
        pi(n__s) = [1]
        pi(n__0) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{11, x1 + 3}
          s_A(x1) = max{8, x1 + 5}
          |0|_A = 1
          p_A(x1) = 0
          n__s_A(x1) = max{9, x1 + 6}
          n__0_A = 2
    
      precedence:
      
        f# = s = |0| = p = n__s = n__0
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(|0|) = []
        pi(p) = []
        pi(n__s) = []
        pi(n__0) = []
    

The next rules are strictly ordered:

  p1

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