YES

We show the termination of the TRS R:

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

and R consists of:

r1: f(|0|()) -> cons(|0|(),n__f(s(|0|())))
r2: f(s(|0|())) -> f(p(s(|0|())))
r3: p(s(X)) -> X
r4: f(X) -> n__f(X)
r5: activate(n__f(X)) -> f(X)
r6: 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: f#(s(|0|())) -> f#(p(s(|0|())))

and R consists of:

r1: f(|0|()) -> cons(|0|(),n__f(s(|0|())))
r2: f(s(|0|())) -> f(p(s(|0|())))
r3: p(s(X)) -> X
r4: f(X) -> n__f(X)
r5: activate(n__f(X)) -> f(X)
r6: activate(X) -> X

The set of usable rules consists of

  r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = x1
        s_A(x1) = ((1,1),(1,1)) x1 + (2,1)
        |0|_A() = (1,1)
        p_A(x1) = ((0,1),(1,1)) x1
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = (0,0)
        s_A(x1) = (1,1)
        |0|_A() = (1,1)
        p_A(x1) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = (0,0)
        s_A(x1) = (1,1)
        |0|_A() = (1,1)
        p_A(x1) = (1,1)
    

The next rules are strictly ordered:

  p1

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