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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      f#_A(x1) = ((1,0),(1,1)) x1
      s_A(x1) = ((1,1),(1,0)) x1
      |0|_A() = (0,1)
      p_A(x1) = ((0,1),(1,1)) x1

The next rules are strictly ordered:

  p1

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