YES

We show the termination of the TRS R:

  f(|0|()) -> cons(|0|())
  f(s(|0|())) -> f(p(s(|0|())))
  p(s(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|()))

and R consists of:

r1: f(|0|()) -> cons(|0|())
r2: f(s(|0|())) -> f(p(s(|0|())))
r3: p(s(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|())
r2: f(s(|0|())) -> f(p(s(|0|())))
r3: p(s(X)) -> X

The set of usable rules consists of

  r3

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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