YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(s(x)) -> f#(p(s(x)))
p2: f#(s(x)) -> p#(s(x))

and R consists of:

r1: f(s(x)) -> s(s(f(p(s(x)))))
r2: f(|0|()) -> |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(x)) -> f#(p(s(x)))

and R consists of:

r1: f(s(x)) -> s(s(f(p(s(x)))))
r2: f(|0|()) -> |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^2
      order: standard order
      interpretations:
        f#_A(x1) = ((0,1),(0,1)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (1,2)
        p_A(x1) = ((1,1),(1,0)) x1
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = (0,0)
        s_A(x1) = ((1,1),(0,0)) x1 + (1,1)
        p_A(x1) = x1 + (0,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = (0,0)
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        p_A(x1) = ((1,1),(1,1)) x1
    

The next rules are strictly ordered:

  p1

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