YES

We show the termination of the TRS R:

  f(s(x)) -> s(f(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#(f(p(s(x))))
p2: f#(s(x)) -> f#(p(s(x)))
p3: f#(s(x)) -> p#(s(x))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      f#_A(x1) = x1 + 1
      s_A(x1) = x1
      f_A(x1) = x1
      p_A(x1) = x1
      |0|_A = 0
    
    2. max/plus interpretations on natural numbers:
    
      f#_A(x1) = max{0, x1 - 3}
      s_A(x1) = max{6, x1 + 4}
      f_A(x1) = x1
      p_A(x1) = max{1, x1 - 4}
      |0|_A = 1
    

The next rules are strictly ordered:

  p1, p2

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