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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{6, x1 + 4}
        s_A(x1) = max{12, x1 + 7}
        f_A(x1) = max{4, x1}
        p_A(x1) = max{5, x1 - 7}
        |0|_A = 5
  
    precedence:
    
      s = f = p > f# = |0|
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = []
      pi(f) = []
      pi(p) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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(f(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(f(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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{7, x1 + 4}
        s_A(x1) = x1 + 5
        p_A(x1) = max{2, x1 - 3}
  
    precedence:
    
      f# = s = p
    
    partial status:
  
      pi(f#) = []
      pi(s) = [1]
      pi(p) = []

The next rules are strictly ordered:

  p1

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