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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1
          s_A(x1) = x1
          f_A(x1) = x1
          p_A(x1) = x1
          |0|_A = 0
    
      precedence:
      
        f# = s = f = p = |0|
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(f) = []
        pi(p) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{7, x1 + 6}
          s_A(x1) = max{5, x1 + 2}
          f_A(x1) = x1
          p_A(x1) = max{3, x1 - 2}
          |0|_A = 1
    
      precedence:
      
        p > f = |0| > f# = s
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(f) = []
        pi(p) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: f#(s(x)) -> f#(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#(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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 2
          s_A(x1) = x1
          f_A(x1) = x1
          p_A(x1) = x1
          |0|_A = 1
    
      precedence:
      
        f# = s = f = p = |0|
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(f) = []
        pi(p) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{0, x1 - 2}
          s_A(x1) = x1 + 4
          f_A(x1) = x1
          p_A(x1) = max{0, x1 - 4}
          |0|_A = 1
    
      precedence:
      
        f = p > |0| > s > f#
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(f) = []
        pi(p) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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