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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{5, x1 + 3}
          s_A(x1) = max{12, x1 + 10}
          |0|_A = 0
          p_A(x1) = max{6, x1 - 9}
    
      precedence:
      
        f# = s = |0| = p
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(|0|) = []
        pi(p) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{11, x1 + 3}
          s_A(x1) = max{8, x1 + 5}
          |0|_A = 1
          p_A(x1) = 0
    
      precedence:
      
        f# = s = |0| = p
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(|0|) = []
        pi(p) = []
    

The next rules are strictly ordered:

  p1

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