YES

We show the termination of the TRS R:

  fac(s(x)) -> *(fac(p(s(x))),s(x))
  p(s(|0|())) -> |0|()
  p(s(s(x))) -> s(p(s(x)))

-- SCC decomposition.

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

p1: fac#(s(x)) -> fac#(p(s(x)))
p2: fac#(s(x)) -> p#(s(x))
p3: p#(s(s(x))) -> p#(s(x))

and R consists of:

r1: fac(s(x)) -> *(fac(p(s(x))),s(x))
r2: p(s(|0|())) -> |0|()
r3: p(s(s(x))) -> s(p(s(x)))

The estimated dependency graph contains the following SCCs:

  {p1}
  {p3}


-- Reduction pair.

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

p1: fac#(s(x)) -> fac#(p(s(x)))

and R consists of:

r1: fac(s(x)) -> *(fac(p(s(x))),s(x))
r2: p(s(|0|())) -> |0|()
r3: p(s(s(x))) -> s(p(s(x)))

The set of usable rules consists of

  r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fac#_A(x1) = x1 + 4
          s_A(x1) = max{22, x1 + 14}
          p_A(x1) = max{2, x1 - 5}
          |0|_A = 23
    
      precedence:
      
        p > |0| > s > fac#
      
      partial status:
    
        pi(fac#) = [1]
        pi(s) = [1]
        pi(p) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fac#_A(x1) = max{14, x1 - 5}
          s_A(x1) = max{13, x1 + 4}
          p_A(x1) = 6
          |0|_A = 14
    
      precedence:
      
        fac# = s = p = |0|
      
      partial status:
    
        pi(fac#) = []
        pi(s) = [1]
        pi(p) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

r1: fac(s(x)) -> *(fac(p(s(x))),s(x))
r2: p(s(|0|())) -> |0|()
r3: p(s(s(x))) -> s(p(s(x)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1) = x1 + 1
          s_A(x1) = x1
    
      precedence:
      
        p# = s
      
      partial status:
    
        pi(p#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1) = x1 + 3
          s_A(x1) = x1 + 1
    
      precedence:
      
        p# > s
      
      partial status:
    
        pi(p#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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