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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        fac#_A(x1) = max{5, x1}
        s_A(x1) = x1 + 5
        p_A(x1) = max{3, x1 - 1}
        |0|_A = 4
  
    precedence:
    
      p > s > fac# = |0|
    
    partial status:
  
      pi(fac#) = [1]
      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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        p#_A(x1) = max{1, x1 - 5}
        s_A(x1) = max{7, x1 + 4}
  
    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.