YES
We show the termination of the TRS R:

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


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

The set of usable rules consists of

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            fac#_A(x1) = max{0, x1 - 19}
            s_A(x1) = max{18, x1 + 3}
            p_A(x1) = max{7, x1 - 3}
            fac#!_A(x1) = max{1, x1 - 16}
            s!_A(x1) = 3
            p!_A(x1) = 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              fac#_A(x1) = x1
              s_A(x1) = x1 + 1
              p_A(x1) = x1
              fac#!_A(x1) = 0
              s!_A(x1) = 1
              p!_A(x1) = x1 + 3
        
  
    partial status:
  
      pi(fac#) = []
      pi(s) = [1]
      pi(p) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.