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, r2

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        fac#_A(x1) = max{3, x1 - 5}
        s_A(x1) = x1 + 9
        p_A(x1) = max{4, x1 - 7}
        fac_A(x1) = 11
        |0|_A = 1
  
    base order 2:
  
      precedence:
      
        fac > |0| > fac# = s > p
      
  
    partial status:
  
      pi(fac#) = []
      pi(s) = []
      pi(p) = []
      pi(fac) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.