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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            fac#_A(x1) = x1 + (1,2)
            s_A(x1) = x1 + (1,2)
            p_A(x1) = ((1,0),(0,0)) x1 + (0,1)
    
      precedence:
      
        fac# = s = p
      
      partial status:
    
        pi(fac#) = []
        pi(s) = []
        pi(p) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            fac#_A(x1) = ((0,0),(1,0)) x1 + (3,1)
            s_A(x1) = x1 + (2,0)
            p_A(x1) = (1,4)
    
      precedence:
      
        fac# = s > p
      
      partial status:
    
        pi(fac#) = []
        pi(s) = []
        pi(p) = []
    

The next rules are strictly ordered:

  p1

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