YES

Problem:
 p(s(x)) -> x
 fac(0()) -> s(0())
 fac(s(x)) -> times(s(x),fac(p(s(x))))

Proof:
 DP Processor:
  DPs:
   fac#(s(x)) -> p#(s(x))
   fac#(s(x)) -> fac#(p(s(x)))
  TRS:
   p(s(x)) -> x
   fac(0()) -> s(0())
   fac(s(x)) -> times(s(x),fac(p(s(x))))
  EDG Processor:
   DPs:
    fac#(s(x)) -> p#(s(x))
    fac#(s(x)) -> fac#(p(s(x)))
   TRS:
    p(s(x)) -> x
    fac(0()) -> s(0())
    fac(s(x)) -> times(s(x),fac(p(s(x))))
   graph:
    fac#(s(x)) -> fac#(p(s(x))) -> fac#(s(x)) -> p#(s(x))
    fac#(s(x)) -> fac#(p(s(x))) -> fac#(s(x)) -> fac#(p(s(x)))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     fac#(s(x)) -> fac#(p(s(x)))
    TRS:
     p(s(x)) -> x
     fac(0()) -> s(0())
     fac(s(x)) -> times(s(x),fac(p(s(x))))
    Usable Rule Processor:
     DPs:
      fac#(s(x)) -> fac#(p(s(x)))
     TRS:
      p(s(x)) -> x
     Bounds Processor:
      bound: 0
      enrichment: match
      automaton:
       final states: {2,1}
       transitions:
        f70() -> 2*
        fac{#,0}(4) -> 1*
        p0(3) -> 4*
        s0(2) -> 3*
        2 -> 4*
      problem:
       DPs:
        
       TRS:
        p(s(x)) -> x
      Qed