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))))
  TDG 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)) -> fac#(p(s(x)))
    fac#(s(x)) -> fac#(p(s(x))) -> fac#(s(x)) -> 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))))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [fac#](x0) = x0 + 0,
      
      [times](x0, x1) = x0 + -5x1 + 2,
      
      [fac](x0) = x0 + 7,
      
      [0] = 1,
      
      [p](x0) = -4x0 + 0,
      
      [s](x0) = 6x0 + 2
     orientation:
      fac#(s(x)) = 6x + 2 >= 2x + 0 = fac#(p(s(x)))
      
      p(s(x)) = 2x + 0 >= x = x
      
      fac(0()) = 7 >= 7 = s(0())
      
      fac(s(x)) = 6x + 7 >= 6x + 2 = times(s(x),fac(p(s(x))))
     problem:
      DPs:
       
      TRS:
       p(s(x)) -> x
       fac(0()) -> s(0())
       fac(s(x)) -> times(s(x),fac(p(s(x))))
     Qed