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))))
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [fac#](x0) = x0 + 0,
    
    [p#](x0) = -2x0 + 0,
    
    [times](x0, x1) = 1x0 + x1 + 0,
    
    [fac](x0) = 4x0 + 0,
    
    [0] = 0,
    
    [p](x0) = -2x0 + 2,
    
    [s](x0) = 3x0 + 3
   orientation:
    fac#(s(x)) = 3x + 3 >= 1x + 1 = p#(s(x))
    
    fac#(s(x)) = 3x + 3 >= 1x + 2 = fac#(p(s(x)))
    
    p(s(x)) = 1x + 2 >= x = x
    
    fac(0()) = 4 >= 3 = s(0())
    
    fac(s(x)) = 7x + 7 >= 5x + 6 = 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