YES

Problem:
 f(s(x)) -> s(s(f(p(s(x)))))
 f(0()) -> 0()
 p(s(x)) -> x

Proof:
 DP Processor:
  DPs:
   f#(s(x)) -> p#(s(x))
   f#(s(x)) -> f#(p(s(x)))
  TRS:
   f(s(x)) -> s(s(f(p(s(x)))))
   f(0()) -> 0()
   p(s(x)) -> x
  Usable Rule Processor:
   DPs:
    f#(s(x)) -> p#(s(x))
    f#(s(x)) -> f#(p(s(x)))
   TRS:
    p(s(x)) -> x
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     p(s(x)) -> x
    interpretation:
     [p#](x0) = -15x0 + 0,
     
     [f#](x0) = -8x0 + 0,
     
     [p](x0) = -7x0 + 8,
     
     [s](x0) = 8x0 + 9
    orientation:
     f#(s(x)) = x + 1 >= -7x + 0 = p#(s(x))
     
     f#(s(x)) = x + 1 >= -7x + 0 = f#(p(s(x)))
     
     p(s(x)) = 1x + 8 >= x = x
    problem:
     DPs:
      
     TRS:
      p(s(x)) -> x
    Qed