YES

Problem:
 f(s(x1)) -> s(s(f(p(s(x1)))))
 f(0(x1)) -> 0(x1)
 p(s(x1)) -> x1

Proof:
 Bounds Processor:
  bound: 0
  enrichment: match
  automaton:
   final states: {2,7,1}
   transitions:
    s0(2) -> 3*
    s0(5) -> 6*
    s0(6) -> 1*
    p0(3) -> 4*
    f0(4) -> 5*
    f40() -> 2*
    00(2) -> 7*
    7 -> 5*
    2 -> 4*
    1 -> 5*
  problem:
   
  Qed