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
  EDG 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
   graph:
    f#(s(x)) -> f#(p(s(x))) -> f#(s(x)) -> p#(s(x))
    f#(s(x)) -> f#(p(s(x))) -> f#(s(x)) -> f#(p(s(x)))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(s(x)) -> f#(p(s(x)))
    TRS:
     f(s(x)) -> s(s(f(p(s(x)))))
     f(0()) -> 0()
     p(s(x)) -> x
    Bounds Processor:
     bound: 0
     enrichment: match-dp
     automaton:
      final states: {1}
      transitions:
       f{#,0}(4) -> 1*
       p0(3) -> 4*
       s0(2) -> 3*
       f60() -> 2*
       2 -> 4*
     problem:
      DPs:
       
      TRS:
       f(s(x)) -> s(s(f(p(s(x)))))
       f(0()) -> 0()
       p(s(x)) -> x
     Qed