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
    Usable Rule Processor:
     DPs:
      f#(s(x)) -> f#(p(s(x)))
     TRS:
      p(s(x)) -> x
     Bounds Processor:
      bound: 1
      enrichment: match
      automaton:
       final states: {5}
       transitions:
        f{#,0}(4) -> 5*
        s0(4) -> 4*
        p0(4) -> 4*
        f{#,1}(12) -> 13*
        p1(11) -> 12*
        s1(10) -> 11*
        4 -> 10*
        10 -> 12*
        13 -> 5*
      problem:
       DPs:
        
       TRS:
        p(s(x)) -> x
      Qed