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
  TDG 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)) -> f#(p(s(x)))
    f#(s(x)) -> f#(p(s(x))) -> f#(s(x)) -> 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
    CDG Processor:
     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
     graph:
      
     Qed