YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(0())) -> p#(s(0()))
   f#(s(0())) -> f#(p(s(0())))
  TRS:
   f(0()) -> cons(0())
   f(s(0())) -> f(p(s(0())))
   p(s(0())) -> 0()
  Usable Rule Processor:
   DPs:
    f#(s(0())) -> p#(s(0()))
    f#(s(0())) -> f#(p(s(0())))
   TRS:
    p(s(0())) -> 0()
   ADG Processor:
    DPs:
     f#(s(0())) -> p#(s(0()))
     f#(s(0())) -> f#(p(s(0())))
    TRS:
     p(s(0())) -> 0()
    graph:
     
    Qed