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()
  TDG 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()
   graph:
    f#(s(0())) -> f#(p(s(0()))) -> f#(s(0())) -> f#(p(s(0())))
    f#(s(0())) -> f#(p(s(0()))) -> f#(s(0())) -> p#(s(0()))
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(s(0())) -> f#(p(s(0())))
    TRS:
     f(0()) -> cons(0())
     f(s(0())) -> f(p(s(0())))
     p(s(0())) -> 0()
    EDG Processor:
     DPs:
      f#(s(0())) -> f#(p(s(0())))
     TRS:
      f(0()) -> cons(0())
      f(s(0())) -> f(p(s(0())))
      p(s(0())) -> 0()
     graph:
      
     Qed