YES

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

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(X)) -> X
  Usable Rule Processor:
   DPs:
    f#(s(0())) -> p#(s(0()))
    f#(s(0())) -> f#(p(s(0())))
   TRS:
    p(s(X)) -> X
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     p(s(X)) -> X
    interpretation:
     [p#](x0) = -2x0 + 0,
     
     [f#](x0) = x0 + 0,
     
     [p](x0) = -1x0 + 0,
     
     [s](x0) = 2x0 + 0,
     
     [0] = 0
    orientation:
     f#(s(0())) = 2 >= 0 = p#(s(0()))
     
     f#(s(0())) = 2 >= 1 = f#(p(s(0())))
     
     p(s(X)) = 1X + 0 >= X = X
    problem:
     DPs:
      
     TRS:
      p(s(X)) -> X
    Qed