YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(0())) -> p#(s(0()))
   f#(s(0())) -> f#(p(s(0())))
   activate#(n__f(X)) -> f#(X)
  TRS:
   f(0()) -> cons(0(),n__f(s(0())))
   f(s(0())) -> f(p(s(0())))
   p(s(0())) -> 0()
   f(X) -> n__f(X)
   activate(n__f(X)) -> f(X)
   activate(X) -> X
  Usable Rule Processor:
   DPs:
    f#(s(0())) -> p#(s(0()))
    f#(s(0())) -> f#(p(s(0())))
    activate#(n__f(X)) -> f#(X)
   TRS:
    p(s(0())) -> 0()
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     p(s(0())) -> 0()
    interpretation:
     [activate#](x0) = -6x0 + 1,
     
     [p#](x0) = 1x0 + 1,
     
     [f#](x0) = 2x0 + 0,
     
     [p](x0) = -4x0 + 1,
     
     [n__f](x0) = 9x0 + 5,
     
     [s](x0) = 2x0 + -16,
     
     [0] = 0
    orientation:
     f#(s(0())) = 4 >= 3 = p#(s(0()))
     
     f#(s(0())) = 4 >= 3 = f#(p(s(0())))
     
     activate#(n__f(X)) = 3X + 1 >= 2X + 0 = f#(X)
     
     p(s(0())) = 1 >= 0 = 0()
    problem:
     DPs:
      
     TRS:
      p(s(0())) -> 0()
    Qed