YES

Problem:
 f(0()) -> s(0())
 f(s(0())) -> s(0())
 f(s(s(x))) -> f(f(s(x)))

Proof:
 DP Processor:
  DPs:
   f#(s(s(x))) -> f#(s(x))
   f#(s(s(x))) -> f#(f(s(x)))
  TRS:
   f(0()) -> s(0())
   f(s(0())) -> s(0())
   f(s(s(x))) -> f(f(s(x)))
  CDG Processor:
   DPs:
    f#(s(s(x))) -> f#(s(x))
    f#(s(s(x))) -> f#(f(s(x)))
   TRS:
    f(0()) -> s(0())
    f(s(0())) -> s(0())
    f(s(s(x))) -> f(f(s(x)))
   graph:
    
   Qed