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)))
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    f(0()) -> s(0())
    f(s(0())) -> s(0())
    f(s(s(x))) -> f(f(s(x)))
   interpretation:
    [f#](x0) = x0 + 0,
    
    [s](x0) = 2x0 + 4,
    
    [f](x0) = x0 + 5,
    
    [0] = 0
   orientation:
    f#(s(s(x))) = 4x + 6 >= 2x + 4 = f#(s(x))
    
    f#(s(s(x))) = 4x + 6 >= 2x + 5 = f#(f(s(x)))
    
    f(0()) = 5 >= 4 = s(0())
    
    f(s(0())) = 5 >= 4 = s(0())
    
    f(s(s(x))) = 4x + 6 >= 2x + 5 = f(f(s(x)))
   problem:
    DPs:
     
    TRS:
     f(0()) -> s(0())
     f(s(0())) -> s(0())
     f(s(s(x))) -> f(f(s(x)))
   Qed