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)))
  ADG 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:
    f#(s(s(x))) -> f#(s(x)) -> f#(s(s(x))) -> f#(s(x))
    f#(s(s(x))) -> f#(s(x)) -> f#(s(s(x))) -> f#(f(s(x)))
   Restore Modifier:
    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)))
    SCC Processor:
     #sccs: 1
     #rules: 1
     #arcs: 2/4
     DPs:
      f#(s(s(x))) -> f#(s(x))
     TRS:
      f(0()) -> s(0())
      f(s(0())) -> s(0())
      f(s(s(x))) -> f(f(s(x)))
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [f#](x0) = x0,
       
       [s](x0) = x0 + 1,
       
       [f](x0) = x0 + 1,
       
       [0] = 1
      orientation:
       f#(s(s(x))) = x + 2 >= x + 1 = f#(s(x))
       
       f(0()) = 2 >= 2 = s(0())
       
       f(s(0())) = 3 >= 2 = s(0())
       
       f(s(s(x))) = x + 3 >= x + 3 = 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