YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(s(x))) -> f#(x)
   f#(s(s(x))) -> f#(f(x))
  TRS:
   f(x) -> s(x)
   f(s(s(x))) -> s(f(f(x)))
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0) = x0,
    
    [s](x0) = x0 + 1,
    
    [f](x0) = x0 + 1
   orientation:
    f#(s(s(x))) = x + 2 >= x = f#(x)
    
    f#(s(s(x))) = x + 2 >= x + 1 = f#(f(x))
    
    f(x) = x + 1 >= x + 1 = s(x)
    
    f(s(s(x))) = x + 3 >= x + 3 = s(f(f(x)))
   problem:
    DPs:
     
    TRS:
     f(x) -> s(x)
     f(s(s(x))) -> s(f(f(x)))
   Qed