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: dim=4
   
   interpretation:
    [f#](x0) = [0 1 0 1]x0,
    
              [0 0 1 0]     [1]
              [0 1 0 1]     [0]
    [s](x0) = [0 0 0 0]x0 + [0]
              [1 0 0 0]     [0],
    
              [0 0 1 0]     [1]
              [0 1 0 1]     [0]
    [f](x0) = [0 0 0 0]x0 + [0]
              [1 0 0 0]     [0]
   orientation:
    f#(s(s(x))) = [1 1 1 1]x + [1] >= [0 1 0 1]x = f#(x)
    
    f#(s(s(x))) = [1 1 1 1]x + [1] >= [1 1 0 1]x = f#(f(x))
    
           [0 0 1 0]    [1]    [0 0 1 0]    [1]       
           [0 1 0 1]    [0]    [0 1 0 1]    [0]       
    f(x) = [0 0 0 0]x + [0] >= [0 0 0 0]x + [0] = s(x)
           [1 0 0 0]    [0]    [1 0 0 0]    [0]       
    
                 [0 0 0 0]    [1]    [0 0 0 0]    [1]             
                 [1 1 1 1]    [1]    [1 1 1 1]    [1]             
    f(s(s(x))) = [0 0 0 0]x + [0] >= [0 0 0 0]x + [0] = s(f(f(x)))
                 [0 0 0 0]    [1]    [0 0 0 0]    [1]             
   problem:
    DPs:
     
    TRS:
     f(x) -> s(x)
     f(s(s(x))) -> s(f(f(x)))
   Qed