YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(x),y) -> f#(x,s(s(x)))
   f#(x,s(s(y))) -> f#(y,x)
  TRS:
   f(s(x),y) -> f(x,s(s(x)))
   f(x,s(s(y))) -> f(y,x)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1) = 4x0 + 1x1,
    
    [f](x0, x1) = 4x0 + 1x1 + -16,
    
    [s](x0) = 2x0 + 0
   orientation:
    f#(s(x),y) = 6x + 1y + 4 >= 5x + 3 = f#(x,s(s(x)))
    
    f#(x,s(s(y))) = 4x + 5y + 3 >= 1x + 4y = f#(y,x)
    
    f(s(x),y) = 6x + 1y + 4 >= 5x + 3 = f(x,s(s(x)))
    
    f(x,s(s(y))) = 4x + 5y + 3 >= 1x + 4y + -16 = f(y,x)
   problem:
    DPs:
     
    TRS:
     f(s(x),y) -> f(x,s(s(x)))
     f(x,s(s(y))) -> f(y,x)
   Qed