YES

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

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