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)
  Usable Rule Processor:
   DPs:
    f#(s(x),y) -> f#(x,s(x))
    f#(x,s(y)) -> f#(y,x)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0, x1) = 2x0 + 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)
    problem:
     DPs:
      
     TRS:
      
    Qed