YES

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

Proof:
 DP Processor:
  DPs:
   f#(s(x),y,y) -> f#(y,x,s(x))
  TRS:
   f(s(x),y,y) -> f(y,x,s(x))
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [f#](x0, x1, x2) = x0 + x1,
    
    [f](x0, x1, x2) = x0 + x1,
    
    [s](x0) = x0 + 1
   orientation:
    f#(s(x),y,y) = x + y + 1 >= x + y = f#(y,x,s(x))
    
    f(s(x),y,y) = x + y + 1 >= x + y = f(y,x,s(x))
   problem:
    DPs:
     
    TRS:
     f(s(x),y,y) -> f(y,x,s(x))
   Qed