YES

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

Proof:
 DP Processor:
  DPs:
   f#(g(x,y),f(y,y)) -> f#(g(y,x),y)
  TRS:
   f(g(x,y),f(y,y)) -> f(g(y,x),y)
  Subterm Criterion Processor:
   simple projection:
    pi(f#) = 1
   problem:
    DPs:
     
    TRS:
     f(g(x,y),f(y,y)) -> f(g(y,x),y)
   Qed