YES

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

Proof:
 DP Processor:
  DPs:
   f#(g(x),y,y) -> f#(x,x,y)
  TRS:
   f(g(x),y,y) -> g(f(x,x,y))
  KBO Processor:
   argument filtering:
    pi(g) = [0]
    pi(f) = 0
    pi(f#) = 0
   weight function:
    w0 = 1
    w(f#) = w(f) = w(g) = 1
   precedence:
    f# ~ f ~ g
   problem:
    DPs:
     
    TRS:
     f(g(x),y,y) -> g(f(x,x,y))
   Qed