YES

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

Proof:
 DP Processor:
  DPs:
   f#(g(x),x,y) -> g#(y)
   f#(g(x),x,y) -> f#(y,y,g(y))
  TRS:
   f(g(x),x,y) -> f(y,y,g(y))
   g(g(x)) -> g(x)
  TDG Processor:
   DPs:
    f#(g(x),x,y) -> g#(y)
    f#(g(x),x,y) -> f#(y,y,g(y))
   TRS:
    f(g(x),x,y) -> f(y,y,g(y))
    g(g(x)) -> g(x)
   graph:
    f#(g(x),x,y) -> f#(y,y,g(y)) -> f#(g(x),x,y) -> f#(y,y,g(y))
    f#(g(x),x,y) -> f#(y,y,g(y)) -> f#(g(x),x,y) -> g#(y)
   EDG Processor:
    DPs:
     f#(g(x),x,y) -> g#(y)
     f#(g(x),x,y) -> f#(y,y,g(y))
    TRS:
     f(g(x),x,y) -> f(y,y,g(y))
     g(g(x)) -> g(x)
    graph:
     
    Qed