YES

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

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