YES

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

Proof:
 DP Processor:
  DPs:
   f#(g(x),y) -> f#(x,y)
  TRS:
   f(x,x) -> a()
   f(g(x),y) -> f(x,y)
  Subterm Criterion Processor:
   simple projection:
    pi(f#) = 0
   problem:
    DPs:
     
    TRS:
     f(x,x) -> a()
     f(g(x),y) -> f(x,y)
   Qed