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