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)
  Usable Rule Processor:
   DPs:
    f#(g(x),y) -> f#(x,y)
   TRS:
    
   Restore Modifier:
    DPs:
     f#(g(x),y) -> f#(x,y)
    TRS:
     f(x,x) -> a()
     f(g(x),y) -> f(x,y)
    SCC Processor:
     #sccs: 1
     #rules: 1
     #arcs: 1/1
     DPs:
      f#(g(x),y) -> f#(x,y)
     TRS:
      f(x,x) -> a()
      f(g(x),y) -> f(x,y)
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [f#](x0, x1) = x0 + 1,
       
       [g](x0) = x0 + 1,
       
       [a] = 0,
       
       [f](x0, x1) = 0
      orientation:
       f#(g(x),y) = x + 2 >= x + 1 = f#(x,y)
       
       f(x,x) = 0 >= 0 = a()
       
       f(g(x),y) = 0 >= 0 = f(x,y)
      problem:
       DPs:
        
       TRS:
        f(x,x) -> a()
        f(g(x),y) -> f(x,y)
      Qed