YES

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

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