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:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0, x1) = x1,
     
     [g](x0) = 1x0 + -2
    orientation:
     f#(x,g(y)) = 1y + -2 >= y = f#(g(x),y)
    problem:
     DPs:
      
     TRS:
      
    Qed