YES

Problem:
 f(g(X)) -> f(X)

Proof:
 DP Processor:
  DPs:
   f#(g(X)) -> f#(X)
  TRS:
   f(g(X)) -> f(X)
  Usable Rule Processor:
   DPs:
    f#(g(X)) -> f#(X)
   TRS:
    
   CDG Processor:
    DPs:
     f#(g(X)) -> f#(X)
    TRS:
     
    graph:
     
    Qed