YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(a())) -> f#(g())
  TRS:
   f(f(a())) -> f(g())
  EDG Processor:
   DPs:
    f#(f(a())) -> f#(g())
   TRS:
    f(f(a())) -> f(g())
   graph:
    
   Qed