YES

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

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