YES

Problem:
 f(f(x)) -> f(x)
 g(0()) -> g(f(0()))

Proof:
 DP Processor:
  DPs:
   g#(0()) -> f#(0())
   g#(0()) -> g#(f(0()))
  TRS:
   f(f(x)) -> f(x)
   g(0()) -> g(f(0()))
  Usable Rule Processor:
   DPs:
    g#(0()) -> f#(0())
    g#(0()) -> g#(f(0()))
   TRS:
    
   TDG Processor:
    DPs:
     g#(0()) -> f#(0())
     g#(0()) -> g#(f(0()))
    TRS:
     
    graph:
     g#(0()) -> g#(f(0())) -> g#(0()) -> g#(f(0()))
     g#(0()) -> g#(f(0())) -> g#(0()) -> f#(0())
    EDG Processor:
     DPs:
      g#(0()) -> f#(0())
      g#(0()) -> g#(f(0()))
     TRS:
      
     graph:
      
     Qed