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()))
  ADG Processor:
   DPs:
    g#(0()) -> f#(0())
    g#(0()) -> g#(f(0()))
   TRS:
    f(f(x)) -> f(x)
    g(0()) -> g(f(0()))
   graph:
    
   Qed