YES

Problem:
 f(X,g(X)) -> f(1(),g(X))
 g(1()) -> g(0())

Proof:
 DP Processor:
  DPs:
   f#(X,g(X)) -> f#(1(),g(X))
   g#(1()) -> g#(0())
  TRS:
   f(X,g(X)) -> f(1(),g(X))
   g(1()) -> g(0())
  TDG Processor:
   DPs:
    f#(X,g(X)) -> f#(1(),g(X))
    g#(1()) -> g#(0())
   TRS:
    f(X,g(X)) -> f(1(),g(X))
    g(1()) -> g(0())
   graph:
    g#(1()) -> g#(0()) -> g#(1()) -> g#(0())
    f#(X,g(X)) -> f#(1(),g(X)) -> f#(X,g(X)) -> f#(1(),g(X))
   EDG Processor:
    DPs:
     f#(X,g(X)) -> f#(1(),g(X))
     g#(1()) -> g#(0())
    TRS:
     f(X,g(X)) -> f(1(),g(X))
     g(1()) -> g(0())
    graph:
     
    Qed