YES

Problem:
 f(s(x)) -> f(g(x,x))
 g(0(),1()) -> s(0())
 0() -> 1()

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