YES

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

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