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()
  TDG 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()
   graph:
    f#(s(x)) -> g#(x) -> g#(s(x)) -> f#(x)
    g#(s(x)) -> f#(x) -> f#(s(x)) -> g#(x)
   Subterm Criterion Processor:
    simple projection:
     pi(g#) = 0
     pi(f#) = 0
    problem:
     DPs:
      
     TRS:
      g(s(x)) -> f(x)
      f(0()) -> s(0())
      f(s(x)) -> s(s(g(x)))
      g(0()) -> 0()
    Qed