YES

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

Proof:
 DP Processor:
  DPs:
   g#(s(x)) -> g#(x)
   g#(s(x)) -> f#(g(x))
  TRS:
   g(0()) -> 0()
   g(s(x)) -> f(g(x))
   f(0()) -> 0()
  TDG Processor:
   DPs:
    g#(s(x)) -> g#(x)
    g#(s(x)) -> f#(g(x))
   TRS:
    g(0()) -> 0()
    g(s(x)) -> f(g(x))
    f(0()) -> 0()
   graph:
    g#(s(x)) -> g#(x) -> g#(s(x)) -> f#(g(x))
    g#(s(x)) -> g#(x) -> g#(s(x)) -> g#(x)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     g#(s(x)) -> g#(x)
    TRS:
     g(0()) -> 0()
     g(s(x)) -> f(g(x))
     f(0()) -> 0()
    Subterm Criterion Processor:
     simple projection:
      pi(g#) = 0
     problem:
      DPs:
       
      TRS:
       g(0()) -> 0()
       g(s(x)) -> f(g(x))
       f(0()) -> 0()
     Qed