YES

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

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