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)))
    Arctic Interpretation Processor:
     dimension: 1
     interpretation:
      [f#](x0) = x0,
      
      [+](x0, x1) = x0 + -9x1,
      
      [g](x0) = 2x0,
      
      [s](x0) = 3x0,
      
      [1] = 1,
      
      [f](x0) = x0,
      
      [0] = 4
     orientation:
      f#(s(x)) = 3x >= x = f#(x)
      
      f(0()) = 4 >= 1 = 1()
      
      f(s(x)) = 3x >= 2x = g(f(x))
      
      g(x) = 2x >= x = +(x,s(x))
      
      f(s(x)) = 3x >= x = +(f(x),s(f(x)))
     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