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()
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    g(0()) -> 0()
    g(s(x)) -> f(g(x))
    f(0()) -> 0()
   interpretation:
    [f#](x0) = x0 + 0,
    
    [g#](x0) = -1x0 + 0,
    
    [f](x0) = 1x0 + 3,
    
    [s](x0) = 2x0 + 4,
    
    [g](x0) = x0 + 1,
    
    [0] = 0
   orientation:
    g#(s(x)) = 1x + 3 >= -1x + 0 = g#(x)
    
    g#(s(x)) = 1x + 3 >= x + 1 = f#(g(x))
    
    g(0()) = 1 >= 0 = 0()
    
    g(s(x)) = 2x + 4 >= 1x + 3 = f(g(x))
    
    f(0()) = 3 >= 0 = 0()
   problem:
    DPs:
     
    TRS:
     g(0()) -> 0()
     g(s(x)) -> f(g(x))
     f(0()) -> 0()
   Qed