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)))
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    
   interpretation:
    [g#](x0) = 0,
    
    [f#](x0) = x0,
    
    [+](x0, x1) = -8x0 + x1 + -16,
    
    [g](x0) = x0 + 4,
    
    [s](x0) = 1x0 + 7,
    
    [1] = 0,
    
    [f](x0) = 0,
    
    [0] = 0
   orientation:
    f#(s(x)) = 1x + 7 >= x = f#(x)
    
    f#(s(x)) = 1x + 7 >= 0 = g#(f(x))
    
    f(0()) = 0 >= 0 = 1()
    
    f(s(x)) = 0 >= 4 = g(f(x))
    
    g(x) = x + 4 >= 1x + 7 = +(x,s(x))
    
    f(s(x)) = 0 >= 7 = +(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