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
   interpretation:
    [g#](x0) = -6x0 + 0,
    
    [f#](x0) = -1x0 + 2,
    
    [+](x0, x1) = 0,
    
    [g](x0) = -5x0 + 1,
    
    [s](x0) = 1x0 + 5,
    
    [1] = 1,
    
    [f](x0) = x0 + 6,
    
    [0] = 6
   orientation:
    f#(s(x)) = x + 4 >= -1x + 2 = f#(x)
    
    f#(s(x)) = x + 4 >= -6x + 0 = g#(f(x))
    
    f(0()) = 6 >= 1 = 1()
    
    f(s(x)) = 1x + 6 >= -5x + 1 = g(f(x))
    
    g(x) = -5x + 1 >= 0 = +(x,s(x))
    
    f(s(x)) = 1x + 6 >= 0 = +(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