YES

Problem:
 f(0()) -> s(0())
 f(s(x)) -> g(s(s(x)))
 g(0()) -> s(0())
 g(s(0())) -> s(0())
 g(s(s(x))) -> f(x)

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