YES

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

Proof:
 DP Processor:
  DPs:
   g#(s(x)) -> f#(x)
   f#(s(x)) -> g#(x)
  TRS:
   g(s(x)) -> f(x)
   f(0()) -> s(0())
   f(s(x)) -> s(s(g(x)))
   g(0()) -> 0()
  Usable Rule Processor:
   DPs:
    g#(s(x)) -> f#(x)
    f#(s(x)) -> g#(x)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [f#](x0) = x0,
     
     [g#](x0) = x0 + -16,
     
     [s](x0) = 1x0 + -2
    orientation:
     g#(s(x)) = 1x + -2 >= x = f#(x)
     
     f#(s(x)) = 1x + -2 >= x + -16 = g#(x)
    problem:
     DPs:
      
     TRS:
      
    Qed