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)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [g#](x0) = -1x0 + 0,
    
    [f#](x0) = 2x0,
    
    [g](x0) = x0 + 3,
    
    [s](x0) = 2x0 + 1,
    
    [f](x0) = 3x0 + -16,
    
    [0] = 0
   orientation:
    f#(s(x)) = 4x + 3 >= 3x + 2 = g#(s(s(x)))
    
    g#(s(s(x))) = 3x + 2 >= 2x = f#(x)
    
    f(0()) = 3 >= 2 = s(0())
    
    f(s(x)) = 5x + 4 >= 4x + 3 = g(s(s(x)))
    
    g(0()) = 3 >= 2 = s(0())
    
    g(s(0())) = 3 >= 2 = s(0())
    
    g(s(s(x))) = 4x + 3 >= 3x + -16 = f(x)
   problem:
    DPs:
     
    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)
   Qed