YES

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

Proof:
 DP Processor:
  DPs:
   f#(f(x)) -> g#(f(x))
   g#(g(x)) -> f#(x)
  TRS:
   f(f(x)) -> g(f(x))
   g(g(x)) -> f(x)
  TDG Processor:
   DPs:
    f#(f(x)) -> g#(f(x))
    g#(g(x)) -> f#(x)
   TRS:
    f(f(x)) -> g(f(x))
    g(g(x)) -> f(x)
   graph:
    g#(g(x)) -> f#(x) -> f#(f(x)) -> g#(f(x))
    f#(f(x)) -> g#(f(x)) -> g#(g(x)) -> f#(x)
   Arctic Interpretation Processor:
    dimension: 1
    interpretation:
     [g#](x0) = x0 + 4,
     
     [f#](x0) = x0,
     
     [g](x0) = 4x0 + 0,
     
     [f](x0) = 4x0 + 4
    orientation:
     f#(f(x)) = 4x + 4 >= 4x + 4 = g#(f(x))
     
     g#(g(x)) = 4x + 4 >= x = f#(x)
     
     f(f(x)) = 8x + 8 >= 8x + 8 = g(f(x))
     
     g(g(x)) = 8x + 4 >= 4x + 4 = f(x)
    problem:
     DPs:
      f#(f(x)) -> g#(f(x))
     TRS:
      f(f(x)) -> g(f(x))
      g(g(x)) -> f(x)
    EDG Processor:
     DPs:
      f#(f(x)) -> g#(f(x))
     TRS:
      f(f(x)) -> g(f(x))
      g(g(x)) -> f(x)
     graph:
      
     Qed