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) = -10x0 + 8,
     
     [f#](x0) = x0,
     
     [g](x0) = 11x0 + 0,
     
     [f](x0) = 12x0 + 9
    orientation:
     f#(f(x)) = 12x + 9 >= 2x + 8 = g#(f(x))
     
     g#(g(x)) = 1x + 8 >= x = f#(x)
     
     f(f(x)) = 24x + 21 >= 23x + 20 = g(f(x))
     
     g(g(x)) = 22x + 11 >= 12x + 9 = f(x)
    problem:
     DPs:
      
     TRS:
      f(f(x)) -> g(f(x))
      g(g(x)) -> f(x)
    Qed