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)
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    f(f(x)) -> g(f(x))
    g(g(x)) -> f(x)
   interpretation:
    [g#](x0) = -11x0 + 0,
    
    [f#](x0) = x0 + -10,
    
    [g](x0) = 12x0 + 11,
    
    [f](x0) = 13x0 + 10
   orientation:
    f#(f(x)) = 13x + 10 >= 2x + 0 = g#(f(x))
    
    g#(g(x)) = 1x + 0 >= x + -10 = f#(x)
    
    f(f(x)) = 26x + 23 >= 25x + 22 = g(f(x))
    
    g(g(x)) = 24x + 23 >= 13x + 10 = f(x)
   problem:
    DPs:
     
    TRS:
     f(f(x)) -> g(f(x))
     g(g(x)) -> f(x)
   Qed