YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [1  0 ]  
   [f](x0) = [-& 1 ]x0,
   
             [0 0]  
   [g](x0) = [0 0]x0
  orientation:
             [1 1]     [1 1]              
   f(g(x)) = [1 1]x >= [1 1]x = g(g(f(x)))
   
             [1 1]     [0 0]              
   f(g(x)) = [1 1]x >= [0 0]x = g(g(g(x)))
  problem:
   f(g(x)) -> g(g(f(x)))
  Bounds Processor:
   bound: 0
   enrichment: match
   automaton:
    final states: {1}
    transitions:
     f20() -> 2*
     g0(4) -> 1*
     g0(3) -> 4*
     f0(2) -> 3*
     1 -> 3*
   problem:
    
   Qed