YES

Problem:
 a(a(x1)) -> b(b(b(x1)))
 b(b(b(b(b(x1))))) -> a(a(a(x1)))

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [b](x0) = 4x0,
   
   [a](x0) = 6x0
  orientation:
   a(a(x1)) = 12x1 >= 12x1 = b(b(b(x1)))
   
   b(b(b(b(b(x1))))) = 20x1 >= 18x1 = a(a(a(x1)))
  problem:
   a(a(x1)) -> b(b(b(x1)))
  Arctic Interpretation Processor:
   dimension: 3
   interpretation:
              [0  -& 0 ]  
    [b](x0) = [-& -& 0 ]x0
              [0  0  0 ]  ,
    
              [0  1  0 ]  
    [a](x0) = [1  0  1 ]x0
              [1  1  -&]  
   orientation:
               [2 1 2]      [0 0 0]                
    a(a(x1)) = [2 2 1]x1 >= [0 0 0]x1 = b(b(b(x1)))
               [2 2 2]      [0 0 0]                
   problem:
    
   Qed