YES

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

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