YES

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

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