YES

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

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