YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [2 0]  
   [b](x0) = [3 1]x0,
   
             [0  1 ]  
   [a](x0) = [-& 0 ]x0
  orientation:
                 [4 5]      [2 0]          
   a(b(a(x1))) = [3 4]x1 >= [3 1]x1 = b(x1)
   
                 [6 4]      [6 4]                
   b(a(b(x1))) = [7 5]x1 >= [7 5]x1 = b(b(b(x1)))
   
              [4 2]      [0  1 ]                
   b(b(x1)) = [5 3]x1 >= [-& 0 ]x1 = a(a(a(x1)))
  problem:
   a(b(a(x1))) -> b(x1)
   b(a(b(x1))) -> b(b(b(x1)))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  0 ]  
    [b](x0) = [-& 1 ]x0,
    
              [0 0]  
    [a](x0) = [0 1]x0
   orientation:
                  [1 2]      [0  0 ]          
    a(b(a(x1))) = [2 3]x1 >= [-& 1 ]x1 = b(x1)
    
                  [0 2]      [0  2 ]                
    b(a(b(x1))) = [1 3]x1 >= [-& 3 ]x1 = b(b(b(x1)))
   problem:
    b(a(b(x1))) -> b(b(b(x1)))
   Arctic Interpretation Processor:
    dimension: 2
    interpretation:
               [0  0 ]  
     [b](x0) = [-& 2 ]x0,
     
               [1  -&]  
     [a](x0) = [0  3 ]x0
    orientation:
                   [1 5]      [0  4 ]                
     b(a(b(x1))) = [2 7]x1 >= [-& 6 ]x1 = b(b(b(x1)))
    problem:
     
    Qed