YES

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

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