YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 3
  interpretation:
             [0  0  0 ]  
   [a](x0) = [-& -& 0 ]x0
             [1  2  -&]  ,
   
             [0  1  -&]  
   [b](x0) = [0  1  -&]x0
             [-& 0  -&]  
  orientation:
                 [2  3  -&]      [2  3  -&]                
   b(b(b(x1))) = [2  3  -&]x1 >= [2  3  -&]x1 = a(a(b(x1)))
                 [1  2  -&]      [1  2  -&]                
   
                 [2  3  0 ]      [2  3  -&]                
   b(a(a(x1))) = [2  3  0 ]x1 >= [2  3  -&]x1 = a(a(b(x1)))
                 [1  2  -&]      [1  2  -&]                
   
              [1  2  0 ]      [0  1  -&]                
   a(a(x1)) = [1  2  -&]x1 >= [0  1  -&]x1 = b(a(b(x1)))
              [1  1  2 ]      [-& 0  -&]                
  problem:
   b(b(b(x1))) -> a(a(b(x1)))
   b(a(a(x1))) -> a(a(b(x1)))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0 0]  
    [a](x0) = [0 0]x0,
    
              [1 0]  
    [b](x0) = [0 1]x0
   orientation:
                  [3 2]      [1 1]                
    b(b(b(x1))) = [2 3]x1 >= [1 1]x1 = a(a(b(x1)))
    
                  [1 1]      [1 1]                
    b(a(a(x1))) = [1 1]x1 >= [1 1]x1 = a(a(b(x1)))
   problem:
    b(a(a(x1))) -> a(a(b(x1)))
   KBO Processor:
    weight function:
     w0 = 1
     w(a) = w(b) = 1
    precedence:
     b > a
    problem:
     
    Qed