YES

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

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