YES

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

Proof:
 Matrix Interpretation Processor: dim=4
  
  interpretation:
             [1 0 0 0]  
             [0 0 1 0]  
   [b](x0) = [1 0 0 0]x0
             [0 0 1 0]  ,
   
             [1 0 1 0]     [1]
             [0 0 0 0]     [0]
   [a](x0) = [0 1 0 1]x0 + [0]
             [0 0 0 0]     [0]
  orientation:
                    [2 0 2 0]     [3]    [2 0 2 0]     [2]                 
                    [0 0 0 0]     [0]    [0 0 0 0]     [0]                 
   a(b(b(a(x1)))) = [2 0 2 0]x1 + [2] >= [2 0 2 0]x1 + [2] = b(a(a(b(x1))))
                    [0 0 0 0]     [0]    [0 0 0 0]     [0]                 
   
                 [2 0 0 0]     [1]    [2 0 0 0]     [1]                 
                 [0 0 2 0]     [0]    [0 0 0 0]     [0]                 
   b(a(b(x1))) = [2 0 0 0]x1 + [1] >= [2 0 0 0]x1 + [0] = a(b(b(b(x1))))
                 [0 0 2 0]     [0]    [0 0 0 0]     [0]                 
  problem:
   b(a(b(x1))) -> a(b(b(b(x1))))
  KBO Processor:
   weight function:
    w0 = 1
    w(a) = 1
    w(b) = 0
   precedence:
    b > a
   problem:
    
   Qed