YES

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

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