YES

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

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