YES

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

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