YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 3
  interpretation:
             [0  0  0 ]  
   [a](x0) = [0  0  1 ]x0
             [-& -& 0 ]  ,
   
             [0  0  -&]  
   [b](x0) = [-& -& 0 ]x0
             [0  1  -&]  
  orientation:
                 [0 1 1]      [0 0 1]                   
   a(b(b(x1))) = [1 1 2]x1 >= [1 1 2]x1 = b(b(a(a(x1))))
                 [0 0 1]      [0 0 1]                   
   
                 [1 1 2]      [0  0  0 ]             
   a(b(a(x1))) = [2 2 3]x1 >= [0  1  -&]x1 = b(b(x1))
                 [1 1 2]      [0  0  1 ]             
  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