YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 3
  interpretation:
             [0  0  -&]  
   [a](x0) = [-& -& 0 ]x0
             [-& 0  -&]  ,
   
             [0  0  1 ]  
   [b](x0) = [0  0  1 ]x0
             [-& -& 0 ]  
  orientation:
                 [0  0  1 ]      [0  0  1 ]                   
   a(a(b(x1))) = [0  0  1 ]x1 >= [0  0  1 ]x1 = b(b(a(a(x1))))
                 [-& -& 0 ]      [-& -& 0 ]                   
   
                 [1 1 2]      [0  0  0 ]                   
   b(a(b(x1))) = [1 1 2]x1 >= [-& 0  -&]x1 = a(a(a(a(x1))))
                 [0 0 1]      [-& -& 0 ]                   
  problem:
   a(a(b(x1))) -> b(b(a(a(x1))))
  Bounds Processor:
   bound: 0
   enrichment: match
   automaton:
    final states: {1}
    transitions:
     f20() -> 2*
     b0(5) -> 1*
     b0(4) -> 5*
     a0(2) -> 3*
     a0(3) -> 4*
     1 -> 3,4
   problem:
    
   Qed