YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0 0]  
   [b](x0) = [0 2]x0,
   
             [0  -&]  
   [a](x0) = [-& -&]x0
  orientation:
                 [0  -&]      [0  -&]                   
   a(b(a(x1))) = [-& -&]x1 >= [-& -&]x1 = a(b(b(a(x1))))
   
                 [2 4]      [0 2]             
   b(b(b(x1))) = [4 6]x1 >= [2 4]x1 = b(b(x1))
  problem:
   a(b(a(x1))) -> a(b(b(a(x1))))
  Bounds Processor:
   bound: 0
   enrichment: match
   automaton:
    final states: {1}
    transitions:
     f20() -> 2*
     a0(5) -> 1*
     a0(2) -> 3*
     b0(4) -> 5*
     b0(3) -> 4*
     1 -> 3*
   problem:
    
   Qed