YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 1 0]     [0]
   [b](x0) = [0 0 1]x0 + [0]
             [0 0 0]     [1],
   
             [1 0 1]  
   [a](x0) = [1 0 1]x0
             [0 0 0]  
  orientation:
                 [2 0 2]     [1]    [2 0 2]     [1]                 
   a(b(a(x1))) = [2 0 2]x1 + [1] >= [2 0 2]x1 + [1] = a(b(b(a(x1))))
                 [0 0 0]     [0]    [0 0 0]     [0]                 
   
                 [1 1 1]     [1]    [1 1 1]     [0]           
   b(b(b(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(b(x1))
                 [0 0 0]     [1]    [0 0 0]     [1]           
  problem:
   a(b(a(x1))) -> a(b(b(a(x1))))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {3}
    transitions:
     a0(3) -> 3*
     b0(3) -> 3*
     b1(11) -> 12*
     b1(14) -> 15*
     b1(10) -> 11*
     a1(9) -> 10*
     a1(12) -> 13*
     13 -> 10,14,3
     3 -> 9*
     15 -> 11*
   problem:
    
   Qed