YES

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

Proof:
 String Reversal Processor:
  a(b(a(x1))) -> a(b(x1))
  b(b(b(x1))) -> b(a(b(x1)))
  a(a(x1)) -> b(b(b(x1)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]     [1]
    [b](x0) = [0 0 0]x0 + [1]
              [0 0 0]     [0],
    
              [1 0 1]     [1]
    [a](x0) = [1 0 1]x0 + [1]
              [0 0 1]     [1]
   orientation:
                  [1 0 1]     [3]    [1 0 0]     [2]           
    a(b(a(x1))) = [1 0 1]x1 + [3] >= [1 0 0]x1 + [2] = a(b(x1))
                  [0 0 0]     [1]    [0 0 0]     [1]           
    
                  [1 0 0]     [3]    [1 0 0]     [3]              
    b(b(b(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(a(b(x1)))
                  [0 0 0]     [0]    [0 0 0]     [0]              
    
               [1 0 2]     [3]    [1 0 0]     [3]              
    a(a(x1)) = [1 0 2]x1 + [3] >= [0 0 0]x1 + [1] = b(b(b(x1)))
               [0 0 1]     [2]    [0 0 0]     [0]              
   problem:
    b(b(b(x1))) -> b(a(b(x1)))
    a(a(x1)) -> b(b(b(x1)))
   Bounds Processor:
    bound: 1
    enrichment: match
    automaton:
     final states: {5,1}
     transitions:
      b1(11) -> 12*
      b1(13) -> 14*
      b1(15) -> 16*
      f20() -> 2*
      b0(6) -> 5*
      b0(2) -> 3*
      b0(4) -> 1*
      b0(3) -> 6*
      a0(3) -> 4*
      a1(12) -> 13*
      14 -> 5*
      16 -> 12*
      4 -> 15*
      2 -> 11*
      1 -> 3,6
    problem:
     
    Qed