YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [1 1]  
   [a](x0) = [0 0]x0,
   
             [0 1]  
   [b](x0) = [0 0]x0
  orientation:
                 [2 2]      [1 1]          
   a(b(b(x1))) = [1 1]x1 >= [0 0]x1 = a(x1)
   
              [2 2]      [1 2]                
   a(a(x1)) = [1 1]x1 >= [1 1]x1 = b(b(b(x1)))
   
                 [2 2]      [2 2]                
   b(b(a(x1))) = [1 1]x1 >= [1 1]x1 = a(b(a(x1)))
  problem:
   a(a(x1)) -> b(b(b(x1)))
   b(b(a(x1))) -> a(b(a(x1)))
  String Reversal Processor:
   a(a(x1)) -> b(b(b(x1)))
   a(b(b(x1))) -> a(b(a(x1)))
   Bounds Processor:
    bound: 1
    enrichment: match
    automaton:
     final states: {5,1}
     transitions:
      a1(12) -> 13*
      a1(14) -> 15*
      a1(16) -> 17*
      b1(17) -> 18*
      b1(13) -> 14*
      f20() -> 2*
      b0(2) -> 3*
      b0(4) -> 1*
      b0(6) -> 7*
      b0(3) -> 4*
      a0(7) -> 5*
      a0(2) -> 6*
      1 -> 17,6
      2 -> 16*
      4 -> 12*
      5 -> 17,6
      15 -> 13,5
      18 -> 12*
    problem:
     
    Qed