YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  a(a(x1)) -> b(a(b(a(x1))))
  b(b(b(b(x1)))) -> a(x1)
  DP Processor:
   DPs:
    a#(a(x1)) -> b#(a(x1))
    a#(a(x1)) -> a#(b(a(x1)))
    a#(a(x1)) -> b#(a(b(a(x1))))
    b#(b(b(b(x1)))) -> a#(x1)
   TRS:
    a(x1) -> x1
    a(a(x1)) -> b(a(b(a(x1))))
    b(b(b(b(x1)))) -> a(x1)
   TDG Processor:
    DPs:
     a#(a(x1)) -> b#(a(x1))
     a#(a(x1)) -> a#(b(a(x1)))
     a#(a(x1)) -> b#(a(b(a(x1))))
     b#(b(b(b(x1)))) -> a#(x1)
    TRS:
     a(x1) -> x1
     a(a(x1)) -> b(a(b(a(x1))))
     b(b(b(b(x1)))) -> a(x1)
    graph:
     b#(b(b(b(x1)))) -> a#(x1) -> a#(a(x1)) -> b#(a(b(a(x1))))
     b#(b(b(b(x1)))) -> a#(x1) -> a#(a(x1)) -> a#(b(a(x1)))
     b#(b(b(b(x1)))) -> a#(x1) -> a#(a(x1)) -> b#(a(x1))
     a#(a(x1)) -> b#(a(b(a(x1)))) -> b#(b(b(b(x1)))) -> a#(x1)
     a#(a(x1)) -> b#(a(x1)) -> b#(b(b(b(x1)))) -> a#(x1)
     a#(a(x1)) -> a#(b(a(x1))) -> a#(a(x1)) -> b#(a(b(a(x1))))
     a#(a(x1)) -> a#(b(a(x1))) -> a#(a(x1)) -> a#(b(a(x1)))
     a#(a(x1)) -> a#(b(a(x1))) -> a#(a(x1)) -> b#(a(x1))
    Arctic Interpretation Processor:
     dimension: 2
     usable rules:
      a(x1) -> x1
      a(a(x1)) -> b(a(b(a(x1))))
      b(b(b(b(x1)))) -> a(x1)
     interpretation:
      [b#](x0) = [1 3]x0 + [0],
      
      [a#](x0) = [3 2]x0 + [0],
      
                [-& 0 ]     [1]
      [b](x0) = [0  1 ]x0 + [0],
      
                [2 1]     [2 ]
      [a](x0) = [1 0]x0 + [-&]
     orientation:
      a#(a(x1)) = [5 4]x1 + [5] >= [4 3]x1 + [3] = b#(a(x1))
      
      a#(a(x1)) = [5 4]x1 + [5] >= [4 3]x1 + [4] = a#(b(a(x1)))
      
      a#(a(x1)) = [5 4]x1 + [5] >= [5 4]x1 + [5] = b#(a(b(a(x1))))
      
      b#(b(b(b(x1)))) = [5 6]x1 + [5] >= [3 2]x1 + [0] = a#(x1)
      
              [2 1]     [2 ]           
      a(x1) = [1 0]x1 + [-&] >= x1 = x1
      
                 [4 3]     [4]    [2 1]     [2]                 
      a(a(x1)) = [3 2]x1 + [3] >= [3 2]x1 + [3] = b(a(b(a(x1))))
      
                       [2 3]     [2]    [2 1]     [2 ]        
      b(b(b(b(x1)))) = [3 4]x1 + [3] >= [1 0]x1 + [-&] = a(x1)
     problem:
      DPs:
       a#(a(x1)) -> b#(a(b(a(x1))))
      TRS:
       a(x1) -> x1
       a(a(x1)) -> b(a(b(a(x1))))
       b(b(b(b(x1)))) -> a(x1)
     Restore Modifier:
      DPs:
       a#(a(x1)) -> b#(a(b(a(x1))))
      TRS:
       a(x1) -> x1
       a(a(x1)) -> b(a(b(a(x1))))
       b(b(b(b(x1)))) -> a(x1)
      EDG Processor:
       DPs:
        a#(a(x1)) -> b#(a(b(a(x1))))
       TRS:
        a(x1) -> x1
        a(a(x1)) -> b(a(b(a(x1))))
        b(b(b(b(x1)))) -> a(x1)
       graph:
        
       Qed