YES

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

Proof:
 String Reversal Processor:
  b(a(a(x1))) -> a(a(b(b(x1))))
  b(a(b(x1))) -> a(a(a(a(x1))))
  DP Processor:
   DPs:
    b#(a(a(x1))) -> b#(x1)
    b#(a(a(x1))) -> b#(b(x1))
   TRS:
    b(a(a(x1))) -> a(a(b(b(x1))))
    b(a(b(x1))) -> a(a(a(a(x1))))
   Arctic Interpretation Processor:
    dimension: 2
    usable rules:
     b(a(a(x1))) -> a(a(b(b(x1))))
     b(a(b(x1))) -> a(a(a(a(x1))))
    interpretation:
               [-& 0 ]     [1]
     [a](x0) = [2  -&]x0 + [1],
     
               [0  2 ]     [3 ]
     [b](x0) = [-& 0 ]x0 + [-&],
     
     [b#](x0) = [-& 1 ]x0 + [0]
    orientation:
     b#(a(a(x1))) = [-& 3 ]x1 + [4] >= [-& 1 ]x1 + [0] = b#(x1)
     
     b#(a(a(x1))) = [-& 3 ]x1 + [4] >= [-& 1 ]x1 + [0] = b#(b(x1))
     
                   [2  4 ]     [5]    [2  4 ]     [5]                 
     b(a(a(x1))) = [-& 2 ]x1 + [3] >= [-& 2 ]x1 + [3] = a(a(b(b(x1))))
     
                   [4 6]     [7]    [4  -&]     [3]                 
     b(a(b(x1))) = [2 4]x1 + [5] >= [-& 4 ]x1 + [5] = a(a(a(a(x1))))
    problem:
     DPs:
      
     TRS:
      b(a(a(x1))) -> a(a(b(b(x1))))
      b(a(b(x1))) -> a(a(a(a(x1))))
    Qed