YES

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

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