YES

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

Proof:
 String Reversal Processor:
  b(b(a(x1))) -> a(x1)
  a(a(x1)) -> b(b(b(x1)))
  a(b(b(x1))) -> a(b(a(x1)))
  DP Processor:
   DPs:
    a#(a(x1)) -> b#(x1)
    a#(a(x1)) -> b#(b(x1))
    a#(a(x1)) -> b#(b(b(x1)))
    a#(b(b(x1))) -> a#(x1)
    a#(b(b(x1))) -> b#(a(x1))
    a#(b(b(x1))) -> a#(b(a(x1)))
   TRS:
    b(b(a(x1))) -> a(x1)
    a(a(x1)) -> b(b(b(x1)))
    a(b(b(x1))) -> a(b(a(x1)))
   TDG Processor:
    DPs:
     a#(a(x1)) -> b#(x1)
     a#(a(x1)) -> b#(b(x1))
     a#(a(x1)) -> b#(b(b(x1)))
     a#(b(b(x1))) -> a#(x1)
     a#(b(b(x1))) -> b#(a(x1))
     a#(b(b(x1))) -> a#(b(a(x1)))
    TRS:
     b(b(a(x1))) -> a(x1)
     a(a(x1)) -> b(b(b(x1)))
     a(b(b(x1))) -> a(b(a(x1)))
    graph:
     a#(b(b(x1))) -> a#(b(a(x1))) -> a#(b(b(x1))) -> a#(b(a(x1)))
     a#(b(b(x1))) -> a#(b(a(x1))) -> a#(b(b(x1))) -> b#(a(x1))
     a#(b(b(x1))) -> a#(b(a(x1))) -> a#(b(b(x1))) -> a#(x1)
     a#(b(b(x1))) -> a#(b(a(x1))) -> a#(a(x1)) -> b#(b(b(x1)))
     a#(b(b(x1))) -> a#(b(a(x1))) -> a#(a(x1)) -> b#(b(x1))
     a#(b(b(x1))) -> a#(b(a(x1))) -> a#(a(x1)) -> b#(x1)
     a#(b(b(x1))) -> a#(x1) -> a#(b(b(x1))) -> a#(b(a(x1)))
     a#(b(b(x1))) -> a#(x1) -> a#(b(b(x1))) -> b#(a(x1))
     a#(b(b(x1))) -> a#(x1) -> a#(b(b(x1))) -> a#(x1)
     a#(b(b(x1))) -> a#(x1) -> a#(a(x1)) -> b#(b(b(x1)))
     a#(b(b(x1))) -> a#(x1) -> a#(a(x1)) -> b#(b(x1))
     a#(b(b(x1))) -> a#(x1) -> a#(a(x1)) -> b#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 12/36
     DPs:
      a#(b(b(x1))) -> a#(b(a(x1)))
      a#(b(b(x1))) -> a#(x1)
     TRS:
      b(b(a(x1))) -> a(x1)
      a(a(x1)) -> b(b(b(x1)))
      a(b(b(x1))) -> a(b(a(x1)))
     Arctic Interpretation Processor:
      dimension: 2
      usable rules:
       b(b(a(x1))) -> a(x1)
       a(a(x1)) -> b(b(b(x1)))
       a(b(b(x1))) -> a(b(a(x1)))
      interpretation:
                 [0 0]     [0]
       [a](x0) = [2 2]x0 + [2],
       
       [a#](x0) = [0 0]x0,
       
                 [0 0]     [0]
       [b](x0) = [2 0]x0 + [2]
      orientation:
       a#(b(b(x1))) = [2 2]x1 + [2] >= [2 2]x1 + [2] = a#(b(a(x1)))
       
       a#(b(b(x1))) = [2 2]x1 + [2] >= [0 0]x1 = a#(x1)
       
                     [2 2]     [2]    [0 0]     [0]        
       b(b(a(x1))) = [4 4]x1 + [4] >= [2 2]x1 + [2] = a(x1)
       
                  [2 2]     [2]    [2 2]     [2]              
       a(a(x1)) = [4 4]x1 + [4] >= [4 2]x1 + [4] = b(b(b(x1)))
       
                     [2 2]     [2]    [2 2]     [2]              
       a(b(b(x1))) = [4 4]x1 + [4] >= [4 4]x1 + [4] = a(b(a(x1)))
      problem:
       DPs:
        a#(b(b(x1))) -> a#(b(a(x1)))
       TRS:
        b(b(a(x1))) -> a(x1)
        a(a(x1)) -> b(b(b(x1)))
        a(b(b(x1))) -> a(b(a(x1)))
      Restore Modifier:
       DPs:
        a#(b(b(x1))) -> a#(b(a(x1)))
       TRS:
        b(b(a(x1))) -> a(x1)
        a(a(x1)) -> b(b(b(x1)))
        a(b(b(x1))) -> a(b(a(x1)))
       EDG Processor:
        DPs:
         a#(b(b(x1))) -> a#(b(a(x1)))
        TRS:
         b(b(a(x1))) -> a(x1)
         a(a(x1)) -> b(b(b(x1)))
         a(b(b(x1))) -> a(b(a(x1)))
        graph:
         a#(b(b(x1))) -> a#(b(a(x1))) -> a#(b(b(x1))) -> a#(b(a(x1)))
        Arctic Interpretation Processor:
         dimension: 2
         usable rules:
          b(b(a(x1))) -> a(x1)
          a(a(x1)) -> b(b(b(x1)))
          a(b(b(x1))) -> a(b(a(x1)))
         interpretation:
                    [2  0 ]     [3 ]
          [a](x0) = [-2 -&]x0 + [-2],
          
          [a#](x0) = [3  -&]x0 + [0],
          
                    [-& 0 ]     [0]
          [b](x0) = [0  -3]x0 + [1]
         orientation:
          a#(b(b(x1))) = [3 0]x1 + [4] >= [1  -&]x1 + [3] = a#(b(a(x1)))
          
                        [2  0 ]     [3]    [2  0 ]     [3 ]        
          b(b(a(x1))) = [-1 -3]x1 + [1] >= [-2 -&]x1 + [-2] = a(x1)
          
                     [4  2 ]     [5]    [-3 0 ]     [1]              
          a(a(x1)) = [0  -2]x1 + [1] >= [0  -3]x1 + [1] = b(b(b(x1)))
          
                        [2  0 ]     [3 ]    [2  0 ]     [3 ]              
          a(b(b(x1))) = [-2 -5]x1 + [-1] >= [-4 -&]x1 + [-2] = a(b(a(x1)))
         problem:
          DPs:
           
          TRS:
           b(b(a(x1))) -> a(x1)
           a(a(x1)) -> b(b(b(x1)))
           a(b(b(x1))) -> a(b(a(x1)))
         Qed