YES

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

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