YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(a(x1))) -> a#(a(x1))
   a#(b(a(x1))) -> b#(a(a(x1)))
   a#(b(a(x1))) -> b#(b(a(a(x1))))
   a#(b(a(x1))) -> a#(b(b(a(a(x1)))))
   a#(b(a(x1))) -> a#(a(b(b(a(a(x1))))))
   b#(a(a(b(x1)))) -> b#(a(b(x1)))
  TRS:
   a(b(a(x1))) -> a(a(b(b(a(a(x1))))))
   b(a(a(b(x1)))) -> b(a(b(x1)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [b#](x0) = [0 0 1]x0,
    
    [a#](x0) = [0 1 1]x0 + [1],
    
              [0 0 0]  
    [b](x0) = [1 0 0]x0
              [1 0 0]  ,
    
              [0 1 1]     [1]
    [a](x0) = [0 0 0]x0 + [1]
              [0 1 1]     [0]
   orientation:
    a#(b(a(x1))) = [0 2 2]x1 + [3] >= [0 1 1]x1 + [2] = a#(a(x1))
    
    a#(b(a(x1))) = [0 2 2]x1 + [3] >= [0 1 1]x1 + [1] = b#(a(a(x1)))
    
    a#(b(a(x1))) = [0 2 2]x1 + [3] >= [0 1 1]x1 + [2] = b#(b(a(a(x1))))
    
    a#(b(a(x1))) = [0 2 2]x1 + [3] >= [1] = a#(b(b(a(a(x1)))))
    
    a#(b(a(x1))) = [0 2 2]x1 + [3] >= [2] = a#(a(b(b(a(a(x1))))))
    
    b#(a(a(b(x1)))) = [2 0 0]x1 + [1] >= [2 0 0]x1 = b#(a(b(x1)))
    
                  [0 2 2]     [3]    [2]                       
    a(b(a(x1))) = [0 0 0]x1 + [1] >= [1] = a(a(b(b(a(a(x1))))))
                  [0 2 2]     [2]    [1]                       
    
                     [0 0 0]     [0]    [0 0 0]     [0]              
    b(a(a(b(x1)))) = [2 0 0]x1 + [2] >= [2 0 0]x1 + [1] = b(a(b(x1)))
                     [2 0 0]     [2]    [2 0 0]     [1]              
   problem:
    DPs:
     
    TRS:
     a(b(a(x1))) -> a(a(b(b(a(a(x1))))))
     b(a(a(b(x1)))) -> b(a(b(x1)))
   Qed