YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(b(a(a(b(x1)))))) -> a#(x1)
   a#(b(b(a(a(b(x1)))))) -> a#(b(a(x1)))
   a#(b(b(a(a(b(x1)))))) -> a#(b(b(a(b(a(x1))))))
   a#(b(b(a(a(b(x1)))))) -> a#(a(b(b(a(b(a(x1)))))))
  TRS:
   a(b(b(a(a(b(x1)))))) -> a(a(b(b(a(b(a(x1)))))))
  Matrix Interpretation Processor: dim=4
   
   interpretation:
    [a#](x0) = [0 0 0 1]x0,
    
              [0 1 1 0]     [0]
              [0 0 0 0]     [1]
    [a](x0) = [0 0 0 0]x0 + [0]
              [1 0 0 1]     [0],
    
              [0 0 0 0]     [0]
              [0 0 0 0]     [1]
    [b](x0) = [1 0 0 0]x0 + [0]
              [0 0 0 1]     [0]
   orientation:
    a#(b(b(a(a(b(x1)))))) = [1 0 0 1]x1 + [1] >= [0 0 0 1]x1 = a#(x1)
    
    a#(b(b(a(a(b(x1)))))) = [1 0 0 1]x1 + [1] >= [1 0 0 1]x1 = a#(b(a(x1)))
    
    a#(b(b(a(a(b(x1)))))) = [1 0 0 1]x1 + [1] >= [1 0 0 1]x1 = a#(b(b(a(b(a(x1))))))
    
    a#(b(b(a(a(b(x1)))))) = [1 0 0 1]x1 + [1] >= [1 0 0 1]x1 = a#(a(b(b(a(b(a(x1)))))))
    
                           [0 0 0 0]     [1]    [0 0 0 0]     [1]                          
                           [0 0 0 0]     [1]    [0 0 0 0]     [1]                          
    a(b(b(a(a(b(x1)))))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = a(a(b(b(a(b(a(x1)))))))
                           [1 0 0 1]     [1]    [1 0 0 1]     [1]                          
   problem:
    DPs:
     
    TRS:
     a(b(b(a(a(b(x1)))))) -> a(a(b(b(a(b(a(x1)))))))
   Qed