YES

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

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