YES

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

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