YES

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

Proof:
 DP Processor:
  DPs:
   b#(b(x1)) -> b#(a(b(x1)))
   b#(b(a(b(x1)))) -> b#(b(x1))
   b#(b(a(b(x1)))) -> b#(a(a(b(b(x1)))))
   b#(b(a(b(x1)))) -> b#(a(b(a(a(b(b(x1)))))))
   b#(a(b(x1))) -> b#(a(a(b(x1))))
   b#(a(a(b(a(b(x1)))))) -> b#(b(x1))
  TRS:
   b(b(x1)) -> b(a(b(x1)))
   b(b(a(b(x1)))) -> b(a(b(a(a(b(b(x1)))))))
   b(a(b(x1))) -> b(a(a(b(x1))))
   b(a(a(b(a(b(x1)))))) -> b(b(x1))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [b#](x0) = [2 1 2]x0,
    
              [0 0 0]  
    [a](x0) = [1 0 0]x0
              [0 0 1]  ,
    
              [0 0 0]     [1]
    [b](x0) = [0 0 0]x0 + [0]
              [2 2 1]     [0]
   orientation:
    b#(b(x1)) = [4 4 2]x1 + [2] >= [4 4 2]x1 + [1] = b#(a(b(x1)))
    
    b#(b(a(b(x1)))) = [4 4 2]x1 + [6] >= [4 4 2]x1 + [2] = b#(b(x1))
    
    b#(b(a(b(x1)))) = [4 4 2]x1 + [6] >= [4 4 2]x1 + [4] = b#(a(a(b(b(x1)))))
    
    b#(b(a(b(x1)))) = [4 4 2]x1 + [6] >= [4 4 2]x1 + [5] = b#(a(b(a(a(b(b(x1)))))))
    
    b#(a(b(x1))) = [4 4 2]x1 + [1] >= [4 4 2]x1 = b#(a(a(b(x1))))
    
    b#(a(a(b(a(b(x1)))))) = [4 4 2]x1 + [4] >= [4 4 2]x1 + [2] = b#(b(x1))
    
               [0 0 0]     [1]    [0 0 0]     [1]              
    b(b(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(b(x1)))
               [2 2 1]     [2]    [2 2 1]     [2]              
    
                     [0 0 0]     [1]    [0 0 0]     [1]                          
    b(b(a(b(x1)))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(b(a(a(b(b(x1)))))))
                     [2 2 1]     [4]    [2 2 1]     [4]                          
    
                  [0 0 0]     [1]    [0 0 0]     [1]                 
    b(a(b(x1))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(a(a(b(x1))))
                  [2 2 1]     [2]    [2 2 1]     [0]                 
    
                           [0 0 0]     [1]    [0 0 0]     [1]           
    b(a(a(b(a(b(x1)))))) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(b(x1))
                           [2 2 1]     [2]    [2 2 1]     [2]           
   problem:
    DPs:
     
    TRS:
     b(b(x1)) -> b(a(b(x1)))
     b(b(a(b(x1)))) -> b(a(b(a(a(b(b(x1)))))))
     b(a(b(x1))) -> b(a(a(b(x1))))
     b(a(a(b(a(b(x1)))))) -> b(b(x1))
   Qed