I am designing a 4-bit comparator with a look ahead unit using a bit slice approach. I have to break the propagation of the Logical expressions for (A<B)i and (A>B)i. The main question is, could I use the following to simplify the boolean expressions.. The i is a subscript meaning the ith bit. The i-1 is a subscript meaning the bit before the ith bit.

(A<B)i = ~Ai*Bi + ~Ai*(A<B)i-1 + Bi*(A<B)i-1

Could I Take ~Ai common from the first two expressions leaving me with

~Ai*(Bi + (A<B)i-1) + Bi(A<B)i-1

Let's call Bi + (A<B)i-1) = Q

And let's call ~Ai = P

I'm left with P*Q + Q

Using the boolean law that Q + Q*P = Q

Can I simplify the expression to just

(A<B)i = Bi + (A<B)i-1

  • $\begingroup$ I would have preferred a simpler notation. $\endgroup$ Commented Feb 17, 2016 at 8:16

1 Answer 1


No, you cannot simplify to $(A<B)_i = B_i + (A<B)_{i-1}$, Take for example $A=11$ and $B = 10$.

Then $(A<B)_2$ = False, $B_2$ = True, and $(A<B)_1$ = False. If you substitute the values above then the equation is not satisfied. The original equation(s) is correct.

$(A<B)_i = \neg A_i B_i + (\neg A_i+B_i)(A<B)_{i-1}$
$(A<B)_1 = \neg A_1 B_1$

In your derivation of simpler formula, $B_i + (A<B)_{i-1} = Q$ but you are incorrectly substituting $B_i (A<B)_{i-1} = Q$ in rightmost term.


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