Simplifying Boolean Expression 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&lt;B)i and (A&gt;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&lt;B)i = ~Ai*Bi + ~Ai*(A&lt;B)i-1 + Bi*(A&lt;B)i-1

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

~Ai*(Bi + (A&lt;B)i-1) + Bi(A&lt;B)i-1

Let's call  Bi + (A&lt;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&lt;B)i = Bi + (A&lt;B)i-1
No, you cannot simplify to $(A&lt;B)_i = B_i + (A&lt;B)_{i-1}$, Take for example $A=11$ and $B = 10$. 

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

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

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