# Simplifying the Boolean expression $A + \bar{A}\bar{B}$?

So I'm trying to simplify the Boolean expression (1) $$A + \bar{A}\bar{B}$$.

I noticed that by Karnaugh maps this is equivalent to $$A+\bar{B}$$, and I also noticed that if I take the complement of (1), I get ~(1)= $$\bar{A}(A + B) = \bar{A}A + \bar{A}B = \bar{A}B$$, so then taking complements again yields (1) = $$A + \bar{B}$$.

But this feels very ad hoc to me, and makes me feel like I'm missing some key point about how to simplify this expression more systematically.

Any thoughts appreciated.

Thanks.

There is a Law of Distributivity of Boolean operation $$\lor$$ (disjunction, which is sometimes denoted by $$+$$) over Boolean operation $$\land$$ (conjunction, which is sometimes simply omitted): $$x \lor (y \land z)= (x \lor y)\land(x \lor z)$$ So, in your case: $$A \lor (\lnot A \land \lnot B) = (A \lor \lnot A) \land (A \lor \lnot B) = A \lor \lnot B$$ For more information please see: Boolean Algebra.