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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.

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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.

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To simplify small expressions, you will use ad hoc methods (in larger expressions, this often lets you achieve _some _ simplification).

For larger cases, you could write or use existing software, which tries to find the “simplest” expression equivalent to a given one, according to your definition of “simplest”. The runtime will grow fast. Actually the question whether an expression can be simplified to “false” is NP-complete (that is practically unsolvable for hard cases), and that is just one very special case of simplifying an expression.

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