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This is a question from my school work. Although i have solved it, i can't seem to do it the way the question wants me to. The question involves you finding the complement of an expression and simplifying the results. The expression was

$$ (y\bar{z} + \bar{x}w)(x\bar{y} + \bar{w}z) $$

While solving it by simplifying than finding the complement was easy, i cant seem to do it by finding the complement first. Any help will be appreciated, thanks in advance!

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    You might want to explicitly state what $+$ and $\cdot$ is here as boolean algebras usually use $\wedge$ etc2017-01-20

1 Answers 1

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The quickest way is perhaps the one based on expansion. We have

$$ f(x,y,\ldots) = (x \wedge f(\top,y,\ldots)) \vee (\neg x \wedge f(\bot,y,\ldots)) \enspace. $$

and

$$ \neg f(x,y,\ldots) = (x \wedge \neg f(\top,y,\ldots)) \vee (\neg x \wedge \neg f(\bot,y,\ldots)) \enspace. $$

In this case, expanding w.r.t. $x$ immediately shows that $f$ is identically false and hence $\neg f$ is identically true.

You can also apply De Morgan's laws and, slowly but almost surely, you'll get the same result. It helps to keep in mind that

$$(a \wedge b) \vee (\neg b \wedge c) = (a \wedge b) \vee (\neg b \wedge c) \vee (a \wedge c) \enspace. $$

After applying De Morgan and distributivity to get a disjunctive normal form (a.k.a. sum of products), the theorem above produces the extra terms that quickly show that the expression is identically true. It's quite a bit more work than with the expansion theorem, though.