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For example if I want to show the equation for $x \rightarrow y$, using truth tables it is the same as:

$$(\neg x \wedge \neg y) \lor (\neg x \wedge y) \lor (x \wedge y)$$

Is there a methodical way to reduce something like this to its simple form $\neg x \lor y$?

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You can prove this way: \begin{align} ((\neg x \wedge \neg y) \lor (\neg x \wedge y) \lor (x \wedge y))&\iff ((\neg x\land(\neg y\lor y))\lor (x \wedge y)) \\ &\iff((\neg x\land 1)\lor (x \wedge y)) \\ &\iff(\neg x\lor (x \wedge y)) \\ &\iff((\neg x\lor x)\land (\neg x\lor y)) \\ &\iff(1\land (\neg x\lor y)) \\ &\iff(\neg x\lor y) \end{align}

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  • $\begingroup$ Wow, I didn't know you could use distributive property like that! $\endgroup$ – Sean Hill Dec 27 '15 at 23:27

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