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Minimize $$(p \land q) \lor (\neg p \lor q)$$ without using a truth table.

I have tried with an online website and it shows $\neg p \lor q$ as the answer. I was able to verify with truth table but I don't get how to approach it.

PS: I have a basic idea of the idempotent associative laws, etc and I am new to discrete mathematics.

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5 Answers 5

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$(p\wedge q)\vee(\neg p\vee q)=(\underbrace{p\vee \neg p}_{1} \vee q)\wedge(q\vee\neg p \vee q)=1\wedge (q\vee \neg p)=\neg p\vee q$

just like J.G. said.

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Conjunction elimination $$(p\land q)\to(p\to q)$$ Disjunction elimination $$(-p\lor q)\to (-p\to -q)$$ Disjunction introduction $$(-p\to -q)\lor(p\to q)\to (-p\lor q) \boldsymbol{\text{or}} (p\lor -q)$$

We'll pick the first for sanity.

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If $p\land q$, then $q$ and in particular $\neg p\lor q$. So $(p\land q)\lor(\neg p\lor q)$ is just $\neg p\lor q$.

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Translating to Boolean algebra,

$$pq + \bar{p} + q = \underbrace{(p+1)}_{=1} q + \bar{p} = q + \bar{p} = \bar{p} + q$$

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$$(p \land q) \lor (\neg p \lor q)\\ =(p\lor\neg p\lor q)\land(q\lor\neg p\lor q)\\ =T\land(\neg p\lor q)\\ =\neg p \lor q$$

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