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Need to proof using laws

$$\lnot(p \land \lnot q) \lor q \equiv \lnot p \lor q$$


$\lnot(p \land \lnot q) \lor q$

$\equiv (\lnot p \lor \lnot(\lnot q)) \lor q\quad$ First De Morgan's law

$\equiv (\lnot p \lor q) \lor q \quad$ Double Negation

I think I have the first to laws right. What else do I need to do to prove the equivalence?

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Yes, you've done fine. Now use the associative law to get $$(\lnot p \lor q) \lor q \equiv \lnot p \lor(q \lor q) \equiv \lnot p \lor q$$

You can call the reason for the last equivalence "repetition" (1): $$q \lor q \equiv q\tag{1}$$ $$q \land q \equiv q\tag{2}$$

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  • $\begingroup$ Idempotence is another name for the ((q∨q)≡q) law. $\endgroup$ – Doug Spoonwood Sep 18 '14 at 19:43

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