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I have one of De Morgan's laws (in propositional logic). I would like to prove the other law from the first using a sequence of equivalences (Resolution).

One is not allowed to use truth tables or the particular De Morgans law which we are trying to prove (obviously).

How can this be done?

$\lnot (A\land B)\equiv \lnot A \lor \lnot B$

$\lnot (A\lor B)\equiv \lnot A \land \lnot B$


a set of resolution equivalences laws can be found on wikipedia

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Hint

Let assume $\lnot (A \land B) \equiv \lnot A \lor \lnot B$.

Then from $\lnot (A \lor B)$ using Double Negation, we get $\lnot (\lnot \lnot A \lor \lnot \lnot B) \equiv \lnot \lnot (\lnot A \land \lnot B)$.

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For future enquirers here is an unformatted solution (using @Mauro's hint) and some relabelling.

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