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Simplifying the given expression led me to $A\lor\neg B$. Here is what I did:

$A \lor [\neg(\neg A)\land B]$ -------> Given

$A \lor [\neg(\neg A)\lor \neg B]$ ------> De Morgan's Law

$A \lor (A\lor \neg B)$ -----------> Law of double negation

$(A \lor A)\lor \neg B$ -----------> Associative Law

$A \lor \neg B$-----------------> Idempotent Law

Now, I want to know if my application of De Morgan's Law (second step) is correct. Thanks!

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  • $\begingroup$ Looks good to me $\endgroup$
    – Eminem
    Jun 17, 2020 at 5:33
  • $\begingroup$ You may also rewrite the last statement as $B\to A$ (but maybe you want to keep it with $\land,\lor,\lnot$ only) $\endgroup$
    – Manlio
    Jun 17, 2020 at 9:21
  • $\begingroup$ Please use mathjax for mathematical formulas. eg the $\vee$ symbol is \vee, the $\neg$ symbol is \neg and the $\wedge$ is \wedge. All you have to do is to surround your formula, e.g A \vee \neg B, with dollar symbols and voilà. : $A \vee \neg B$ :) $\endgroup$
    – Olivier Roche
    Jun 17, 2020 at 9:26
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    $\begingroup$ @OlivierRoche I use \land (logic and) for $\land$ and \lor (logic or) for $\lor$. Easier to remember for me. $\endgroup$
    – Henno Brandsma
    Jun 17, 2020 at 9:44
  • $\begingroup$ @HennoBrandsma Nice, I wasn't aware of these aliases :) $\endgroup$
    – Olivier Roche
    Jun 17, 2020 at 10:20

1 Answer 1

Reset to default
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Everything is fine, but you may simplify even further : by definition of $\to$, the last line is equivalent to $$B \to A$$

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  • $\begingroup$ Oh my! You're right! I just proved their logical equivalence through truth table! Thanks a lot! $\endgroup$
    – romeoPH
    Jun 17, 2020 at 9:35

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