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I would like to ask

not(A and not B) = (not A or B) by De Morgan's laws

Is the above statement true?

Because I can not find examples of applying De Morgan's laws to that kind of statement. It usually applies to not(A and B) not(A or B).

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  • $\begingroup$ Yes, it is true. $\endgroup$ – yugikaiba Dec 28 '20 at 6:00
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    $\begingroup$ Yes, it’s true. If $\varphi$ and $\psi$ are any well-formed formulas, then so are $\neg(\varphi\land\psi)$ and $\neg\varphi\lor\neg\psi$, and they are logically equivalent. In your case take $\varphi=A$ and $\psi=\neg B$, and you get $$\neg(A\land\neg B)\equiv\neg A\lor\neg(\neg B)\,,$$ and since $\neg(\neg B)\equiv B$, you end up with $$\neg(A\land\neg B)\equiv\neg A\lor B\,.$$ $\endgroup$ – Brian M. Scott Dec 28 '20 at 6:01

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