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I am solving a problem that is asking to provide an equivalent for $\neg ((a \wedge b) \Leftrightarrow c)$ and the equivalent shouldn't contain $\wedge$ (and), $\Rightarrow$ (implication) or $\Leftrightarrow$ (iff). What I have worked out is:

$\neg(\neg((\neg a \vee \neg b) \vee \neg c) \vee \neg(\neg(\neg a \vee \neg b) \vee c))$

I wonder based on the requirement stated above, is there a shorter equivalent for my answer?

Update

$\oplus$ is not allowed in the requirement either.

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Outgoing from requirements can you use $\oplus$? if yes, then, as we know, that $\neg(a\Leftrightarrow b) = a \oplus b$, then we can obtain $$\neg ((a \wedge b) \Leftrightarrow c) = \neg (\neg a \lor\neg b) \oplus c$$

Update.

Accordingly to your update I wrote my one: not shorter, slightly different, but logically equivalent yours one $$\neg \big[(a \wedge b) \Leftrightarrow c\big] = \\ =\neg \big[\big((a \wedge b) \Rightarrow c \big) \land \big(c \Rightarrow (a \wedge b)\big) \big]=\\ =\neg \big[ \big(\neg (a \wedge b) \lor c \big) \land \big( \neg c \lor (a \wedge b) \big) \big]=\\ =\neg \big[ \big(\neg a \lor\neg b \lor c \big) \land \big(\neg c \lor \neg(\neg a \lor\neg b) \big) \big]=\\ = \neg\big(\neg a \lor\neg b \lor c \big) \lor \neg\big( \neg c \lor \neg(\neg a \lor\neg b) \big) $$

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  • $\begingroup$ Thanks for your answer, but I think no I can't use XOR either, I update the question. $\endgroup$ – Coder Dec 17 '20 at 9:52

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