I recently have started practicing problems on Discrete Mathematics by Keneth H.Rosen. There seems to be an interesting problem on Propositional Equivalence which states that when is $s^{*} = s$. Where s is a compound proposition and $s^{*}$ is the dual of $s$.
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1$\begingroup$ $s^*$ is the formula obtained from $s$ by exchanging the symbols $\vee$ and $\wedge$? $\endgroup$– xyzCommented May 15, 2021 at 2:45
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2 Answers
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One solution I was thinking of is s* = s whenever s* <-> s is a tautology. Correct me if I am wrong.
Looking for suggestions.
Thanks!
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$\begingroup$ Isn't that just what = means here? So it doesn't seem to say much in a way. I think the question is looking for an explanation in the shape of the formula itself. Is (p and not p) or q an example of such a formula? $\endgroup$ Commented Jun 28, 2021 at 17:08
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$\begingroup$ Yes, I proved it by applying demorgan's law on $\neg p \vee \neg q$ gives $\neg (p \wedge q)$, so $(\neg p \vee \neg q) \leftrightarrow \neg (p \wedge q)$ is indeed Tautology. $\endgroup$– tbhaxorCommented Mar 5 at 17:57
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If $s$ has any occurrences of $∧,∨, T, or F$, then the process of forming the dual will change it. Then $s^∗=s$ if and only if $s$ is one propositional variable.
N.B: I have recently started learning discrete mathematics and solved this problem a few days ago. :)
