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This is from a Discrete Mathematics term test question.

Q4. Which of the following statements is/are logically equivalent to $p \iff q$?

Note: $\sim$ means the negation of

(I) $(\sim p \lor q) \land (p \lor \sim q)$

(II) $(\sim p \land \sim q) \lor (p \land q)$

(III) $(\sim p \lor \sim q) \land (p \lor q)$

(IV) $(\sim p \land q) \lor (p \land \sim q)$

The answer is (I) and (II). I could easily get the first one since $p \iff q \equiv (p \to q) \land (q \to p)$.

As this is MCQ, I definitely avoid truth tables. I also want to avoid using a lot of equivalent laws to save time and reduce sources of error. However for (II), I cannot think of simple way other than applying distributive law into 4 terms like into $$((\sim p \land \sim q) \lor p) \land ((\sim p \land \sim q) \lor q)$$ then another round of distributive law to simplify to $$((\sim p \lor p) \land (p \lor \sim q)) \land ((\sim p \lor q) \land (\sim q \lor q))$$, which feels very tedious to me especially since I might mess the ordering up since some of these properties might not be commutative. Does anyone have a solution to this problem?

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  • $\begingroup$ $p \iff q$ is true only when both $p$ and $q$ have the same truth values. It is fairly easy to see that option II satisfies that. $\endgroup$ – Anurag A Sep 28 '18 at 4:57
  • $\begingroup$ So is the general idea to get the possible set of truth values that the original statement allows, for this case (p,q) only can be (T,T) and (F,F) , then do a substitution test to see whether it evaluates as True? $\endgroup$ – Prashin Jeevaganth Sep 28 '18 at 5:02
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Using double distribution : \begin{align}(\lnot p \land \lnot q)\lor(p\land q) &\equiv (\lnot p \lor p) \land (\lnot p \lor q) \land (\lnot q \lor p) \land (\lnot q \lor q) \\ &\equiv (\lnot p \lor q) \land (\lnot q \lor p) \\ &\equiv p\iff q \end{align}.

Hope that help!

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  • $\begingroup$ That is what I wanted to avoid actually... $\endgroup$ – Prashin Jeevaganth Sep 28 '18 at 5:06
  • $\begingroup$ Only way I know how without using a logic table. Sometimes, this is consider as an equivalence for $\iff$. $\endgroup$ – Zamarion Sep 28 '18 at 5:12
  • $\begingroup$ I mean brute force is one thing, but I think I want to free up the carelessness factor during examination pressure $\endgroup$ – Prashin Jeevaganth Sep 28 '18 at 5:13
  • $\begingroup$ Yes, I understand, having done this course myself. Sometimes there kind of computation can get tedious, but this one is only three step. Anyway, sorry I couldn't help much. $\endgroup$ – Zamarion Sep 28 '18 at 5:18

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