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I want to know why $p \leftrightarrow q$ is equivalent to $(p \wedge q) \vee (\neg p \wedge \neg q)$? Without using the truth table.

Thanks all

Clive Newstead
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user2849967
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    I've formatted your post in $\LaTeX{}$ and changed 'equal' to 'equivalent' (the distinction matters!) – Clive Newstead Oct 27 '13 at 16:36
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    By distributivity $(p\land q)\lor (\neg p\land\neg q)$ is equivalent to $(p\lor \neg q)\land (p\lor\neg q)\land (\neg p\lor q)\land (q\lor \neg q)$. You can drop $p\lor\neg p$ and $q\lor \neg q$ and observe that $p\lor\neg q$ is $q\to p$, $\neg p\lor q$ is $p\to q$, hence we have $(p\to q)\land (q\to p)$, i.e. $p\leftrightarrow q$ – Hagen von Eitzen Oct 27 '13 at 16:37
  • @Hagen thank you, I like your proof but I think there is a little mistake :) – user2849967 Oct 27 '13 at 17:29

2 Answers2

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If $p$ is true then $p\iff q$ tells us that $q$ is true as well. Also if $q$ is true then $p\iff q$ tells us that $p$ is true as well. So it cannot be that exactly one of them is true. They are both true or both not true.

drhab
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Just think about the statement.

$p \leftrightarrow q$: This says that $p$ occurs only if $q$ occurs and that $q$ happens only if $p$ does. Meaning, that either they both happen or nothing happens at all.

But look at my last sentence. They BOTH happen OR NEITHER happens. They both happen is $p \wedge q$. They both DON'T happen is $\neg p \wedge \neg q$. So either they both happen or they both don't is $$(p \wedge q) \vee (\neg p \wedge \neg q)$$

mathematics2x2life
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