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Given that:

$P \leftrightarrow Q$

How can I prove:

$\neg P \leftrightarrow \neg Q$

It's pretty obvious, but the best I can come up with is "well just look at the truth table, damn it". Is there a more formal way to show this? Is there a name for this rule?

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    In propositional logic; nothing is more fundamental than a truth table, because that is how things are defined. However some methods may be equally fundamental, and while a truth table is formal; often Authors will want to use something else because they do not consider them to be as formal. – 2017-01-30
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    Also, if you want an English description of "if and only if", you can just consider it to mean "When one is true both are true or when one is false both are false". From that it is quite obvious that negation of the atomic statements will mean the same thing. – 2017-01-30

2 Answers 2

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$P \leftrightarrow Q \Leftrightarrow$

$(P \rightarrow Q) \land (Q \rightarrow P) \Leftrightarrow$ (contraposition)

$(\neg Q \rightarrow \neg P) \land (\neg P \rightarrow \neg Q) \Leftrightarrow$

$\neg P \leftrightarrow \neg Q$

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    Is there a name for this equivalence? I found it useful in a proof that the compliments of two isomorphic graphs are themselves isomorphic. I used it without knowing how to cite it – 2017-01-30
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Assuming you are working in classical propositional logic, one way that does not rely on the full truth table of the equivalence, but still relies on some truth table, is as follows:

  • Decompose $P \iff Q$ in $(P \implies Q) \land (Q \implies P)$ and then show that for all propositions $P$ and $Q$, $(P \implies Q) \implies (\neg Q \implies \neg P)$.

  • Remark that $P \implies Q$ and $\neg P \lor Q$ have the same truth table (or you may have $P \implies Q$ defined as $\neg P \lor Q$). From this, $\neg \neg Q \lor \neg P \implies Q \lor \neg P$, since from the law of excluded middle you get $\neg \neg Q \implies Q$. This gives you $(P \implies Q) \implies (\neg Q \implies \neg P)$, which is the building block you needed.