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I have a question about bi conditional equivalencies and how to prove that they are logically equivalent. The question states, show that $\neg\;(p \leftrightarrow q)$ and $p \leftrightarrow\neg\; q$ are logically equivalent.

I tried multiple approaches using several laws from left side or right side but I can't find the right path that leads to the right answer. Any ideas? Please help

Thanks in advance!!!

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    Draw up a truth table? – 2017-01-15

4 Answers 4

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Recall that $p \def\if{\leftrightarrow}\def\imp{\rightarrow}\if q$ is equivalent to $(p\imp q) \land (q \imp p)$, hence $$ \neg (p\if q)\iff \neg(p\imp q) \lor \neg(q\imp p) $$ As $(p \imp q)$ is $\neg p \lor q$, we have $$ \neg(p\imp q) \iff p \land \neg q $$ therefore $$ \neg (p\if q) \iff (\neg p\land q)\lor (\neg q \land p) $$ distributivity of $\lor$ over $\land$ and $p \lor \neg p \iff 1$ gives $$ \neg (p\if q) \iff (\neg p \lor \neg q)\land (p\lor q)\iff (p\imp\neg q) \land(\neg q \imp p) $$ Therefore $$ \neg (p\if q) \iff (\neg q \if p) $$

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    Martini, the right side of the final answer should be (p↔¬q) an not (¬p↔q). I am a little confused. – 2017-01-16
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    Both are equivalent, corrected. – 2017-01-16
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$p \leftrightarrow q$ is true iff $p$ and $q$ have the same truth-value, and thus:

$\neg(p \leftrightarrow q)$ iff

$p$ and $q$ do not have the same truth-value iff

$p$ and $q$ have opposite truth-values iff

$p$ and $\neg q$ have the same truth-value iff

$p \leftrightarrow \neg q$

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One can quite easily create a truth table, because there are only four cases:

P Q P↔Q ¬(P↔Q) ¬Q P↔¬Q 
T T  T    F    F   F
T F  F    T    T   T
F T  F    T    F   T
F F  T    F    T   F

The fourth and sixth column are equal, proving the statement in your question.

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As Henning has said in the comments, the easiest way would be to draw up a truth table. If you are opposed to this though, then consider the fact that $p\def\if{\leftrightarrow}\def\imp{\rightarrow}\if q$ is logically equivalent to $$(\neg p\lor q)\land(\neg q\lor p)\dots(1)$$ (in first order logic this is actually how the biconditional symbol is defined in the metatheory). Thus $\neg (p\def\if{\leftrightarrow}\def\imp{\rightarrow}\if q)$ is logically equivalent to $$(p\land\neg q)\lor (q\land \neg p)$$ via De Morgan's Law. Using substitution this is equivalent to, $$((p\land\neg q)\lor q)\land ((p\land\neg q)\lor \neg p).$$ Using substitution again shows that this is equivalent to $$((q\lor\neg q)\land(q\lor p))\land ((p\lor \neg p)\lor (\neg q \lor \neg p)).$$ Finally applying the negation and identity laws shows that this is equivalent to $$(q\lor p)\land(\neg q \lor \neg p)$$ Now compare this with $p \def\if{\leftrightarrow}\def\imp{\rightarrow}\if \neg q$, which we obtain by substituting $\neg q$ in the place of $q$ in $(1)$.