Discrete question. Prove that two propositions are logically equal. 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!!!
 A: 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
$$ \neg (p\if q) \iff (\neg q \if p) $$
A: $p \leftrightarrow q$ 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
$p \leftrightarrow \neg q$
A: 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.
A: 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 $(1)$. 
