Equivalency of these two logical statements

Question

I need some help figuring out how these two logical statements are equivalent.

p <--> ~q == (~p && q) OR (p && -q)

I made a truth table and found that they are no equivalent BUT they are suppose to be equivalent. I tried to break both of the statements down but I'm kind of stuck here. If someone could help me out, that'd be great!

Thank you in advance.

Edit: I have to prove this using the laws of equivalence. I have attached it below. Laws of equivalence

Answer 1

Hint

To prove it with equivalences, we have to start "unpacking"" the LHS :

$(p \to \lnot q) \land (\lnot q \to p)$

and then apply Material implication to get :

$(\lnot p \lor \lnot q) \land (q \lor p)$.

Answer 2

Let's walk through how to build this truth table. I will use $\wedge$ to mean "and," $\vee$ to mean "or." To start with, we simply need all the possible combinations of truth values of $p$ and $q,$ giving us: $$\begin{array}{c|c}p & q\\\hline T & T\\T & F\\F & T\\F & F\end{array}$$

Next, it's rather simple to determine the truth values of $\sim p$ and $\sim q$: $$\begin{array}{c|c|c|c}p & q &\sim p & \sim q\\\hline T & T & F & F\\T & F & F & T\\F & T & T & F\\F & F & T & T\end{array}$$

Now, $p\wedge\sim q$ will be true precisely when both $p$ and $\sim q$ are true, giving us: $$\begin{array}{c|c|c|c|c}p & q &\sim p & \sim q & p\wedge\sim q\\\hline T & T & F & F & F\\T & F & F & T & T\\F & T & T & F & F\\F & F & T & T & F\end{array}$$

Similarly, we have:

$$\begin{array}{c|c|c|c|c|c}p & q &\sim p & \sim q & p\wedge\sim q &q\wedge\sim p\\\hline T & T & F & F & F & F\\T & F & F & T & T & F\\F & T & T & F & F & T\\F & F & T & T & F & F\end{array}$$

Now, $(p\wedge\sim q)\vee(q\wedge\sim p)$ will be true whenever at least one of $p\wedge\sim q$ and $q\wedge\sim p$ is true, and so we have:

$$\begin{array}{c|c|c|c|c|c|c}p & q &\sim p & \sim q & p\wedge\sim q &q\wedge\sim p & (p\wedge\sim q)\vee(q\wedge\sim p)\\\hline T & T & F & F & F & F & F\\T & F & F & T & T & F & T\\F & T & T & F & F & T & T\\F & F & T & T & F & F & F\end{array}$$

Finally, $p\iff\sim q$ is true precisely when $p$ and $\sim q$ are either both true or both false, so we have:

$$\begin{array}{c|c|c|c|c|c|c|c}p & q &\sim p & \sim q & p\wedge\sim q &q\wedge\sim p & (p\wedge\sim q)\vee(q\wedge\sim p)& p\iff\sim q\\\hline T & T & F & F & F & F & F & F\\T & F & F & T & T & F & T & T\\F & T & T & F & F & T & T & T\\F & F & T & T & F & F & F & F\end{array}$$

If you have questions about any of these steps, please let me know, and I'll try to elaborate. Hopefully, seeing where your truth table differs will help you to figure out what went wrong.