# Discrete Maths Logically Equivalent [closed]

Just have a question on logically equivalency specifically the logical equivalency. Question find a simple expression using $p$ and $q$, logically equivalent to $q \iff(\neg p\lor \neg q)$.

Im not really sure but could it be: $p \iff (\neg q \land p)$?

Thanks.

## closed as off-topic by 5xum, skyking, user91500, JMP, Claude LeiboviciAug 29 '17 at 12:19

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, skyking, user91500, JMP, Claude Leibovici
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• Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Aug 29 '17 at 7:13
• Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 5xum Aug 29 '17 at 7:13
• Sorry didn't input it correctly,its an OR symbol. – John Aug 29 '17 at 7:17
• Undestood. But you still need to adress the concernst from the rest of my comments. – 5xum Aug 29 '17 at 7:18
• What have you tried to solve this? Are you just guessing whether it could be $p\iff (\neg q \land p)$? I hope not - you must have a reason to believe so. Also please use MathJAX - we've helped you a bit getting started by editing your question, but there is a tutorial: math.meta.stackexchange.com/questions/5020/… – skyking Aug 29 '17 at 7:24

No, that's not correct. A systematic way to do these kind of things is to use truth tables:

$$\begin{matrix} p & q & q\iff(\neg p\lor \neg q)\\ \hline\\ f & f & f \\ f & T & T \\ T & f & f \\ T & T & f \\ \end{matrix}$$

which we see that is $\neg p\land q$.

Or put in text: if you consider the case where $q$ is false then $\neg p \lor \neg q$ is obviously true so for the equivalence to hold $q$ must be true. Similarily we see that if $q$ is true then so must $\neg p\lor \neg q$ and $\neg q$ is false so $\neg p$ is true, that is $p$ is false. This means $q\land\neg p$.

Your guess however if $q$ is true then $\neg q$ is false and therefore so is $\neg q\land p$ which fails at being equivalent to $p$ if $p$ is true. Which also can be seen using truth table:

$$\begin{matrix} p & q & p\iff(\neg q \land p)\\ \hline\\ f & f & T \\ f & T & T \\ T & f & f \\ T & T & T \\ \end{matrix}$$

You have two options to solve this:

Option $1$: write a truth table. That is, fill the following table:

$$\begin{array}{c|c|c} p & q & \neg p & \neg q & (\neg p \lor \neg q) & q\iff(\neg p \lor\neg q)\\\hline T&T&&&\\ T&F\\ F&T\\ F&F \end{array}$$

Then, use the first two columns and the last column to come up with a reasonable explanation. For example, if you get something like

$$\begin{array}{c|c|c} p & q & q\iff(\neg p \lor\neg q)\\\hline T&T& T\\ T&F&F\\ F&T&F\\ F&F&F \end{array}$$

then you know the answer is $p \land q$.

Option 2:

Use known identities. For example, use:

$$x\iff y\text{ is the same as }(x\land y)\lor(\neg x\land \neg y)$$

to transform your statement into something involving only $\neg, \lor, \land$.

Then, use DeMorgan's laws, the law of double negation, and distributivity of $\land$ and $\lor$ to simplify the expression.