# Biconditional Proposition using Identities [closed]

I'm revising for a Mathematics exam for my Computer Science degree and found this question on a past paper. I've looked at the propositional logic laws and cannot see how they can be applied to the propositions to show that the 2 statements are equal. I would attempt to answer this question but I honestly have no idea where to start with this. I'd appreciate any help with this question and on answering questions similar to this.

## closed as off-topic by Graham Kemp, B. Mehta, Xander Henderson, HK Lee, Claude LeiboviciMay 8 '18 at 11:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "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." – Graham Kemp, B. Mehta, Xander Henderson, HK Lee, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

$$p \leftrightarrow q \overset{Definition}\equiv$$

$$(p \land q) \lor (\neg p \land \neg q) \overset{Distribution}\equiv$$

$$(p \lor \neg p) \land (p \lor \neg q) \land (q \lor \neg p) \land (q \lor \neg q) \overset{Complement} \equiv$$

$$T \land (p \lor \neg q) \land (q \lor \neg p) \land T \overset{Identity}\equiv$$

$$(p \lor \neg q) \land (q \lor \neg p)\overset{Commutation}\equiv$$

$$(p \lor \neg q) \land (\neg p \lor q)$$

• Sorry should have included this earlier, but these are the rules of which can be used: snag.gy/O5SUAL.jpg - is it still possible using these laws? – Sean2148 May 7 '18 at 21:35
• Yes you should have mentioned the usable laws earlier. You should also not use an anonymised image server which may be blocked by some internet providers (eg mine). – Graham Kemp May 7 '18 at 22:57
• @Sean2148 Yes.When I do Distribution in one step you may want to do it in two: $(p \land q) \lor (\neg p \land \neg q) \overset{Distribution Or}\equiv ((p \land q) \lor \neg p) \land ((p \land q) \lor \neg q) \overset{Distribution Or}\equiv (p \lor \neg p) \land (p \lor \neg q) \land (q \lor \neg p) \land (q \lor \neg q)$ ... what I call Complement is Simplification Or, and what I call Identity is Simplification And – Bram28 May 8 '18 at 14:22