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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.

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

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

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

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

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

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

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

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  • $\begingroup$ 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? $\endgroup$ – Sean2148 May 7 '18 at 21:35
  • $\begingroup$ 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). $\endgroup$ – Graham Kemp May 7 '18 at 22:57
  • $\begingroup$ @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 $\endgroup$ – Bram28 May 8 '18 at 14:22

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