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The question asks me to find a simpler statement that is logically equivalent with $ q \iff (\neg p \lor ¬q)$. I believe this should be $\neg p \land q$ and am required to prove it with logic laws. I'm really struggling with find the right logic laws that take me to the answer.

This is what I have currently:

Starting with LHS: $ q \iff (\neg p \lor ¬q) \equiv \neg p \land q $

Then, using a known equivalence:

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

Then following De Morgan's Laws

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

Then the double negation law

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

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closed as off-topic by 5xum, Claude Leibovici, Shailesh, José Carlos Santos, J. M. is a poor mathematician Aug 30 '17 at 0:02

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  • $\begingroup$ This question is suspiciously similar to math.stackexchange.com/questions/2409613/…... $\endgroup$ – 5xum Aug 29 '17 at 8:09
  • $\begingroup$ @5xum just much worse in terms of question-writing skills $\endgroup$ – Kenny Lau Aug 29 '17 at 8:10
  • $\begingroup$ @KennyLau Not really, if you saw the original version of the linked question... $\endgroup$ – 5xum Aug 29 '17 at 8:11
  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Aug 29 '17 at 8:21
  • $\begingroup$ Very encouraging to a new user, thanks guys. I apologise for not using Math Jax. I'm guessing the other person and I are probably trying to finish the same assignment at the last minute. I will work through your solutions over there and come back here if I'm still stuck. $\endgroup$ – Derpm Aug 29 '17 at 8:23
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I can continue where you left off:

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

Then you use complement law (ie $\neg q \land q \equiv \bot)$ together with associative and commutative law:

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

And anihilation and identity (ie $\bot\land p\equiv\bot)$ and $\phi\lor\bot\equiv\phi$):

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

Distributive law:

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

And finally complement and identity:

$q \land \neg p$

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  • $\begingroup$ Thanks so so so much, this makes perfect sense. $\endgroup$ – Derpm Aug 29 '17 at 9:54
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Here is a hint in the form of an analytic tableau:

enter image description here.

You start with the negation of $$(q\leftrightarrow(\lnot p\lor \lnot q))\leftrightarrow(\lnot p\land q)$$ then proceed to use a few contradiction-hunting rules to eliminate all possibilities.

Each path ends in a contradiction, so the tableau is closed and thus the result follows.

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  • $\begingroup$ I'm sorry but I don't really understand how that works at all $\endgroup$ – Derpm Aug 29 '17 at 8:57
  • $\begingroup$ Perhaps this will help. $\endgroup$ – Shaun Aug 29 '17 at 8:59

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