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)) $


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

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." – 5xum, Claude Leibovici, Shailesh, José Carlos Santos, J. M. is a poor mathematician
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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

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:

$q \land \neg p$

  • $\begingroup$ Thanks so so so much, this makes perfect sense. $\endgroup$ – Derpm Aug 29 '17 at 9:54

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.

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.