Propisitional logic exam questions and answers I'm going over exam questions, since my exam is hours away. I'd be extremely grateful if you could check out my answers and evaluate them.


Hopefully you guys can see the truth table. Also, i have a question for one of you people who know this type of logic. If i come across a formula that has a not outside the bracket and then e.g. not p how do i do the truth table for this, not not ? does it go back to its regular value?
 A: $\rm a)$ A quick check using W|A shows that your truth table is indeed correct.
$\rm b)$ The question asks you to formulate certain proposition only in terms of certain connectives which are $\lnot$ and $\lor$. Obviously, the way one should attack this problem is to recall a theorem which links for instance $p\land q$ to $p\lor q$. You may recall that $$\neg (p\land q)\equiv \neg p\lor\neg q,$$ taking the negation of both sides yields, $$\neg(\neg (p\land q))\equiv p\land q\equiv \neg(\neg p\lor\neg q),$$ and you're done! The same thing for $p\Rightarrow q$, except that now it will be really obvious knowing that $p\Rightarrow q\equiv \lnot p\lor q$.
$\rm c)$ The second one is correct, however the first one isn't. Consider the situation in which the sentence “When the front and back doors are closed” $($ i.e. $(p\land q)$ is false$)$ and when the light is turned off $(r$ is true$)$, then $r\Rightarrow(p\lor q)$ will be false, however looking at the original statement “When the front and back doors are closed then the light is off”, we expect it to be incorrect only when the front and back doors are actually closed, but the light is turned on. If we even look at the structure of the sentence “When the front and back doors are closed then the light is off”, we see that it has the form “When $P$ then $Q$”, this is clearly the same as “If $P$ then $Q$”, so the correct formulation of that sentence would rather be $(p\land q)\Rightarrow r$.

If I come across a formula that has a not outside the bracket and then e.g. $ \lnot p$ how do I do the truth table for this, not not?

If it's something like $\lnot(\lnot p)\land q$ then you may use the fact that $\lnot(\lnot p)\equiv p$, so it would be unnecessary to make another column containing the possible truth values for $\lnot p$. However if it's something like $\lnot(\lnot p\lor q)\land r$, then you may either write it as $(\lnot(\lnot p)\land\lnot q)\land r$, and then use 
$\lnot(\lnot p)\equiv p$, but you'll then have to make a specific column for $\lnot q$, or you may instead make a column for $\lnot p$ and then use it to find $(\lnot p\lor q)$, and then $\lnot(\lnot p\lor q)$, ...etc.
