0
$\begingroup$

So I made a truth table, and found that the simplest proposition is

p <-> q

My question is whether it is certain this is the simplest form, because nothing else could fit that truth table simpler than the biconditional right?

Also is ¬p ↔ ¬q the contrapositive of p <-> q>

I feel like this is really obvious, but I don't have enough knowledge to know if I'm being ignorant.

$\endgroup$
1
$\begingroup$

It kind of depends on what the question means by 'simplest', but in terms of number of symbols, you cannot do better: the truth of the statement clearly depends on $p$: if $p$ is true, then you get a different truth-value for the statement than when $p$ is false. Same goes for $q$. So, the statement needs both a $p$ and a $q$ ... and therefore also at elast one operator. So, your statement, that has a single $p$, a single $q$, and one operator is 'simplest' in terms of number of symbols.

An equally 'simple' statement would be $q \leftrightarrow p$ of course.

Finally, is $\neg p \leftrightarrow \neg q$ the contrapositive of $p \leftrightarrow q$? Well, it's a little unusual to talk about contrapositives of biconditionals, as typically this is used for one-way conditionals: the contrapositive of $p \rightarrow q$ is $\neg q \rightarrow \neg p$. But, as such, if there is such a thing as a contrapositive of a bi-conditional, I would expect it to be $\neg q \leftrightarrow \neg p$ ... although that is of course equivalent to $\neg p \leftrightarrow \neg q$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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