# Find a simplest proposition that is logically equivalent to ¬p ↔ ¬q (don't need a solution,need explanation).

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.

## 1 Answer

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 $$\neg p \leftrightarrow \neg q$$.