# Are these two simple logic statements equivalent?

I've got two statements:

1. Set X always contains 3.
2. Set X set never contains not 3.

My question: Are these two statements logically equivalent?

I ask this question because of an argument between my brother... Maybe language plays a role.

• In set language you have "$X$ contains $3$", i.e. $3 \in X$ and "$X$ does not contain $3$", i.e. $3 \notin X$. Commented Mar 7, 2017 at 19:37
• "Not $3$" makes no sense; we have to negate the "verb" (contain) and not the "object" (the number 3). Commented Mar 7, 2017 at 19:37
• You could take not 3 as the complement of 3, so it never contains any number other than 3. This doesn't mean it contains 3 though. Commented Mar 7, 2017 at 19:40
• In addition to the above, what do you mean by "always" and "never" here? Is $X$ something that varies with time? Commented Mar 7, 2017 at 19:51

"Set $X$ always contains $3$," and "Set X never not contains 3" are equivalent. It is the property of dual negation of the modal quantifier. $$\Box\,(3\in X) ~\iff~ \neg \Diamond\, (3\notin X)$$
However, "Set X never contains not 3" is not the same thing at all.$$\neg\Diamond\,\exists x~(x\neq 3~\wedge~ x\in X)$$
Consider $X=\{3, 4\}$. This is plausible under each of the first two statements, but is implausible under the third.