No, Cornman, it's as simple as the associative property of set intersection. Ditto for $A\cup (B \cup C) \equiv (A\cup B) \cup C).$ In any case, you answered a dupe.

I dont understand. Associativity of intersection is to prove here? I know that this is a dube, but I still think that my answer helps the TS better. It is more personal.

Please see: en.wikipedia.org/wiki/…, @Cornman. I'm surprised you don't already understand this.

I still dont understand. You have to prove a rule to use it, no?

What should I understand? The linked article lists a bunch of of properties? So what? Still have to prove them.

You haven't proven anything: hint, try element chasing, in which one uses their knowledge of propositional logic, so that $(x \in A \land x \in B) \land x \in C \equiv x\in A \land (x\in B \land x \in C)$. This is not just a property; it is an identity. Your answer proves nothing. More so, because you never acknowledge associativity. No need to reinvent the wheel. I will delete all my comments here, if you actually use element chasing in a proof. No need to nuclear bomb fly; this question is tagged in Freshman level discrete math, and elementary set theory.

The associativity comes from logic.

I dont get what you are trying to say. Are you treating associativity of sets as some sort of axiom or definition? You have to prove it. This has nothing to do with "nuclear bomb fly" or what ever you are trying to imply with that. It is a property of sets that has to be proven. Check Section 4 of Halmos for example. He stats all the properties of your linked wikipedia article as "easily proved facts". I do not know what you are trying to say here. Honestly, you are just wrong.

That's precisely what I just said. Good you understand my previous comment. my comment above was to your "associativity comes from logic." Dah, but using element chasing in sets, like I did the my penultimate comment, that's why associativity holds in strict strings of intersecting sets, also in such a string of strictly unions of sets. I'm done here, Please prove this, as suggested, despite the fact that you've answered a very poor question with no research effort, and low quality. I came here only to suggest what an appropriate answer might look like. Take it or leave it.

No, you have not said that. The associativity of the logical operator $\wedge$ is also something that can be proven and is not some sort of axiom, or what ever you are trying to argue. I also dont understand what you mean with "You havent proven anything". My answer is supposed to give a guidline on what there is to do, and not a proof.

Yes I did: math.stackexchange.com/questions/4409368/…. Do I have to write out word for word the additional words for you? Associativity of \land, deconstructing sets into element chasing, appealing to $p\land (q \land \r)$ by associativity, is equivalent to $(p\land q) \land r$, so .... Thou doth protest too much, Cornman.

With associativity comes from logic, I meant that the statement $(A\wedge B)\wedge C$ is logically equivalent to $A\wedge (B\wedge C)$. And this can be proven and then comes into account in the prove of the set associativity. Yes. So what exactly is now wrong with my answer above?

Also it is very hard to argue with someone who always edits his comments, before my comment is finished.

I'm done here. @Cornman.

Thank you that you stop harassing me for your misunderstanding of my answer above, and the linked wikipedia article.

I edit out typos, period. Thats a courtesy. comments have five seconds of a window to edit. If you can't hold your horses for five seconds, before responding, that's your problem, not mine. Your last comment: Lol! Thanks for the laugh.

xD you didnt fix only typos, but added whole paragraphs to your comments.

Have a nice day xxxxxx .... oops, Cornman.

What ever bothers you at the moment, I wish you nothing but the best for your future. ^-^ Cheers.

Watching this comment chain unfold was very entertaining. Amwhy was clearly trolling in Cornman in this exchange and throws in the lie followed up by gaslighting in the end (always a kicker right before you leave a discussion). If this was not an elaborate troll, I fear for the safety of anyone in his immidiate vicinity.

This type of exchange is typical of amWhy. Don't take it too personally. I have been on the receiving end of similar exchanges myself from them in the past. If anything, I'm surprised they didn't lash out about using the wrong pronouns for them at you. In any event, both sides are correct about this... the answer here didn't prove anything, but that's because it didn't set out to... just to give a starting point for a proof. In the end, as with most things, definitions and axioms are a good place to start from.