# Specific question about this set theory proof **URL:** \ **Category:** Factual Questions\ **Created:** [September 22, 2015, 2:12pm UTC](https://boards.straightdope.com/t/specific-question-about-this-set-theory-proof/732206 "2015-09-22T14:12:13Z")\ **Posts on this page:** 6\ **Page:** 1
**Author:** ![D18](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@D18](https://boards.straightdope.com/u/D18)\ **Post date:** [September 22, 2015, 2:12pm UTC](https://boards.straightdope.com/t/specific-question-about-this-set-theory-proof/732206/1 "2015-09-22T14:12:13Z")
I actually follow [this proof](https://www.easycalculation.com/theorems/associative-law.php), but have a question about the endgame. Why can’t mission accomplished be declared at this line: x ∈ A ∩ (B ∩ C) (third line above equation 1) At that point, x has been shown to be in both (A ∩ B) ∩ C and A ∩ (B ∩ C), so I would have thought one could write therefore (A ∩ B) ∩ C = A ∩ (B ∩ C). QED. Instead, the proof goes on to show that (A ∩ B) ∩ C and A ∩ (B ∩ C) are subsets of each other, and therefore are equal. (Yes, I understand this, so no need to explain why this is so.) I just don’t see why it is necessary. ---
**Author:** ![TATG](https://avatars.discourse-cdn.com/v4/letter/t/50afbb/32.png) [@TATG](https://boards.straightdope.com/u/TATG)\ **Post date:** [September 22, 2015, 2:21pm UTC](https://boards.straightdope.com/t/specific-question-about-this-set-theory-proof/732206/2 "2015-09-22T14:21:56Z")
> [@D18](#): > > At that point, x has been shown to be in both (A ∩ B) ∩ C and A ∩ (B ∩ C) Not quite. x has shown to be in (A ∩ B) ∩ C and A ∩ (B ∩ C) _on the assumption_ that x is in (A ∩ B) ∩ C. (Thus the next step is saying that _if_ x is in (A ∩ B) ∩ C, then it is in A ∩ (B ∩ C).) ---
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [September 22, 2015, 3:09pm UTC](https://boards.straightdope.com/t/specific-question-about-this-set-theory-proof/732206/3 "2015-09-22T15:09:10Z")
Remember equality is read left to right and right to left. The first part shows that if x starts out in (A ∩ B) ∩ C then it is also in A ∩ (B ∩ C) but that does nothing to show us that if it is in A ∩ (B ∩ C) that it is also in (A ∩ B) ∩ C. In other words, if (A ∩ B) ∩ C is a proper subset of A ∩ (B ∩ C), the first part of the proof works but the two sets are not equal. ---
**Author:** ![D18](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@D18](https://boards.straightdope.com/u/D18)\ **Post date:** [September 22, 2015, 5:14pm UTC](https://boards.straightdope.com/t/specific-question-about-this-set-theory-proof/732206/4 "2015-09-22T17:14:31Z")
Gotcha ya. Thank you. When I did my relearning of high-school math, in the simple proofs that were covered, it was enough simply to get some manifestation of the LHS to equal some manifestation of the RHS without working backwards. ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [September 22, 2015, 10:27pm UTC](https://boards.straightdope.com/t/specific-question-about-this-set-theory-proof/732206/5 "2015-09-22T22:27:25Z")
> [@D18](#): > > Gotcha ya. Thank you. When I did my relearning of high-school math, in the simple proofs that were covered, it was enough simply to get some manifestation of the LHS to equal some manifestation of the RHS without working backwards. But step 1 did not show they were equal; it showed only one inclusion. It would be like showing that x \<= y. You would then show that y \<= x and conclude x = y. ---
**Author:** ![Senegoid](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/senegoid/32/6606_2.png) [@Senegoid](https://boards.straightdope.com/u/Senegoid)\ **Post date:** [September 22, 2015, 10:28pm UTC](https://boards.straightdope.com/t/specific-question-about-this-set-theory-proof/732206/6 "2015-09-22T22:28:00Z")
Why don’t they teach set theory like algebra, starting with some basic axioms and working up from there? Set theory has almost the same basic axioms as conventional algebra: 1. Intersection is commutative: A ∩ B = B ∩ A 2. Intersection is associative: (A ∩ B) ∩ C = A ∩ (B ∩ C) 3. Union is commutative: A ∪ B = B ∪ A 4. Union is associative: (A ∪ B) ∪ C = A ∪ (B ∪ C) Ah, but which is distributive over the other? This gets interesting. They are BOTH distributive over the other! 1. Intersection is distributive over union: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) 2. Union is distributive over intersection: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) These rules exactly parallel the logical operations of conjunction and disjunction (I can never remember which is which), if you just replace ∩ with ∧ and ∪ with ∨.