Proofs question

Question

I have a question about set theory that and proofs that I was hoping you could help with. The goal is the prove or disprove that:

$A\setminus (A \cap B) = A\setminus B$

So far I have:

$A\setminus (A \cap B)$ is equivalent to $A \cap (A \cap B)^\mathsf{c}$

From De Morgan's law, we can derive that this is equivalent to: $A \cap A^\mathsf{c} \cup B^\mathsf{c}$

Or, if my logic is correct: $\varnothing \cup B^\mathsf{c}$

This is where I am getting stuck, as I'm not sure where to go from here. Thank you very much.

It si not $ \emptyset \cup B^c$. It is $\emptyset \cup (A\cap B^{c})$ — geetha290krm, Dec 05 '22 at 23:47
Overthinking. A simpler approach is that to prove that set $R$ equals set $S$, you let $x$ be any element in $R$ and you let $y$ be any element in $S$. Then you prove that $x$ is an element of $S$ and that $y$ is an element of $R$. Collectively, you have then proven that $$\left{~R \subseteq S ~~\text{and}~~ S \subseteq R ~\right} \implies R = S.$$ — user2661923, Dec 05 '22 at 23:55
"From demorgans law, we can derive that this is equivalent to: A ∩ $A^c$ ∪ $B^c$" - This is not quite correct: you should get $A\cap(A^\mathsf{c}\cup B^\mathsf{c})$, with parentheses. — Akiva Weinberger, Dec 06 '22 at 00:24
Lessee without jumping the parenthesis. $A\cap (A\cap B)^c = A\cap (A^c\cup B^c)$. We do NOT have any rule that says $M\cap (N\cup P)=(M\cap N)\cup P$. That is false. We have someoneorother's rule though that $M\cap(N\cup P)=(M\cap N)\cup (M\cap N)$ though. SO continuing $(A\cap A^c)\cup (A\cap B^c)=\emptyset \cup (A\cap B^c) = A\cap B^c$ and... that's that.... $A\cap B^c = A\setminus B$. — fleablood, Dec 06 '22 at 02:58
@user2661923 Although I agree the element chasing is easier for these that is no reason to shun formal laws and derivation. If you do a simple (and it is simple) derivation it is important to know where the error was. The OP was doing a great job until they "jumped the parenthesis gun" and wrote $A\cap (A\cap B)^c$ as (without parenthesis) $A\cap A^c \cup B^c$ and wrongly interpretted this as $(A\cap A^c)\cup B^c$. We owe it to the OP to point out where the simple (but fatal) error lay. — fleablood, Dec 06 '22 at 03:03
@fleablood I agree that element chasing totally ignores the OP's question. This is why I would not post my remark as an answer. However, I personally don't see the utility in developing skill at navigating through (even moderately) complicated set theory expressions. The alternative approach of simply using a table to examine (for example) the $2^n$ possibilities when $n$ sets are involved is so foolproof and simple, that unraveling complicated set theory expressions seems pointless. ...see next comment — user2661923, Dec 06 '22 at 03:54
@fleablood My rule of thumb is that if I immediately see an easy way to unravel the set theory expression, then that is what I will do. Otherwise, I will immediately turn to one form of element chasing or another. Either trying to show for example that $R \subseteq S$ and $S \subseteq R$ or constructing a table with $2^n$ entries, and plowing through them, quickly and efficiently. This is what I intended by the statement : "Overthinking". That is, for this particular OP, at this point in his developing intuition, I regard his method of attack as inferior. — user2661923, Dec 06 '22 at 03:58

RyRy the Fly Guy · Accepted Answer · 2022-12-05T23:59:21.433

$A - (A \cap B)$

by definition of set difference

$\Leftrightarrow A \cap (A \cap B)^C$

by DeMorgan's Law

$\Leftrightarrow A \cap (A^C \cup B^C)$

I believe this is where you got stuck in your proof. From here, you incorrectly derived $\emptyset \cup B^C$ because you dropped the parantheses after applying DeMorgan's Law, computing $A \cap A^C$. However, the parantheses keeps $A^C \cap B^C$ together. You proceed by applying the distributive law...

$\Leftrightarrow (A \cap A^C) \cup (A \cap B^C)$

by negation law

$\Leftrightarrow \emptyset \cup (A \cap B^C)$

by identity law

$\Leftrightarrow A \cap B^C$

by definition of set difference

$\Leftrightarrow A-B$

Therefore, $A - (A \cap B) \Leftrightarrow A-B$

Proofs question

1 Answers1