Prove: $A\cap (B - C) = (A \cap B) - (A \cap C)$

Question

Prove: $A \cap (B - C) = (A \cap B) − (A \cap C)$

I can understand this using Venn Diagrams, however I am struggling to translate this into a formal proof.

Answer 1

The most common strategy for proving that sets $S=T$ is to show that $S\subset T$ and $T\subset S$. And the most common strategy for proving $S\subset T$ is to prove that $(x\in S) \implies (x\in T)$ for any element $x$ in the universe of the problem. So let's do that!

First, let $x\in A\cap (B-C)$ be given. This means that $x\in A$ and $x\in B-C$, meaning that $x\in B$ and $x\notin C$. In other words, $x\in A\cap B$ but $x\notin A\cap C$, so $x\in (A\cap B)-(A\cap C)$.

Conversely, let $x\in(A\cap B)-(A\cap C)$ be given. Then $x\in A\cap B$ and $x\notin A\cap C$. Since $x\in A\cap B$, $x\in A$ and $x\in B$. But $x\notin A\cap C$, so it must be that $x\notin C$. Since $x\in B$ and $x\notin C$, $x\in B-C$. Therefore, $x\in A\cap (B-C)$.

Answer 2

To show that two sets are equal you normaly show the two relations $\subseteq$ and $\supseteq$.

This is how it is done, where I write it down explanatory and not how I would do so in a mathematical proof.

So we first want to show that $A\cap (B-C)\subseteq (A\cap B)-(A\cap C)$

So let $x\in A\cap (B-C)$. By definition of $\cap$ this means that $x\in A$ and $x\in B-C$. By definition of $-$ we have that $B-C$ implies $x\in B$ and $x\notin C$.

Now we have $x\in A$ and $x\in B$. So $x\in A\cap B$. (By definitino of $\cap$) Since $x\in A$ and $x\notin C$, it is $x\notin A\cap C$. (By definition of $\cap$)

So $x\in (A\cap B)-(A\cap C)$ (By defintion of $-$)

Now show $A\cap (B-C)\supseteq (A\cap B)-(A\cap C)$.

Start like this: Let $x\in (A\cap B)-(A\cap C)$. Then ...

The proof is done (exactly) like above.

Answer 3

$$A \cap (B-C) = A \cap (B \cap C^\complement)$$ while

$$(A\cap B) - (A \cap C) = (A \cap B) \cap (A \cap C)^\complement = \\ (A \cap B) \cap (A^\complement \cup C^\complement) (\text{ de Morgan) } =\\ (A \cap B \cap A^\complement) \cup (A \cap B \cap C^\complement) = A \cap B \cap C^\complement$$

so we have equality (the first component of the final union is empty).

Answer 4

Here it is an approach based on the definition of difference and the De Morgan Laws:

\begin{align*} (A\cap B) - (A\cap C) & = (A\cap B)\cap(\overline{A\cap C}) = (A\cap B)\cap(\overline{A}\cup\overline{C})\\ & = (A\cap B\cap\overline{A})\cup(A\cap B\cap\overline{C})\\ & = A\cap B\cap\overline{C} = A\cap(B - C) \end{align*}