1. ## Intersection/Union Proofs

Let A, B, C be sets.

Here's my attempt:

Suppose $x \in A - (B \cap C)$ then $x \in A$ and $x \notin B$ and $x \notin C$

So, x is in A not B or C, therefore (is this explanation sufficient?):

$x \in (A-B) \cup (A-C)$

$
A - (B \cap C) \subseteq (A-B) \cup (A-C)
$

Conversely, let $x \in (A-B) \cup (A-C)$, which means $x \in A$ and $x \notin B$ or $x \in A$ and $x \notin C$

(I'm not sure what to explain here!)

Therefore, $x \in A - (B \cap C)$

Hence: $(A-B) \cup (A-C) \subseteq A - (B \cap C)$

I appreciate it if anyone could help me with this. Thanks.

2. $A - (B \cap C) = A \cap (B \cap C)^{c} = A \cap (B^{c} \cup C^{c}) = (A \cap B^{c}) \cup (A \cap C^{c}) = (A-B) \cup (A-C)$

3. Originally Posted by Roam
Let A, B, C be sets.

Here's my attempt:

Suppose $x \in A - (B \cap C)$ then $x \in A$ and $x \notin B$ and $x \notin C$

So, x is in A not B or C, therefore (is this explanation sufficient?):

$x \in (A-B) \cup (A-C)$

$
A - (B \cap C) \subseteq (A-B) \cup (A-C)
$

Conversely, let $x \in (A-B) \cup (A-C)$, which means $x \in A$ and $x \notin B$ or $x \in A$ and $x \notin C$

(I'm not sure what to explain here!)

Therefore, $x \in A - (B \cap C)$

Hence: $(A-B) \cup (A-C) \subseteq A - (B \cap C)$

I appreciate it if anyone could help me with this. Thanks.
You gotta be careful about those parentheses, that left set means x is in A, but not in both B and C. It could be in B but not C and still be in $A-(B\cap C)$. You should probably break it into three cases:x in A and
1) in neither B nor C. $\Rightarrow x \in A-B$ and $x \in A-C$
2) in B but not C $\Rightarrow x \in A-C$
3) in C but not B $\Rightarrow x \in A-B$
In all three cases we see it is in the set on the right.

See if you can go the other way, hopefully now that you understand what that left set actually is, you can get it. Otherwise the proof above me is spot on, but thought I would go through the proof the way you were trying to do it.

4. Hello, Roam!

Let $A, B, C$ be sets.

Show that: . $A - (B \cap C) \;=\;(A - B) \cup (A - C)$

. . $\begin{array}{ccc}
A - (B \cap C) && \text{Given} \\ \\
A \cap \overline{(B \cap C)} && \text{d{e}f. Subtraction} \\ \\
A \cap (\overline{B} \cup \overline{C}) && \text{DeMorgan's Law} \\ \\
(A \cap \overline B) \cup (A \cap \overline C) && \text{Distributive} \\ \\
(A - B) \cup (A - C) && \text{d{e}f. Subtraction}
\end{array}$