Prove A-(BnC) = (A-B)U(A-C), explain with an example

Question

asked Jul 18, 2023 34.2k views

1 Answer

Valerybodak · Answer 1 · 2023-07-25T03:15:47+0000

Answer:

Prove set equality by showing that for any element
,
$x \in (A \backslash (B \cap C))$ if and only if
$x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Explanation:

Proof for
$[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element
:

Assume that
$x \in (A \backslash (B \cap C))$ . Thus,
$x \in A$ and
$x \\ot \in (B \cap C)$ .

Since
$x \\ot \in (B \cap C)$ , either
$x \\ot \in B$ or
$x \\ot \in C$ (or both.)

If
$x \\ot \in B$ , then combined with
$x \in A$ ,
$x \in (A \backslash B)$ .
Similarly, if
$x \\ot \in C$ , then combined with
$x \in A$ ,
$x \in (A \backslash C)$ .

Thus, either
$x \in (A \backslash B)$ or
$x \in (A \backslash C)$ (or both.)

Therefore,
$x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for
$[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that
$x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either
$x \in (A \backslash B)$ or
$x \in (A \backslash C)$ (or both.)

If
$x \in (A \backslash B)$ , then
$x \in A$ and
$x \\ot \in B$ . Notice that
$(x \\ot \in B) \implies (x \\ot \in (B \cap C))$ since the contrapositive of that statement,
$(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore,
$x \\ot \in (B \cap C)$ and thus
$x \in A \backslash (B \cap C)$ .
Otherwise, if
$x \in A \backslash C$ , then
$x \in A$ and
$x \\ot \in C$ . Similarly,
$x \\ot \in C \!$ implies
$x \\ot \in (B \cap C)$ . Therefore,
$x \in A \backslash (B \cap C)$ .