Proof:A ∩ (B∪C) ⊆ (A∩B) ∪ (A∩C)

From CS2800 wiki
[math]A \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} (B\href{/cs2800/wiki/index.php/%E2%88%AA}{∪}C) \href{/cs2800/wiki/index.php/%E2%8A%86}{⊆} (A\href{/cs2800/wiki/index.php/%E2%88%A9}{∩}B) \href{/cs2800/wiki/index.php/%E2%88%AA}{∪} (A\href{/cs2800/wiki/index.php/%E2%88%A9}{∩}C) [/math]
Proof:
Choose arbitrary sets [math]A [/math], [math]B [/math], and [math]C [/math]. We want to show to show that every [math]x \href{/cs2800/wiki/index.php/%E2%88%88}{∈} A \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} (B \href{/cs2800/wiki/index.php/%E2%88%AA}{∪} C) [/math] is also in [math](A \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} B) \href{/cs2800/wiki/index.php/%E2%88%AA}{∪} (A \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} C) [/math]. Choose an arbitrary [math]x [/math] in the left hand side. Then [math]x ∈ A [/math] and [math]x ∈ B \href{/cs2800/wiki/index.php/%E2%88%AA}{∪} C [/math]. This means either (1) [math]x ∈ B [/math] or (2) [math]x ∈ C [/math]. In case (1), we have [math]x \href{/cs2800/wiki/index.php/%E2%88%88}{∈} A [/math] and [math]x \href{/cs2800/wiki/index.php/%E2%88%88}{∈} B [/math], so [math]x \href{/cs2800/wiki/index.php/%E2%88%88}{∈} A \href{/cs2800/wiki/index.php/%E2%88%A9}{∩} B [/math], and thus [math]x \href{/cs2800/wiki/index.php/%E2%88%88}{∈} \href{/cs2800/wiki/index.php/RHS}{RHS} [/math]. Case (2) is similar. In either case, [math]x \href{/cs2800/wiki/index.php/%E2%88%88}{∈} RHS [/math], as required.