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If [math]P [/math] and [math]Q [/math] are propositions, then "[math]P [/math] or [math]Q [/math]" is a proposition (written [math]P \href{/cs2800/wiki/index.php/%E2%88%A8}{∨} Q [/math]); it is true if [math]P [/math] is true, or if [math]Q [/math] is true (or both).

Note that this is somewhat different from the use of "or" in colloquial English; if both [math]P [/math] and [math]Q [/math] are true, we still consider [math]P [/math] or [math]Q [/math] to be true. This saves us work: we can prove [math]P [/math] or [math]Q [/math] by just proving [math]P [/math]; we don't have to also disprove [math]Q [/math].

To prove "[math]P [/math] or [math]Q [/math]", you can either prove [math]P [/math], or you can prove [math]Q [/math] (your choice!)

If you know that "P or Q" is true for some statements P and Q, and you wish to show a third statement R, you can do so by separately considering the cases where P is true and where Q is true. If you are able to prove R in either case, then you know that R is necessarily true.

This technique is often referred to as case analysis.

To disprove "[math]P [/math] or [math]Q [/math]", you must both disprove [math]P [/math] and disprove [math]Q [/math]. Put another way, the logical negation of "[math]P [/math] or [math]Q [/math]" is "not [math]P [/math] and not [math]Q [/math]".