From CS2800 wiki
Revision as of 13:52, 10 February 2018 by {{GENDER:Mdg39|[math]'"2}} [/math]'"7
(<math>1) </math>2 | <math>3 (</math>4) | <math>5 (</math>6)

If [math]P [/math] and [math]Q [/math] are propositions, then "[math]P [/math] and [math]Q [/math]" is a proposition (written [math]P \href{/cs2800/wiki/index.php/%E2%88%A7}{∧} Q [/math]); it is true if both [math]P [/math] is true and [math]Q [/math] is true.

To prove "[math]P [/math] and [math]Q [/math]", you can separately prove [math]P [/math] and then prove [math]Q [/math].

If you have already proved (or assumed) [math]P [/math] and [math]Q [/math], you can conclude [math]P [/math]. You can also conclude [math]Q [/math].

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