Answer to Question #346072 in Discrete Mathematics for Nam Sae

Question

Show that (~𝒑∨𝒒)∧(𝒑∧~𝒒) is a contradiction.

Answer 1

$(\sim p\lor q)\land (p\land \sim q)$

$(\sim p\lor q)\land p\land \sim q$ , associative law

$\big((\sim p\land p)\lor(q\land p)\big)\land \sim q$ , distributive law

$(F\lor (q\land p))\land \sim q$ , negation law

$(q\land p)\land \sim q$ , identity law

$(q\land \sim q)\land p$ , associative law

$F\land p$ , negation law

$F$ , domination law

It is a contradiction