Question

Show algebraically(using laws) that: ¬(p∧q)∨ p ≡ TRUE In your answer make sure that you (1) Show your steps (2) Do not skip any steps and (3) Justify each step with the name of the law being used.

Answer 1

We start by applying De Morgan's law:

$\neg (p \wedge q) \vee p \equiv [(\neg p) \vee (\neg q)] \vee p.$

We now use the fact that the disjunction is commutative:

$[(\neg p) \vee (\neg q)] \vee p \equiv p \vee [(\neg p) \vee (\neg q)].$

We now use the fact that the disjunction is distributive:

$p \vee [(\neg p) \vee (\neg q)] \equiv [p \vee (\neg p)] \vee (\neg q).$

Using the law of excluded middle, we have $p \vee (\neg p) \equiv \textrm{TRUE}$, so that:

$[p \vee (\neg p)] \vee (\neg q) \equiv \textrm{TRUE} \vee (\neg q).$

We finally use the fact that TRUE is the absorbing element of the disjunction:

$\textrm{TRUE} \vee (\neg q) \equiv \textrm{TRUE}.$