Logically equivalence

Question

I was asked to show that these two are logically equivalent, however, I am having a hard time using the logic laws and somehow get rid of the "r" proposition. Thanks!

(p → q) ∧ (¬q ∧ (r ∨ ¬q)) and ¬(q ∨ p)

Answer 1

$(p \rightarrow q) \wedge [\neg q \wedge (r \vee \neg q)]$

$\Leftrightarrow (\neg p \vee q) \wedge [\neg q \wedge (r \vee \neg q)]$ by implication rule

$\Leftrightarrow (\neg p \vee q) \wedge [\neg q \wedge (\neg q \vee r)]$ by commutative rule

$\Leftrightarrow (\neg p \vee q) \wedge \neg q$ by absorption rule

$\Leftrightarrow \neg q \wedge (\neg p \vee q)$ by commutative rule

$\Leftrightarrow (\neg q \wedge \neg p) \vee (\neg q \wedge q)$ by distributive rule

$\Leftrightarrow (\neg q \wedge \neg p) \vee F$ by negation law

$\Leftrightarrow \neg q \wedge \neg p$ by identity law

$\Leftrightarrow \neg (q \vee p)$ by DeMorgan's Rule

Hence, $(p \rightarrow q) \wedge [\neg q \wedge (r \vee \neg q)] \Leftrightarrow \neg (q \vee p)$

Answer 2

Use:

Absorption

$P \land (P \lor Q) \Leftrightarrow P$

So, with that: $\neg q \land (r \lor \neg q)$ just becomes $\neg q$ ... and your $r$ is gone!

If you don't have Absorption, and have to use some more elementary equivalence principles:

$\neg q \land (r \lor \neg q) \Leftrightarrow$

$(\neg q \lor \bot) \land (r \lor \neg q) \Leftrightarrow$

$\neg q \lor (\bot \land r) \Leftrightarrow$

$\neg q \lor \bot \Leftrightarrow$

$\neg q$