I used the left side to attempt to prove it was T. Chart of Laws
(p ∧ q) → q ≡ T
(p ∧ q) → q
Using the first line in table 7
¬(p ∧ q) v q
D'Morgans Law
(¬p v ¬q) v q
Associative Law
¬p v (¬q v q)
Negation Law
¬p v T
Domination Law
T ≡ T
I was wondering if this is right since I felt that you can't do that with domination law. If anyone could double check thank you.
Yes that's completely correct.
The domination law states that $q \lor T$ is always true for any statement $q$. So it's true especially for any negated statement $\neg p \equiv q$.
Another way to think about this is to consider:
$$(\neg p \lor T) \equiv \neg\neg (\neg p \lor T) \equiv \neg(p \land F) \equiv \neg F \equiv T$$
that is using the double negation law and then applying De Morgan's laws one time to use the domination law directly with the $\land$ operator.