Proof of laws of logic.

Question

I'm really confused with laws of logic unlike truth tables. Here, i'm trying to prove:

$$(\lnot p \lor \lnot q)\implies \lnot q \equiv p \lor\lnot q$$

LHS:

\begin{align}&&(\lnot p \lor \lnot q)⇒ \lnot q\\ \equiv&& \lnot(\lnot p \lor \lnot q) \lor \lnot q \tag{by Implication law}\\ \equiv&&p \lor q \lor \lnot q \tag{by Double Negation law}\end{align}

I'm stuck till there where I have no idea on how to get rid of $q$ as it seemed like I'm already close to the answer.

Answer 1

It is enough to prove that $$(\lnot p\lor\lnot q)\to(\lnot q\leftrightarrow(p\lor\lnot q))$$ is a tautology; that is, that the tableau of its negation is closed; but

is indeed closed. (Each path contains a contradiction.) Hence $$(\lnot p \lor \lnot q)\implies \lnot q \equiv p \lor\lnot q.$$

Answer 2

$\neg (\neg p \vee \neg q) \vee \neg q \equiv\\ (p\color{red}{\wedge} q) \vee \neg q \equiv\\ (p\vee\neg q)\wedge (q \vee \neg q) \equiv\\ (p\vee\neg q) \wedge \text{True} \equiv (p\vee\neg q) $

Answer 3

$$ \begin{array} {c|c} \lnot p \lor \lnot q)\implies \lnot q&Premise\\ \hline \lnot(\lnot p \lor \lnot q)\lor \lnot q&Implication\\ \hline (p\land q)\lor \lnot q&De Morgan\\ \hline (p\lor \lnot q)\land(q\lor \lnot q)&Distribution\\ \hline (p\lor \lnot q)\land1&Excluded\;Middle\\ \hline p\lor \lnot q&Redundant\;Truth \end{array} $$

using Propositional Calculus.