Prove tautology by using boolean laws $\neg q \to \neg(q\wedge(p\to\neg q))$

Question

$$\neg q \to \neg(q\wedge(p\to\neg q))$$

Please help me to prove if it's tautology or not by using the logic law.

Answer 1

Using $p\to q = \neg p \vee q $ and distributive laws, $$\neg{q} \to \neg(q\wedge(p\to \neg q))$$ $$=\neg q \to \neg (q\wedge(\neg p \vee \neg q))$$ $$=\neg q \to \neg ((q\wedge\neg p)\vee(q\wedge\neg q))$$ $$=\neg q \to \neg((q\wedge \neg p)\vee \mathrm{False})$$ $$=\neg q \to \neg(q\wedge\neg p)$$ $$=\neg q \to (\neg q \vee p)$$ $$=q\vee(\neg q \vee p)$$ $$=(q\vee\neg q)\vee(q\vee p)$$ $$=\mathrm(True)\vee(q\vee p)$$ $$=\mathrm{True}$$

Answer 2

Using DeMorgan's law on the conclusion, we see that it becomes $\neg q \lor \neg(p\to q)$.

Since $\neg q \to \neg q$, then surely $\neg q \to (\neg q \lor A),$ where $A$ is any propositional statement.

So your statement is a tautology, since it's always true.