If $\Sigma=\lbrace(\neg\alpha),(\neg\beta),(\alpha\vee\beta)\rbrace$, show that $\Sigma$ is inconsistent.

Question

If $\Sigma=\lbrace(\neg\alpha),(\neg\beta),(\alpha\vee\beta)\rbrace$, show that there is some formula $\theta$ such that $\theta$ and $(\neg\theta)$ can be deduced from $\Sigma$ (i.e., that $\Sigma$ is inconsistent)

I understand I must come up with a formula that can be deduced from this set of hypotheses, but I am unsure how to begin thinking of a formula that would work in this scenario.

EDIT: I was able to solve, thank you!

Answer 1

Since $\alpha\lor\beta$ is equivalent to $(\lnot \alpha)\to \beta$, by $\lnot\alpha$ and modus ponens, we have $\beta$. But $\lnot\beta$ by hypothesis, a contradiction.

See, also, disjunctive syllogism:

$$\frac{\Sigma\vdash \lnot\alpha, \Sigma\vdash (\alpha\lor \beta)}{\Sigma\vdash \beta}.$$

Answer 2

I am unsure how to begin thinking of a formula that would work in this scenario.

Hint: Given $\neg \alpha$ and $\neg \beta$, what can you infer about the truth of of $\alpha \lor \beta$? What can then be said about $\neg (\alpha \lor \beta)$ under $\Sigma$? Can we (if so, how?) deduce $\alpha \lor \beta$ from $\Sigma$?
Combine these two observations to show that both $\alpha \lor \beta$ and $\neg (\alpha \lor \beta)$ can be deduced from $\Sigma$.