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!

What have you tried? Have you read the sentences you are given aloud and thought about what you might deduce? — Ross Millikan, Feb 26 '20 at 00:03
@RossMillikan I am not given any sentences. I simply know what $\Sigma$ is and must come up with formula $\theta$ through my intuition — JOSEPH GAFFNEY, Feb 26 '20 at 00:06
$\Sigma$ is a set of sentences. $\alpha$ represents a sentence and $\lnot \alpha$ is its negation. Try to prove some things from them. — Ross Millikan, Feb 26 '20 at 00:16
@JOSEPHGAFFNEY You are being asked to demonstrate that $\Sigma$ is inconsistent. Now, if indeed these three formula form an inconsistent set, then any two among them will entail the negation of the third. So select any two and show that they do. — Graham Kemp, Feb 26 '20 at 00:30
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score 3 · Answer 1 · answered Feb 26 '20 at 00:05

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 $\Sigma$.

Thank you so much, I was able to get the solution from this! — JOSEPH GAFFNEY, Feb 26 '20 at 01:12

Shaun · Accepted Answer · 2020-02-26T00:12:37.460

2

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:

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

edited Feb 26 '20 at 00:12

answered Feb 26 '20 at 00:07

Shaun

44,997

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

2 Answers2