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It is a well-known observation that in every non-empty bar there always is a customer that can rightfully shout "When I drink, everybody drinks." In the first-order logic this is expressed by the sentence$$\exists x(x = x) \supset \exists x(D(x) \supset \forall y D(y)).$$What would a formal proof of this sentence in the first-order calculus look like?

The textbook I am using does not have a whole lot of examples, and it would be enlightening to have a detailed worked out example to have in mind when thinking about related things.

Any help would be appreciated.

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    There are many logical systems, and the proof you're looking for would depend on the particular logical system. Which one are you using? You could list the axioms and previously-proved theorems in your question. Also, the sentence you show is not a good representation of your English sentence. Why do you have the first part? (That may be needed in your particular logical system, which doubly shows that you need to clarify your system.) Finally, even in a bar you know that sentence need not be true, so it is not provable--not in the usual logical systems and not without more axioms on D(). – 2017-01-07

1 Answers 1

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I'll prove it with Natural Deduction.

Assuming the "standard" semantics, with non-empty domain for the interpretation, the antecedent $\exists x (x=x)$ is valid; thus, I'll prove only the validity of the consequent :

$\exists x (Dx \to \forall y Dy)$.

Proof :

1) $\lnot\exists x (Dx \to \forall y Dy)$ --- assumed [a]

2) $\lnot Dx$ --- assumed [b]

3) $Dx$ --- assumed [c]

4) $\bot$ --- from 2) and 3) by $(\lnot \text E)$

5) $\forall y Dy$ --- from 4) by $(\bot \text E)$

6) $Dx \to \forall y Dy$ --- from 3) and 5) by $(\to \text I)$, discharging [c]

7) $\exists x (Dx \to \forall y Dy)$ --- from 6) by $(\exists \text I)$

8) $\bot$ --- from 1) and 7) by $(\lnot \text E)$

9) $Dx$ --- from 2) and 8) by $(\text {DN})$ (Double Negation), discharging [b]

10)$\forall y Dy$ --- from 9) by $(\forall \text I)$

11) $Dx \to \forall y Dy$ --- from 10) by $(\to \text I)$

12) $\exists x (Dx \to \forall y Dy)$ --- from 11) by $(\exists \text I)$

13) $\bot$ --- from 1) and 12) by $(\lnot \text E)$

$\exists x (Dx \to \forall y Dy)$ --- from 1) and 13) by $(\text {DN})$, discharging [a].