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I have an infinite set of variables X and I want to state that the property that there is a unique variable in X with value 2.

For a finite set, I would write the first-order logic formula: $$ (x_0 = 2 \wedge x_1 \neq 2 \wedge ... \wedge x_n\neq2) \vee (x_0 \neq 2 \wedge x_1 = 2 \wedge ... \wedge x_n\neq2) \vee ... $$

For the infinite set X, I could do the same but using infinite conjunction/disjunction.

Alternatively, using second-order logic, I could (if I understand correctly) quantify over X stating. $$ \exists x\in X. (x=2 \wedge (\forall y\in X. (x=y \vee y\neq 2))) $$

  1. Is there a name for first-order logic with infinite conjunction/disjunction?
  2. Does the example demonstrate that this logic is then more expressive than first-order logic with finite/binary conjunction/disjunction?

I am clearly not a logician so please don't overwhelm me too much :D

Ronnie Marksch
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    First-order logic has quantifiers. I don't see what is second-order about $\exists x \in X (x = 2 \land (\forall y \in X (x \neq y \rightarrow y \neq 2)))$. – Nika Mar 14 '20 at 20:03
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    @Nika $X$ here is the set of constant symbols... You need SOL to be able to quantify on it – Holo Mar 14 '20 at 20:03
  • @ℋolo I guess I'll stick around and learn something new.. Why would we need SOL to be able to quantify over a set? What is special about the set of constant symbols? I thought that SOL allows quantification over sets of sets, whereas FOL allows quantification over only sets; I could just be completely misguided but I don't see any sets of sets here. – Nika Mar 14 '20 at 20:08
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    @Nika because $X$ is not part of the theory, it is a meta-mathematic object. I now edit my answer to talk about this – Holo Mar 14 '20 at 20:09
  • @ℋolo If $X$ is not part of the theory, then why are we worried about a(n infinite) conjunction over $X$? – Nika Mar 14 '20 at 20:13
  • @Nika because you can do finite conjunction over X, so why not thinking about infinite? – Holo Mar 14 '20 at 20:17
  • @ℋolo If we aren't working in the theory, what is wrong with just saying "There is a unique variable that is equal to 2" in plain ole English? Also, if regular FOL doesn't allow you to quantify over certain sets, how would allowing infinite conjunctions (which is what I am assuming infinitary logic is doing) fix that issue? – Nika Mar 14 '20 at 20:27
  • @Nika the sentence "There is a unique variable that is equal to 2" is not a FO sentence, even if you are able to say it in plain English. If you are working on a meta-theory, this sentence is okay, but usually we are not working in that context(at least without specifically saying so). FOL does allow you quantify over Terms of the theory/elements of the model, set of language symbols are not a term. – Holo Mar 14 '20 at 20:33
  • @ℋolo Actually first-order logic doesn't let you quantify over terms of the theory - simply over elements of the model. (If you could quantify over terms you could pin down $\mathbb{N}$, for example.) – Noah Schweber Mar 14 '20 at 20:39
  • @NoahSchweber oops, I confused myself without bounded quantifiers in set theory... Yeah, you are correct. – Holo Mar 14 '20 at 20:44

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The name you are looking for is infinitary logic.

And yes, this is a simple example to show that first order logic cannot do everything.

To makes things a bit more formal: in logic "$X$" is a set of "constant symbols", or "0-ary function symbol", each element of $X$ is an element of your language, and not some set/number.

If you have a set $Y$ of elements of the model(for example, $Y$ is a set of integers), then FOL is enough: $∃y∈Y(y=2)$. But because $X$ is a set of elements of the language, and not the model you are working in you cannot talk about $X$ directly in FOL.

Holo
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    So that logic would then be the infinitary logic $L_{\omega_1,\omega}$? If $\omega$ is the first infinite ordinal, is $\omega_1$ the second infinite ordinal $\omega+1$? Is there some paper that introduced this particular logic the first time? Is that logic really equivalent to second-order logic? – Ronnie Marksch Mar 15 '20 at 07:32