Talk:Nonfirstorderizability

Moving some material here

I removed the following from the article. Feel free to clean it up and reintroduce it. dbtfz^talk 02:38, 25 April 2006 (UTC)

Consider the sentence:

(2) Some of Fianchetto's men went into the warehouse unaccompanied by anyone else.

Quine argued that this could be captured in first-order logic as follows:

(3) (There was at least one man x such that)(x went into the warehouse and x was Fianchetto's man, and (For every man y)(If y accompanied x then y was one of Fianchetto's men)).

(3) $\exists x\mathrm {Man} (x)\wedge \mathrm {WentIntoWarehouse} (x)\wedge \mathrm {FianchettosMan} (x)\Rightarrow ((\forall y\mathrm {Man} (y)\wedge \mathrm {Accompanied} (y,x))\Rightarrow \mathrm {FianchettosMan} (y)).$

Boolos, however, argues that "anyone else" should--at least, in some contexts--be read as referring not to "anyone who wasn't one of Fianchetto's men" but rather to "anyone who wasn't one of them", i.e. the "some men" referred to in the first half of the sentence. Since "these men" have not been given any distinguishing predicate (keep in mind that they may not have been the only ones to enter the warehouse, and presumably are not the only men of Fianchetto's), they can be referred to collectively and exclusively only by some sort of quantification. This can be done by speaking of (most commonly) the set of them, (more rarely) the mereological fusion of them, or (as Boolos urges, on the grounds that it better captures the intuitive meaning), by reading "some men" as a second-order quantification. Hence:

(4) (There were some men, X, (of whom there was at least one, x), such that)(For every man y)(if y was among X, then (y went into the warehouse and (for every man, z)(if z accompanied y then z was among the X))).

If somebody could put standard logical symbolism in here it'd help a lot.

Remove the less readable form of (3) ... or keep both?

confused about the statement given

I would appreciate a brief clarification explaining why the given statement is false for the standard model of arithmetic. I'm sure I'm missing something, but it seems to me that there DOES exist such an X in the standard model: the set of all numbers 0, 1, 2, .... —Preceding unsigned comment added by 141.209.169.66 (talk) 23:25, 11 March 2008 (UTC)

explanation

Suppose N is a model of true arithmetic and that the sentence holds in N. If X includes any standard number, n, then b/c X is closed under predecessors it includes 0,1,2,...,n. Clearly A00 holds and according to the sentence this should imply 0!=0. So X can't include any standard number. It's easy to see that the set of all non-standard numbers does satisfy the sentence provided that there is a non-standard number. Charliescherer (talk) 15:26, 27 November 2009 (UTC)

Is the translation of the Geach-Kaplan sentence correct?

Isn't the formalized sentence true also when there's only one critic in the set X (who doesn't of course admire anyone)? Intuitively speaking, there should be at least two critics in X (wich is of course easily modified). —Preceding unsigned comment added by 83.150.115.123 (talk) 23:20, 28 November 2010 (UTC)

You have a point

I think so to but I would not consider myself experienced enough on the topic to make the alteration. But looking up the example given at http://plato.stanford.edu/entries/plural-quant/

∃S(∃u.u∈S & ∀u(u∈S →Cu) & ∀u∀v(u∈S & Auv → v∈S & u≠v))

you see that they have added ...& u≠v. But I don't get the ∀u(u∈S →Cu) part. What is C in this case?

Also when I boil down the formula to pseudo linguistics (which I'm not sure is correct) the alternative of there being one critic only admiring himself seems plausible.

       There exist one or more critics where
               there exist one or more persons who admire where
                       the admirer is a critic admiring a critic
               and
               there exists one or more persons where
                       the person is not a critic
               and
               for every person who admire
               for every person being admired
                       the admirer is a critic admiring a critic
                       the admirer is a critic not admiring a critic or non critic
                       the admirer is a non critic admiring or not admiring a critic or non critic

If you where to rewrite the current example it would be:

$\exists X(\exists x,y(Xx\land Xy\land Axy)\land \exists x\neg Xx\land \forall x\,\forall y(Xx\land Axy\rightarrow Xy\land x\neq y))$

— Preceding unsigned comment added by 83.248.247.202 (talk) 16:32, 4 October 2016 (UTC)

New explanation, for the improved illustration.

Suppose N is a model of true arithmetic and that the sentence holds in N. X is nonempty, so it must include some numbers in N; if any of these are standard then X includes all standard naturals, since X is closed under both successor and predecessor. So X either contains all standard naturals (but not all of N), or no standard naturals (but it still contains some elements of N). So N cannot be the standard naturals; but in any other model we could take X to be the set of all nonstandard naturals (or the set of all standard naturals), and get a model of the sentence. Ben Standeven (talk) 00:50, 17 June 2011 (UTC)

If Axy is understood to mean "x admires y," and the universe of discourse is the set of all critics, then a reasonable translation of the sentence into second order logic is:

$\exists X(\exists x,y(Xx\land Xy\land Axy)\land \exists x\neg Xx\land \forall x\,\forall y(Xx\land Axy\rightarrow Xy))$

That this formula has no first-order equivalent can be seen as follows. Substitute the formula (y = x + 1 v x = y + 1) for Axy. The result,

what if a sentence contains more than one relation? or a ternary one? the motivation for this substitution is unclear to me

$\exists X(\exists x,y(Xx\land Xy\land (y=x+1\lor x=y+1))\land \exists x\neg Xx\land \forall x\,\forall y(Xx\land (y=x+1\lor x=y+1)\rightarrow Xy))$

states that there is a nonempty set which is closed under the predecessor and successor operations and yet does not contain all numbers.

what allows you to restrict x to numbers? suppose i define superman, batman and spiderman to be cyclically predecessors and successors, the set contains the 3 of the superheros yet no numbers, what implies X is a set of numbers?

Thus, it is true in all nonstandard models of arithmetic but false in the standard model.

euh?

Since no first-order sentence has this property, the result follows.

QED by reference without reference, I would like to see proof of the fact that no first-order theory sentence has a property that is true in "all" non-standardmodels yet false in standard model of arithmetic and that this applies on non arithmetic sentences as well. — Preceding unsigned comment added by 213.49.89.185 (talk) 02:35, 21 May 2012 (UTC)