Talk:Nonfirstorderizability

message: I removed the following from the article.  Feel free to clean it up and reintroduce it.  dbtfztalk 02:38, 25 April 2006 (UTC)
thread_title:  Moving some material here 

message: 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)
thread_title:  confused about the statement given 

message: 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)
thread_title:  explanation 

message: 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)
replies: text: \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 \ne y))
— Preceding unsigned comment added by 83.248.247.202 (talk) 16:32, 4 October 2016 (UTC)
thread_title:  Is the translation of the Geach-Kaplan sentence correct? 

message: 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 \ne y))
— Preceding unsigned comment added by 83.248.247.202 (talk) 16:32, 4 October 2016 (UTC)
thread_title:  You have a point