tag:blogger.com,1999:blog-4987609114415205593.post4661048049465451294..comments2024-03-28T13:40:26.497+00:00Comments on M-Phi: On What "Is" IsJeffrey Ketlandhttp://www.blogger.com/profile/01753975411670884721noreply@blogger.comBlogger26125tag:blogger.com,1999:blog-4987609114415205593.post-57040361054292889292021-02-02T10:33:00.979+00:002021-02-02T10:33:00.979+00:00hello.. I read your blog and i must say that is re...hello.. I read your blog and i must say that is really great.keep sharing and if you are facing technical issues of Norton antivirus visit at <a href="https://contactklantenservicenederlands.com/bellen-avast-telefoonnummer.html/" rel="nofollow">norton ondersteuning</a>norton bellen nederlandhttps://www.blogger.com/profile/18428964846882688658noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-42835093113759241852020-11-05T11:59:54.429+00:002020-11-05T11:59:54.429+00:00Hi..I read out your blog and that is so informativ...Hi..I read out your blog and that is so informative for me.Keep sharing always. You must visit on another blog which is very helpful for us <a href="https://support-nummerdeutschland.com/mcafee-support-deutschland.html" rel="nofollow"> Mcafee deinstallieren<br /></a><br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-56210355582911494122020-11-05T11:32:53.738+00:002020-11-05T11:32:53.738+00:00Unbelievable blog! This blog provides a brief intr...Unbelievable blog! This blog provides a brief introduction which is very helpful for me. Instagram is the most usable platform in the world because of its latest features but the user some time confronts some issues on Instagram. For more information, you can visit <a href="https://tekninenasiakaspalvelu.com/instagram-tukinumero-suomi.html" rel="nofollow">Instagram-tili</a>.Instagram tukihttps://www.blogger.com/profile/01715269611793141770noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-60387414395655407822020-11-05T11:01:20.882+00:002020-11-05T11:01:20.882+00:00Hi, Thank you for sharing such a good and valuable...Hi, Thank you for sharing such a good and valuable information,It is very important for me. Gmail is the worldwide used email service but sometimes user faces some problems in it. If you want to get some information about the Gmail then you can visit <a href="https://tekninenasiakaspalvelu.com/gmail-tukinumero-suomi.html" rel="nofollow">Gmail asiakaspalvelu</a>.<br />Gmail tuki suomihttps://www.blogger.com/profile/03332253924113399989noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-32281818705509843372020-11-05T09:56:35.272+00:002020-11-05T09:56:35.272+00:00Your blog is very informative and interesting to r...Your blog is very informative and interesting to read, finally, I found exactly what I searching for. There are lots of users of Macfee antivirus in the world because of its features and easy interface. If you want to explore more interesting facts about Mcafee antivirus or want to resolve your technical issues then must visit <a href="https://contactklantenservicenederlands.com/bellen-mcafee-telefoonnummer.html" rel="nofollow">helpdesk Mcafee</a>. <br />Mcafee ondersteuninghttps://www.blogger.com/profile/02735746914133699950noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-28790959390354396782020-11-05T07:00:02.002+00:002020-11-05T07:00:02.002+00:00Your blog is very informative, finally, I found ex...Your blog is very informative, finally, I found exactly what I want. Paypal is an excellent service for online payments but lots of its users confront issues while they access Paypal. If you want to resolve your problems then must visit <a href="https://contactklantenservicenederlands.com/bellen-paypal-telefoonnummer.html" rel="nofollow">Paypal contact</a>.Paypal Nederlandhttps://www.blogger.com/profile/10441019087095996665noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-33593319511094124502011-08-19T07:31:51.935+01:002011-08-19T07:31:51.935+01:00Hi Tristan,
Right, all turns on how fine-grained ...Hi Tristan,<br /><br />Right, all turns on how fine-grained the individuation of languages is. So, we might call the view I'm sympathetic to Ultra-Fine Individuation:<br /><br />L is identical to L' <br />iff L and L' are identical in syntax, phonology, semantics and pragmatics.<br /><br />I can't, however, give any definite argument for this highly theoretical position about language individuation! The argument is to consider our intuitions about small changes (including modal ones) in L. <br /><br />String/expression individuation is much less fine-grained than language individuation though, because the same string can belong to different languages. So, a string doesn't have any particular meaning/referent intrinsically attached, as it were.Jeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-82043326668074415662011-08-19T04:57:48.932+01:002011-08-19T04:57:48.932+01:00Hi Jeff,
OK, you've brought home to me that m...Hi Jeff,<br /><br />OK, you've brought home to me that my claim - that the co-reference analysis of identity, read in a certain way, is false - vaguely involved a certain way of way of individuating languages and "content". <br /><br />And to be sure, this way is less fine-grained than the one you lean towards.<br /><br />At this point, I'm wondering whether there is any clear disagreement. I'm very pluralistic about different ways of individuating contents and interpreted expressions - I think your way is coherent, but I'm inclined to think that ordinary uses of the relevant expressions tend to invoke a less fine-grained conception.<br /><br />So, part of what I'm wondering is whether you have some sort of claim in the back of your mind for your way - e.g. that it is the only reasonable way of going, is useful for X or Y, fits best with usage (or some region thereof), etc. Any thoughts?<br /><br />There's a way of individuating languages I'm interested in which is pretty fine-grained, but stops short of your conception in not requiring - loosely speaking - identity of reference/extension for identity of language, but still requiring identity of sense (in some sense!). (This hangs together with my <a href="http://sprachlogik.blogspot.com/2011/05/three-true-semantic-externalisms.html" rel="nofollow">semantic externalism</a>, and I use it to make sense of the necessary <i>a posteriori</i>.)<br /><br />On that approach to individuating expressions-in-languages, what I said initially still seems right to me.Tristan Hazehttp://sprachlogik.blogspot.com/noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-79550421585435339502011-08-17T22:49:10.137+01:002011-08-17T22:49:10.137+01:00Hi Jeff,
Thanks, that was really helpful! I'm...Hi Jeff,<br /><br />Thanks, that was really helpful! I'm much clearer now on what the constraints are, i.e. what count's as a circular proof in this context. Need to think more about it!<br /><br />Also, thanks for the reference to your RSL paper, will certainly read when I get chance.<br /><br />Best,<br />SamSam Robertshttp://bbk.academia.edu/SamRobertsnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-11823709169080872222011-08-17T03:11:57.124+01:002011-08-17T03:11:57.124+01:00Sam, quick PS on my comment on your last comment, ...Sam, quick PS on my comment on your last comment, which mentions separability. In the RSL paper, Definition 2.6, I consider defining indiscernibility in terms of sets & relations "separating" objects $a$ and $b$. For a set $X$, it's the usual topological notion ($a \in X \wedge b \notin X$). For an $n$-ary relation $R$ (with $n>1$), it's defined using permutations of arguments (as in the clauses of the Hilbert-Bernays-Quine indiscernibility formula). <br />E.g., if $R$ is a ternary relation and $\underline{d} =(d_1,d_2)$ is an ordered pair then, then;<br />$(R,\underline{d})$ separates $a$ and $b$<br />iff <br />$(Rad_1d_2 \leftrightarrow \neg Rbd_1d_2) \vee (Rd_1ad_2 \leftrightarrow \neg Rd_1bd_2) \vee (Rd_1d_2a \leftrightarrow \neg Rd_1d_2b)$. <br /><br />Then: $a$ and $b$ are separable iff some set $X$, or some $(R, \underline{d})$ separate $a$ and $b$.<br /><br />(All relative to some structure, of course.)<br /><br />At the end of the article, Section 4.3, I relate the simplest kind of separability (i.e., $T_0$) of a topological space $S=(D,U)$ to definability of identity in the corresponding first-order structure $\mathcal{M}_S$. Under a certain side condition, $\mathcal{M}_S$ is Quinian iff $S$ is $T_0$.Jeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-76497309617076155982011-08-16T20:15:16.242+01:002011-08-16T20:15:16.242+01:00Hi again Tristan,
I think the distinction you mak...Hi again Tristan,<br /><br />I think the distinction you make at the end of your comment corresponds to the distinction between 2 kinds of meta-statement from a semantic metatheory for an interpreted language,<br /><br />(1) "Phosphorus = Hesperus" is true-in-English iff the denotation-in-English of "Phosphorus" is the denotation-in-English of "Hesperus".<br /><br />(2) "Phosphorus = Hesperus" expresses-in-English the proposition that Phosphorus = Hesperus.<br /><br />The first, (1), uses the clause (for atomic sentences) from the extensional semantic (inductive) definition of truth for English; while (2) should be a theorem of some intensional semantic meta-theory for English - stating what propositions are expressed (in English) by what strings.<br /><br />I agree that (1) doesn't state the propositional content of "Phosphorus = Hesperus". It does, however, allow us to unwind the truth condition, eventually to <br /><br />(3) "Phosphorus = Hesperus" is true-in-English iff Phosphorus = Hesperus.<br /><br />(assuming axioms for what "Phosphorus" and "Hesperus" denote-in-English.)<br /><br />The view I'm sympathetic is that semantic claims like (1), (2) and (3) express necessities.<br /><br />Cheers,<br /><br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-72609951323761935292011-08-16T19:47:05.459+01:002011-08-16T19:47:05.459+01:00Hi Sam,
Thanks - that makes perfect sense.
Oops!...Hi Sam,<br /><br />Thanks - that makes perfect sense.<br /><br />Oops! You're right. I'm replying too quick and trying to get the proof of PII sorted out - the principle you use isn't extensionality.<br />The proof you give uses "if $x$ and $y$ are distinct, they belong to distinct sets" and not "if $x$ and $y$ are distinct, they have distinct elements". I.e., $\forall z (x \in z \rightarrow y \in z) \rightarrow x = y$. <br /><br />But surely that does function as a definition of identity (in set theory, with urelements, even)? I think the same would hold for the modal semantic approach you mention too, "if $x$ and $y$ are distinct then $x$ and $y$ could be distinctly named". Again, that seems to be a definition of identity. (For me, a fact can be a definition.)<br /><br />If so, such definitions give us a formula $\phi(x,y)$ such that $x = y \leftrightarrow \phi(x,y)$. Then we can prove PII, as above.<br /><br />Cheers,<br /><br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-56520976472555465392011-08-16T17:07:07.319+01:002011-08-16T17:07:07.319+01:00Hi Jeff,
Thanks so much for this! (And no, you do...Hi Jeff,<br /><br />Thanks so much for this! (And no, you don't give the impression of complaining!)<br /><br />So, I wasn't assuming that extensionality held in the set theory. Indeed, I assumed that I was working with some facts about sets with urelements, and so even if I had assumed extensionality, it wouldn't give us an explicit definition of identity in general (only identity for sets). <br /><br />The point I had in mind was that there are lots of facts about sets concerning identity. For instance, it is a fact about sets that if $x\not= y$, then there is some set which separates them (we need not assume that the set in question is a singleton). And the question is which of these facts would it be circular\non-circular to appeal to in proving that the second-order reduction is adequate. <br /><br />Another example of facts that the reductionist could appeal to might (?) be the following. It is a fact about abstract semantics that if $x\not= y$, then we could introduce a name $n$, which names $x$ but does not name $y$: formally: $\Diamond^L \exists n[Name(n,x) \wedge \neg Name(n,y)]$. Again, we can then apply the relevant instance of second-order comprehension to get our conclusion.<br /><br />Here, again, it doesn't look like we need or need to presuppose an explicit definition of identity. <br /><br />Hope that makes sense!<br /><br />Best,<br />SamSam Robertshttps://www.blogger.com/profile/07195370003468343442noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-70228538725549986242011-08-16T04:30:16.910+01:002011-08-16T04:30:16.910+01:00Hi Tristan,
"In the second sense, I think it...Hi Tristan,<br /><br />"In the second sense, I think it's wrong that Hesperus = Phosphorus iff 'Hesperus' and 'Phosphorus' codenote, for the name 'Hesperus' for instance might have denoted some other object."<br /><br />This makes an assumption about the modal status of semantic axioms, and I'm sceptical about it. The view I'm sympathetic to is this. <br /><br />First, denotation (of a string of letters) is language relative. For it is meaningless to say, of the finite sequence ("c", "a", "t") that it denotes anything, simpliciter. Rather the finite sequence ("c", "a", "t") denotes-in-English the set of cats. Uninterpreted strings lack semantic properties.<br /><br />So, we need to include this language relativity when properly writing down sentences of the semantic meta-theory for an interpreted object language L. For example,<br /><br />(1) "snow is white" is true-in-English iff snow is white.<br />(2) "Hesperus" denotes-in-English Hesperus<br /><br />(This is even more apparent when the object language is not homophonically translated into the metalanguage.)<br /><br />Next, given this language relativity, one can argue that languages are very finely individuated. The slightest syntactic, semantic or pragmatic change leads to a new language. If that's right, then semantic axioms expresses (in the metalanguage) necessary truths!<br /><br />A fuller argument is this: there is no possible world where "Hesperus" denotes-in-English the moon. There is a possible world where "Hesperus" denotes-n-L the moon, but L is not English. The same goes for what a string "says" or "expresses". If S is a string, then what it says or expresses is always relative to the language L it belongs to.<br /><br />So, to consider your example,<br /><br />(3) Hesperus = Phosphorus iff "Hesperus" denotes-in-English what "Phosphorus" denotes-in-English<br /><br />On the view I just described, (3) expresses (in the metalanguage) a necessary truth. Clearly we do have some L such that "Hesperus" denotes-in-L Alpha Centauri and "Phosphorus" denotes-in-L Betelgeuse. But L isn't English.<br /><br />(There is a complication here: there is, strictu dictu, no such thing as English. Rather, there are countless idiolects, all of which are so similar, lexically, syntactically, semantically, pragmatically, that we idealize them as a single overarching entity. But - particularly in the present discussion - this can be very misleading.)<br /><br />On this view, languages do not modally (or even temporally) vary.<br /><br />An alternative view would try to introduce some modal notion of "counterpart of a language".Jeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-51721252584944799582011-08-16T02:35:06.419+01:002011-08-16T02:35:06.419+01:00Hi Jeff,
Thanks for responding. You're right ...Hi Jeff,<br /><br />Thanks for responding. You're right that the main question in that paper was 'What's special about identity?'. If I re-write it, maybe as a book chapter, I'll make that clearer.<br /><br />That's a cool point about there being just four kinds of permutation-invariant relation. That's definitely part of what's special about identity.<br /><br />However, I think one has to be careful to separate the name-view of identity (Geach's view, perhaps Frege's view in the Begriffschrift) from the more minimal claim that 'a = b' is true iff 'a' and 'b' codenote.<br /><br />That last claim, interpreted in a certain way, everyone would presumably agree with. <br /><br />But talk of truth-conditions is ambiguous: here, we're talking about the conditions in which the statement has a certain meaning and says something true, but there is another notion of truth-conditions central to philosophy, where we just take what the actual statement actually says, and talk about the conditions in which what it says is true is true.<br /><br />In the second sense, I think it's wrong that Hesperus = Phosphorus iff 'Hesperus' and 'Phosphorus' codenote, for the name 'Hesperus' for instance might have denoted some other object.Tristan Hazehttps://www.blogger.com/profile/18008340011384137776noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-71758812288617361932011-08-16T00:03:11.747+01:002011-08-16T00:03:11.747+01:00So, continuing, ... if $T$ is a second-order theor...So, continuing, ... if $T$ is a second-order theory containing an explicitly definition of $=$ (by $\phi(x,y)$ say), we get,<br /><br />1. $T \vdash \forall x\forall y(x = y \leftrightarrow \phi(x,y))$ (defn of $=$)<br /><br />2. $T \vdash \forall y \exists X \forall x(Xx \leftrightarrow \phi(x,y))$ (instance of Comp)<br /><br />3. $T \vdash \forall y \exists X \forall x(Xx \leftrightarrow x = y)$<br /><br />4. $T \vdash \exists X \forall x(Xx \leftrightarrow x = a)$<br /><br />5. $T \vdash \forall x(Ax \leftrightarrow x = a)$<br /><br />6. $T \vdash Aa$<br /><br />7. $T, a \neq b \vdash \neg Ab $<br /><br />8. $T, a \neq b, \vdash Aa \wedge \neg Ab$<br /><br />9. $T \vdash a \neq b \rightarrow (Aa \wedge \neg Ab)$<br /><br />10. $T \vdash a \neq b \rightarrow \exists X(Xa \wedge \neg Xb)$<br /><br />11. $T \vdash x \neq y \rightarrow \exists X(Xx \wedge \neg Xy)$<br /><br />12. $T \vdash \forall x \forall y(\forall X(Xx \rightarrow Xy) \rightarrow x = y)$<br /><br />And the proof uses an instance of comprehension at line 3, containing $=$.<br /><br />Cheers,<br /><br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-66307275507083413032011-08-15T23:10:06.528+01:002011-08-15T23:10:06.528+01:00Hi Sam,
At line 9 above in the "meta-derivat...Hi Sam,<br /><br />At line 9 above in the "meta-derivation", the (parametrized) Comprehenson instance asserts the existence of the haecceity of $a$. I then call this property $A$ at line 10. Then, using the properties of $=$, $T$ proves both $Aa$ and (assuming $a \neq b$, $\neg Ab$.<br /><br />But this works because $=$ is explicitly defined in $T$ by $\phi(x,y)$ to begin with!<br />So, line 9 may as well be:<br /><br />$T \vdash \exists X \forall x(Xx \leftrightarrow x = a)$.<br /><br />Cheers,<br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-1904950564062562032011-08-15T22:59:56.961+01:002011-08-15T22:59:56.961+01:00Let me post this rather technical and boring comme...Let me post this rather technical and boring comment just to give a clearer idea. Suppose $L$ is a second-order language, with identity as a primitive (i.e., including the deductive rules for =). Suppose $T$ is a theory in $L$.<br /><br />Lemma: Suppose $T$ contains Comprehension and an explicit definition of $=$.Then $T$ proves PII.<br /><br />1. $T \vdash \forall x\forall y(x = y \leftrightarrow \phi(x,y))$<br /><br />2. $T \vdash \forall y \exists X \forall x(Xx \leftrightarrow \phi(x,y))$<br /><br />3. $T \vdash \forall x \phi(x,x)$<br /><br />4. $T \vdash \forall x \forall y(\phi(x,y) \rightarrow \phi(y,x))$<br /><br />5. $T \vdash \phi(a,a)$<br /><br />6. $T \vdash \phi(b,a) \rightarrow \phi(a,b)$<br /><br />7. $T, a \neq b \vdash \neg \phi(a,b)$<br /><br />8. $T, a \neq b \vdash \neg \phi(b,a)$<br /><br />9. $T \vdash \exists X \forall x(Xx \leftrightarrow \phi(x,a))$<br /><br />10. $T \vdash \forall x(Ax \leftrightarrow \phi(x,a))$<br /><br />11. $T \vdash Ab \leftrightarrow \phi(b,a)$<br /><br />12. $T \vdash Aa \leftrightarrow \phi(a,a)$<br /><br />13. $T \vdash Aa$<br /><br />14. $T, a \neq b \vdash \neg Ab $<br /><br />15. $T, a \neq b, \vdash Aa \wedge \neg Ab$<br /><br />16. $T \vdash a \neq b \rightarrow (Aa \wedge \neg Ab)$<br /><br />17. $T \vdash a \neq b \rightarrow \exists X(Xa \wedge \neg Xb)$<br /><br />18. $T \vdash x \neq y \rightarrow \exists X(Xx \wedge \neg Xy)$<br /><br />19. $T \vdash \forall x \forall y(\forall X(Xx \rightarrow Xy) \rightarrow x = y)$Jeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-64770986204345427592011-08-15T22:12:49.813+01:002011-08-15T22:12:49.813+01:00Hi Sam,
Yes, that is pretty much the dialectic on ...Hi Sam,<br />Yes, that is pretty much the dialectic on this stuff over the last thirty years or so.<br /><br />Your proof is cool, sure.<br />(Do I give the impression of complaining at lot? Cue Morgenbesser joke: "Why is there something rather than nothing?" "Even if there was nothing, you'd still be complaining".)<br /><br />The proof works within a (first-order) theory where identity is already explicitly defined, by the axiom of extensionality,<br />$\forall z(x \in z \leftrightarrow y \in z) \rightarrow x = y$.<br />A similar proof would start with the theory of a strict linear order, $<$, and then we have a theorem,<br />$(\neg (x < y) \wedge \neg (y < x)) \rightarrow x = y)$. Suppose $a \neq b$. Then either $a < b$ or $b < a$. By comprehension,<br />$\exists X \forall x(Xx \leftrightarrow (x < b \vee b < x))$. <br />So, by the above, $Xa$. But $\neg Xb$, since $<$ is irreflexive. So, $\exists X (Xa \wedge \neg Xb)$. Qed. <br /><br />Maybe I'm being a bit unclear here, though. <br />We want to consider a theory lacking an explicit definition of identity and then introduce a definition of identity somehow. So, suppose $T$ in $L$ is a first-order theory where we don't assume $=$ as a primitive in $L$. Can we introduce a definition of identity which somehow doesn't presuppose $=$?<br /><br />Cheers,<br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-82199480154048394462011-08-15T17:18:34.252+01:002011-08-15T17:18:34.252+01:00Hi Jeff,
Thanks!
So, the dialectic is something ...Hi Jeff,<br /><br />Thanks!<br /><br />So, the dialectic is something like this (?):<br /><br />If one means to interpret $\forall X$ as a quantifier over legit physical properties, then there are examples of distinct objects which have all the same properties (or so someone like Cortes thinks -- thanks for this reference!).<br /><br />On the other hand, if we are more liberal with our interpretation of $\forall X$, we find that in order to prove the reduction successful, we need to appeal to instance of the comprehension schema which include $=$, and this is somehow circular. <br /><br />I suppose it would be nice to know more about what the rules are here (i.e. what a non-circular proof might look like -- whether the constraint of a non-circular proof is at all fair once made clear). <br /><br />For instance, here's a simple (silly) proof which does not appeal to instances of comprehension including $=$. <br /><br />If $x\not=y$, then $\exists z(x\in z \wedge y\not\in z)$. By second-order comprehension on $v \in z$ we get: $\exists X(Xx \wedge \neg Xy)$ as required!<br /><br />Of course, I assume you won't be happy with this either, but it would be nice to know why.<br /><br />Best,<br />Sam<br /><br />Ps, yes, but I hadn't thought about typing in LaTeX myself until now!Sam Robertshttps://www.blogger.com/profile/07195370003468343442noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-83457459235458264832011-08-14T20:44:22.498+01:002011-08-14T20:44:22.498+01:00Hi Tristan, thanks for sending the link to your pa...Hi Tristan, thanks for sending the link to your paper. I had a brief read through.<br /><br />One question you have there concerns what is so special about identity? Well, the binary relation of identity on a set D is the set {(x,x): x in D}. So, it's simply the diagonal of the set D. Why is it special? Why is it a logical relation? Tarski gave a standard answer to this. So long as |D| > 1, the identity relation on D is one of exactly 4 binary relations on D that is *invariant under any permutation of D*. These are<br />(i) the empty relation (i.e., the emptyset).<br />(ii) the universal relation (i.e., the complement of (i)).<br />(iii) the identity relation (i.e., the diagonal).<br />(iv) the distinctness relation (i.e., complement of (iii)).<br /><br />Suppose E is a binary atomic predicate in L, and E(t1, t2) is an atomic sentence, where t1 and t2 are L-terms. Suppose I is an interpretation of L. <br />Then the standard truth definition for atomic formulas gives us:<br /><br />(a) E(t1, t2) is true in I iff (I(t1), I(t2)) is in I(E).<br /><br />where I(t) is the denotation of t in I, and I(E) is the relation that E denotes in I. Suppose next that I(E) is the identity relation. Then:<br /><br />(b) E(t1, t2) is true in I iff I(t1) = I(t2).<br /><br />And this is Frege's view in his famous 1892 article: the sentence "Phosphorus = Hesperus" is true iff the denotation of "Phosphorus" is the denotation of "Hesperus".<br /><br />We freely use the notion of identity in the metatheory. So, we do not impose any demand of reduction.<br /><br />Cheers,<br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-38314158084333922702011-08-14T20:07:44.304+01:002011-08-14T20:07:44.304+01:00Hi Sam, agree on both your point.
I think the sec...Hi Sam, agree on both your point.<br /><br />I think the second-order Leibnizian reduction is ok - it works, so long as the range of "AX" is right. That is, = is strict indiscernibility. But there is a kind of circularity involved, because the properties required - haecceities - presuppose identity: in lambda notation, the property of being $x$ is:<br />$\lambda_y(y = x)$. <br /><br />In the phil physics literature, there are examples where people allege that PII is violated (usually by considering quantum indiscernibility - i.e., symmetrized n-particle states). E.g., see <br />Cortes, Alberto. 1976: "Leibniz's Principle of the Indentity of Indiscernibles: A False Principle", Philosophy of Science 43.<br /><br />http://philpapers.org/rec/CORLPO<br /><br />To this, I always say "PII isn't really violated: rather, it's just a case where the property quantifiers are restricted". <br /><br />So, then the question comes down to: is the property of being $x$ a physically legitimate property? I would say , "Yes". I think that Simon Saunders's view is that being $x§ isn't a genuine physical property.<br /><br />Cheers, <br />Jeff<br /><br />PS - does LaTeX view right in your web browser?Jeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-87806118365595933532011-08-14T12:15:27.870+01:002011-08-14T12:15:27.870+01:00Hi Jeff,
Thanks for the reply!
I wonder what you...Hi Jeff,<br /><br />Thanks for the reply!<br /><br />I wonder what you think about the following two responses:<br /><br />(1) It's not at all clear what the existence of non-standard models tells us. For instance, there are non-standard countable models of ZF, but that doesn't undermine my belief that there are uncountably many sets. Similarly, I might say that the Henkin models don't (and shouldn't) undermine my belief that `AX.....' works as a reduction of identity. To undermine that belief, what one would want is an example of clearly distinct x and y which (actually) share all of the same properties (in the sense of `properties' that Phi Phys people have in mind).<br /><br />(2) Couldn't we simply claim that there has been a confusion over the term `property' -- that someone proposing the reduction really meant `AX' to be interpreted as `Axx' or even as ranging over Fregean concepts?<br /><br />Thanks again,<br />Sam<br /><br />Ps, I assume that their notion of property rules out unordered pairs as well (otherwise the reduction would work in all three membered domains)?Sam Robertshttp://bbk.academia.edu/SamRobertsnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-47247057071626109282011-08-14T06:26:09.271+01:002011-08-14T06:26:09.271+01:00(This is the second time in a short while that I&#...(This is the second time in a short while that I've posted a link to some of my work in a comment here, but I can't help it. It's bound to stop soon, since I haven't actually done that much work.)<br /><br />A couple of years ago I wrote a paper in which I argue that identity statements should be regarded as being fundamentally different from other relational statements. Not only was I urging that 'the identity relation is primitive', but that identity statements are of their own logical form, irreducible to the others. (Some things Strawson said in his book <i>Subject and Predicate in Logic and Grammar</i> are very similar to what I was urging. More recently, Recanati's work.)<br /><br />This sort of formulation no longer does it for me, and I've come to abandon the idea of arguing against the orthodox view that identity is a relation between objects - but still I think that this doesn't do enough to mark the special nature of identity statements. This in turn affects (or infects) our thinking about modality.<br /><br /><a href="http://sites.google.com/site/tristanhaze/OnIdentityStatements.pdf" rel="nofollow">On Identity Statements: Against the ascriptional views</a>Tristan Hazehttps://www.blogger.com/profile/18008340011384137776noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-30678389711834146222011-08-13T22:52:50.677+01:002011-08-13T22:52:50.677+01:00Hi Sam, agree entirely!
But some in the PII physi...Hi Sam, agree entirely! <br />But some in the PII physics literature don't like haecceities - they're not legit physical properties. So, the countermodel gives a very simple example of what happens when the range of the 2nd order quantifier doesn't include all unit sets.<br />Cheers,<br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.com