Nick Thomas
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Registered User
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May 3 |
awarded | ● Nice Question |
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Feb 22 |
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Status of the 196 conjecture? Aaron: Sorry I missed your comment! That's an exciting idea! Could you possibly state your conjecture more explicitly? Are you trying to give a necessary and sufficient condition for $s(x)$ to be a palindrome? |
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Feb 12 |
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Status of the 196 conjecture? added 86 characters in body |
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Feb 12 |
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Status of the 196 conjecture? Aaron: Shoot! I will try to see where I have gone wrong. |
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Feb 12 |
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Status of the 196 conjecture? deleted 14 characters in body |
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Feb 12 |
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Math major at 36 Thank you, Andre! |
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Feb 12 |
answered | Math major at 36 |
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Feb 12 |
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Status of the 196 conjecture? Fixed error in proof.; added 72 characters in body |
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Feb 12 |
answered | Status of the 196 conjecture? |
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Jan 19 |
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Compactness-like property for universal generalization? Goldstern: Ah, I understand. That's a useful insight; thank you! |
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Jan 19 |
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Compactness-like property for universal generalization? Goldstern: My trouble is figuring out what other relations between the models might be relevant here. (Obviously I'll post if I figure that out.) Unfortunately, I do not understand the part in quotation marks. :-/ (Care to explain more?) $\phi$ does not mention the well-order. Francois: Thanks for the suggestion! I am going to play with it and see if it gets me anywhere. |
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Jan 19 |
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Compactness-like property for universal generalization? Andres: That's an excellent question, and it shows that what I'm asking for can't be done in general. In my specific problem, $\phi(x)$ has a form which excludes that case. But it seems clear that I haven't asked the right question, because I haven't included enough constraints to yield a solvable problem. I will see if I can repair my question; and in the meantime, thanks for your help! |
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Jan 19 |
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Compactness-like property for universal generalization? deleted 8 characters in body |
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Jan 19 |
asked | Compactness-like property for universal generalization? |
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Dec 30 |
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Constructible models of New Foundations? Andreas: Excellent, thanks! |
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Dec 30 |
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Constructible models of New Foundations? @Andreas: Thanks; I'll edit the question. I'd like to use the trick you described for eliminating parameters in a proof I'm writing. Is it something you thought of in this thread, and if so may I cite your comments here? Thank you! |
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Dec 30 |
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Constructible models of New Foundations? added 56 characters in body |
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Dec 29 |
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Constructible models of New Foundations? @Andreas: That looks right to me. Thank you for working this out with me. So perhaps what I really want to ask for is simply a model of NF wherein every set is definable by a stratified formula without parameters. Does that seem sensible? |
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Dec 28 |
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Constructible models of New Foundations? @Andreas: So your thought is that every definable set would already be defined in the first stage? That seems like a distinct possibility. If $a$ is a definable parameter defined by $\psi(x)$, replace $x \in a$ with $\psi(x)$, and $a \in x$ with... ? Let me know if you have an answer there. As for your question about stratification, truly no clue. |
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Dec 28 |
awarded | ● Nice Question |
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Dec 28 |
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Constructible models of New Foundations? Also (sorry about the fourth comment!): can anybody explain what Holmes' result (if correct) says (if anything) about the relative consistency of NF and, say, ZF or ZFC? (Should this be a new question?) |
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Dec 28 |
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Constructible models of New Foundations? The iterative conception here is as follows. Stage 0 is sets definable by formulas without parameters. Stage $n+1$ is sets definable by formulas with parameters from stage $n$. |
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Dec 28 |
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Constructible models of New Foundations? @Andreas and @Francois: You both raise similar and important issues with the way I've framed my question. In particular, you're right, Andreas, that the requirement I stated is trivially fulfilled. Thanks for pointing that out. Here's a possible alternate formulation. Given a model $M$, call a set $x \in M$ "definable" iff $x$ is defined by a (stratified) formula, with quantifiers ranging over $M$, whose parameters are definable. Do you think that makes sense of the question? |
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Dec 28 |
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Constructible models of New Foundations? @Andres: it sounds like you've essentially answered my question, with the answer being "nobody's done it;" correct? Also, thanks for sharing the announcement of that exciting result! @Ben: Yeah, but you can assume that a model exists and then play with it to make new models. Maybe you already know this, and maybe I'm missing something. But e.g., assume that a model of ZFC exists, and use said model to build a model where the continuum hypothesis fails, and you've effectively proven that if ZFC is consistent (a model exists) then ZFC doesn't prove CH (there is a model of ZFC where CH fails). |
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Dec 27 |
asked | Constructible models of New Foundations? |
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Dec 27 |
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Understanding Specker’s disproof of the axiom of choice in New Foundations Thanks for the correction! I believe Randall Holmes suggested the same thing, so this is an error on my part. Thanks also for your other correction. |
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Dec 25 |
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Existential instantiation in Hilbert-style deduction systems Andrej, thanks for writing that! I am now confident that I was not failing to grasp anything. And I believe that I am talking about existential variables, not constants, assuming you're using those words in the way with which I am familiar. |
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Dec 23 |
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Understanding Specker’s disproof of the axiom of choice in New Foundations Randall, thanks for the corrections! |
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Dec 23 |
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Understanding Specker’s disproof of the axiom of choice in New Foundations deleted 21 characters in body |
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Dec 23 |
revised |
Understanding Specker’s disproof of the axiom of choice in New Foundations added 7 characters in body |
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Dec 23 |
awarded | ● Teacher |
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Dec 23 |
revised |
Understanding Specker’s disproof of the axiom of choice in New Foundations edited body; added 3 characters in body |
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Dec 23 |
answered | Understanding Specker’s disproof of the axiom of choice in New Foundations |
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Dec 22 |
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Understanding Specker’s disproof of the axiom of choice in New Foundations Andreas, thanks for the excellent explanation! |
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Dec 22 |
revised |
Understanding Specker’s disproof of the axiom of choice in New Foundations added 95 characters in body |
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Dec 22 |
asked | Understanding Specker’s disproof of the axiom of choice in New Foundations |
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Dec 20 |
awarded | ● Commentator |
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Dec 20 |
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Existential instantiation in Hilbert-style deduction systems Emil, thanks for clearing up those remaining issues! I will take a look at Shoenfield. |
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Dec 20 |
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Existential instantiation in Hilbert-style deduction systems I realize, also, that I have not given enough information for you to verify my reasoning; let me know if you would like to have that. |
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Dec 20 |
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Existential instantiation in Hilbert-style deduction systems variable $y$ (regardless of whether it is fresh). In that situation, it is a meta-theorem that we may deduce $\forall x (\phi(x))$ from $T$. But having that $\exists x (\phi(x))$ is deducible from $T$ does not automatically fulfill this condition, because in that case we can only deduce $\phi(y)$ for fresh (i.e., not yet mentioned) $y$, not for arbitrary $y$. To be quite honest I did not think about any of this until you brought it up, and it's entirely possible that you have found an error in my system, in which case I would be indebted to you. However I do not yet see an error. (...) |
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Dec 20 |
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Existential instantiation in Hilbert-style deduction systems Andrej, looking at what I have written there, I see that it does not at all explain how this doesn't let us go from $\exists x (\phi(x))$ to $\forall x (\phi(x))$. I will try again. A given deduction from $T$ may include any number of new variables which we did existential instantiation into. Once a variable has been instantiated into (or mentioned free at all) at a given point in a proof, it is no longer "fresh," and we may not instantiate into it again. However, consider a situation where at any point in a proof from $T$, we may write steps which deduce $\phi(y)$ for an arbitrary (...) |
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Dec 20 |
awarded | ● Supporter |
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Dec 20 |
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Existential instantiation in Hilbert-style deduction systems Andrej, (1) you're right, the opening sentence is in error. I have edited the post accordingly. (2) I was speaking loosely, but in case I've failed to grasp something, can you explain the difference between the meaning of an open statement and the meaning of its universal closure? (3) In the system in question, universal introduction is a meta-theorem, which goes roughly like this: if $T \vdash \phi(y)$ for any variable $y$, then $T \vdash \forall x (\phi(x))$. (There are additional complications stemming from the fact that the logic is non-classical.) I can give you the paper if you wish. |
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Dec 20 |
revised |
Existential instantiation in Hilbert-style deduction systems added 328 characters in body |
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Dec 19 |
awarded | ● Scholar |
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Dec 19 |
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Existential instantiation in Hilbert-style deduction systems Emil: OK, I understand. Thank you for your help! I am marking this as the answer. I would be curious to see fuller proofs of Lemmas 1 and 1'. I am also curious to know what restrictions need to be placed on the deduction theorem. Not using the generalization rule is sufficient for DT to hold; but can we give a condition that is necessary and sufficient? And what would be best of all, can you point me to a source which treats these questions? I've had difficulty finding sources which address these sorts of nitty-gritty details, especially for Hilbert systems. Thank you! |
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Dec 18 |
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Existential instantiation in Hilbert-style deduction systems Emil, thanks for the thorough and helpful response! I need to play with your math more before I understand fully, but I wanted to thank you for writing. I will be getting back to you once I have finished playing. |
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Dec 18 |
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Existential instantiation in Hilbert-style deduction systems I am gathering, from your response and Emil's response, that this is a somewhat nonstandard way to do things. I believe that it does work, as in a previous project I built a sound and complete deduction system using this approach. In any case, thanks for informing me! |
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Dec 18 |
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Existential instantiation in Hilbert-style deduction systems Andrej, Thanks for writing! That is a good concern to raise. The rule I mention assumes a semantics in which free variables are not implicitly universally quantified (as in the semantics given below by Emil Jeřábek), but instead have something like implicit existential quantification (as in the "complicated semantics" I described). So then $\phi(y) \vdash \forall x (\phi(x))$ isn't valid in general. |
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Dec 18 |
awarded | ● Student |
