bio | website | science.marshall.edu/mummertc |
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location | Marshall University | |
age | 36 | |
visits | member for | 5 years, 2 months |
seen | Jun 25 at 21:17 | |
stats | profile views | 2,104 |
I work in mathematical logic. My main areas of interest are arithmetic, reverse mathematics, computability, and proof theory.
Jun 5 |
awarded | Enlightened |
Jun 4 |
awarded | Nice Answer |
Jun 3 |
comment |
Does Peano's existence theorem admits a constructive proof?
It is not clear to me that Walter's paper is really about constructive mathematics; I think it is about something weaker, in the way that classical mathematicians sometimes use the word "constructive", which is a much broader sense of the term than is used in constructive mathematics. |
Jun 3 |
revised |
Does Peano's existence theorem admits a constructive proof?
added 1390 characters in body |
Jun 3 |
answered | Does Peano's existence theorem admits a constructive proof? |
May 19 |
awarded | Nice Answer |
May 10 |
comment |
Is second-order ZFC categorical with regard to its proper class models
In particular, for people who have some favored set theoretic background (such as a "platonisitic $V$" consisting of "all sets"), the argument goes through in their background like any other. So rather than being more restrictive, the viewpoint that second-order logic is only defined relative to a set-theoretic background is more general. The source of categoricity is not only the use of second-order semantics; it is also in the choice of a particular set-theoretic background in which to interpret those semantics. @Andy |
May 10 |
comment |
Is second-order ZFC categorical with regard to its proper class models
The real challenge is to decide how you want to formalize the question. As Joel David Hamkins' answer explains, if we choose an arbitrary model of $V$ of ZFC as out metatheory to give a definition of full second-order semantics, then that model $V$ will think that $V$ is itself the only class model of full second-order ZFC. Does this really show what you wanted to show? |
Apr 25 |
comment |
A question on complexity notation
Because every $\Pi^1_n$ formula is $\Delta^2_0$, the comprehension schemes are ordinarily nested. I suppose you could view the latter as "improper" instances of the former scheme. But, to try to make the comprehension schemes disjoint, how would you separate the "properly" $\Pi^2_1$ instances of comprehension from the "improper" ones (making up words). That seems to involve more than just syntactic analysis, while the ordinary definition of the schemes is syntactic. For example, if $\phi$ is $\Pi^1_n$ and $\psi(X^2)$ is logically valid we might consider $(\exists X^2)[\psi(X) \land \phi]$. |
Apr 18 |
awarded | Yearling |
Apr 3 |
awarded | Nice Answer |
Mar 19 |
comment |
This modal logic semantics is not S5, but is it something else well-known?
I think this may be related to your question: arxiv.org/abs/1401.0648 . It is not quite the same because we consider models that do not have all possible interpretations, but there seems to be some similarity on a quick reading of your question. |
Feb 28 |
comment |
Completeness of a set of propositional formulas
I have voted to put this on-hold because it is not about research level mathematics - this question would be a better fit on mathematics.stackexchange.com. In any event: when we have a finite set of variables and a finite set of formulas, it is trivially decidable whether the set is complete, by using truth tables. When the set of formulas may be infinite, it is not decidable whether a given set is complete, by a simple diagonalization argument. |
Feb 22 |
comment |
Uncomputability of the identity relation on computable real numbers
Thanks. The linked paper is "Computability and analysis: the legacy of Alan Turing" by Avigad and Brattka, arxiv.org/pdf/1206.3431v2.pdf . The citation they give is: H. Gordon Rice. Recursive real numbers. Proceedings of the American Mathematical Society, 5:784–791, 1954. |
Jan 21 |
revised |
Total formulae in a theory equivalent to $\Delta_0$ formulae in the theory?
edited tags; edited title |
Jan 4 |
comment |
Axiomatic ZFC Set Theory
Of course, it also matters exactly how replacement is stated. If it is stated in the form sometimes called "collection", it no longer implies the comprehension axioms (this form of replacement only says that the image of the function is a subset of some set). But if we already include the comprehension axioms, then it makes no difference which form of replacement is included. To see why this is particularly relevant: the axiom of replacement stated in Kunen's standard book is the "collection" form which does not imply the comprehension scheme. |
Nov 27 |
awarded | Nice Answer |
Oct 11 |
comment |
Forcing is intuitionistic
Some modern accounts also incorporate classical logic into the definition of forcing by beginning with a limited set of connectives (e.g. including $\lnot$, $\land$, and $\forall$, but not $\exists$ and not $\lor$), and then assuming the other connectives are given by their classical definitions, which are not intuitionistically correct. |
Oct 11 |
answered | Forcing is intuitionistic |
Sep 30 |
awarded | Explainer |