bio | website | science.marshall.edu/mummertc |
---|---|---|
location | Marshall University | |
age | 36 | |
visits | member for | 4 years, 11 months |
seen | 23 hours ago | |
stats | profile views | 2,074 |
I work in mathematical logic. My main areas of interest are arithmetic, reverse mathematics, computability, and proof theory.
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 |
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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 |
Sep 25 |
comment |
An interpretation of not-Con(PA)
Of course, some of the steps may also be nonstandard instances of axiom schemes, such as nonstandard instances of the induction scheme or nonstandard tautologies. |
Sep 14 |
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Deduction theorem
+1. The solution of Negri and Hakli is elegant - they redefine the meaning of $\vdash$ so that what they have is not quite the standard Hilbert system (because they have changed the N rule to a weaker rule), but which does have a deduction theorem. Of course the original Hilbert system, with the full N rule, does not satisfy the deduction theorem, in that it admits $A \vdash \Box A$ as a derived rule but not $\vdash A \to \Box A$. Presumably one could apply a similar method to certain systems for first order logic as well. |
Aug 24 |
awarded | Nice Answer |
Aug 24 |
comment |
Deduction theorem
@bellpeace: the original system is sound for a particular class of models, namely those in which $A \to B$ holds. Whenever we add new rules of inference, we have to restrict the set of models to those for which the new rules are sound. Of course the system I described, with the additional inference rule, is not complete, and no example that answers the question can be complete in that sense. |
Jul 12 |
awarded | Enlightened |
Jul 12 |
awarded | Nice Answer |
Jul 9 |
revised |
Infinite decreasing sequence by the Turing jump
added 826 characters in body |
Jul 9 |
answered | Infinite decreasing sequence by the Turing jump |
Jun 5 |
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Is the Invariant Subspace Problem arithmetic?
Here "equivalent" means "provably equivalent in a sufficiently weak system", which is more or less what the question is asking. Of course every true statement is uninerestingly equivalent to $0=0$. |
Jun 5 |
comment |
Is the Invariant Subspace Problem arithmetic?
Do we know that the statement in question is true in the setting of computable analysis? If it is not, then that immediately prevents it from being equivalent to any arithmetical formula, because every true arithmetical formula is satisfied by the model of second-order arithmetic with the standard natural numbers and only computable sets. Actually, every true $\Pi^1_1$ formula is true in that model. This is a standard method for showing that particular theorems are not expressible by excessively simple formulas. |
May 18 |
comment |
Necessity of omega-models in second order arithmetic
I am not aware of any reverse math research in which the base theory includes true first-order arithmetic. But there is another point: the induction schemes in reverse mathematics all include set parameters. This is the case even for schemes like B$\Sigma^0_2$. Formulas of true first order arithmetic don't include set parameters. |