6,101 reputation
12040
bio website science.marshall.edu/mummertc
location Marshall University
age 36
visits member for 4 years, 5 months
seen 9 hours ago
I work in mathematical logic. My main areas of interest are arithmetic, reverse mathematics, computability, and proof theory.

2d
comment 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
comment 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.
May
18
answered Necessity of omega-models in second order arithmetic
May
18
revised Are there “non-constructive” sets in second-order arithmetic?
added 2880 characters in body
May
17
answered Are there “non-constructive” sets in second-order arithmetic?
May
17
comment Proof complexity of two directions of equivalency?
This sort of question comes up very naturally, but it is often hard to find precisely the notion we want to capture. The length of a formal proof is not a very interesting metric, because we don't normally look at formal proofs anyway (so the length is meaningless for practice) and because the length depends as much on the proof system as on the theorem being proved. Reverse Mathematics can capture the set-existence axioms required for each direction. But trying to capture how "easy" or "natural" each direction is to prove is a challenge that has not been solved.
May
15
comment Does formalizing math require search and creativity, or is it near-mechanical?
In his paper about the formalization of the prime number theorem, Avigad reported that once he was up to speed it took about a day to formalize a page of mathematics in Isabelle. I recommend that paper for its discussion of the formalization process. repository.cmu.edu/philosophy/31
May
5
comment What new primitive recursive functions are needed to reconcile Turing time complexity with Gödel time complexity?
@Joel: Yes, exactly. Some other models have a different, unchangeable tape symbol to mark the left edge of the tape.
May
5
comment What new primitive recursive functions are needed to reconcile Turing time complexity with Gödel time complexity?
@Joel: By the way, there is a technical reason to start the machine on the first input symbol, which is to accommodate machines that have a one-sided (i.e. semi-infinite) input tape, which are common in the computational complexity literature. These need to know where the left end of the tape is, because in the usual framework they have no way to find it otherwise.
May
5
revised What new primitive recursive functions are needed to reconcile Turing time complexity with Gödel time complexity?
added 512 characters in body
May
5
comment What new primitive recursive functions are needed to reconcile Turing time complexity with Gödel time complexity?
@Joel: Silly Turing machines - everyone has their own idea how they work! In my mind they need to start on the first symbol of input. This is also (likely) why the author said that $h(x) = x-1$ takes $O(|x|)$ steps; it would also take only one step if the machine started on the last digit. I'll edit the answer, however.
May
5
comment What new primitive recursive functions are needed to reconcile Turing time complexity with Gödel time complexity?
@Dan Turetsky: as I point out in my answer, the question is somewhat trivially unsolvable in that context.