Timeline for Deduction theorem
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Feb 7, 2015 at 8:32 | answer | added | kakaz | timeline score: 1 | |
| Sep 12, 2014 at 20:46 | answer | added | user37159 | timeline score: 1 | |
| Aug 24, 2014 at 2:13 | comment | added | zpavlinovic | Deduction theorem also does not hold in epistemic logic. | |
| Aug 17, 2013 at 0:26 | comment | added | Doug Spoonwood | @abo I have to wonder what you mean by "interesting" when you say that [A => (B => A)] (under detachment and uniform substitution) is not interesting as an axiomatic system. If you know how that system works, you can "observe" that anytime you have an axiom set for a logic L with [A => (B => A)] as either an axiom or theorem, you can form at least a countably infinite class C of axiom sets (given countably infinite variables), where each member of C axiomitizes L also, such that no axiom set belonging to C has a member in common with any other axiom set. | |
| Jun 28, 2013 at 2:28 | comment | added | Sniper Clown | Short answer is paraconsistent logic | |
| Jun 28, 2013 at 0:05 | comment | added | Doug Spoonwood | @EmilJeřábek But, what if the only derivations permitted in the Hilbert System are the theses which are detachable from other theses in the system? | |
| Jun 28, 2013 at 0:02 | answer | added | Doug Spoonwood | timeline score: -2 | |
| Jun 25, 2013 at 3:02 | review | First posts | |||
| Jun 25, 2013 at 11:22 | |||||
| Jun 5, 2013 at 20:27 | answer | added | Rafał Gruszczyński | timeline score: 7 | |
| Jun 3, 2013 at 3:10 | vote | accept | Octavio | ||
| Jun 3, 2013 at 3:09 | vote | accept | Octavio | ||
| Jun 3, 2013 at 3:10 | |||||
| Jun 3, 2013 at 3:09 | vote | accept | Octavio | ||
| Jun 3, 2013 at 3:09 | |||||
| May 30, 2013 at 17:38 | answer | added | J Marcos | timeline score: 15 | |
| May 30, 2013 at 16:08 | comment | added | François G. Dorais | You're right Emil, what I wrote doesn't make sense to me this morning. I don't know if I was just too tired to think or thinking something other than what I wrote. | |
| May 30, 2013 at 11:57 | comment | added | Emil Jeřábek | @François: How come? A is derivable from A by a one-line proof consisting of just A, even if your system has no logical axioms or rules. More generally, every Hilbert system defines a Tarski-style consequence relation, irrespective of the presence of any particular rules. | |
| May 30, 2013 at 5:31 | comment | added | abo | @Francois. I wasn't aware I was loosening the rules. Looking at Mendelson, he defines a formal axiomatic theory for the propositional calculus with three axioms. Keep only the first of the three, which is A => (B => A). Then A => A isn't provable, but at least according to Mendelson's definition, it is a formal axiomatic theory (just not an interesting one). Point taken that there are some systems where not even A follows from A. | |
| May 30, 2013 at 0:44 | history | edited | The User |
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| May 30, 2013 at 0:33 | answer | added | Carl Mummert | timeline score: 26 | |
| May 29, 2013 at 23:29 | comment | added | Noah Schweber | I think for this question to make sense, we need a definition of "axiomatic system." For example, personally I would consider taking the deduction theorem as part of the definition of "axiomatic system;" I could in principle be talked out of it, but it seems like a reasonable requirement to make. | |
| May 29, 2013 at 22:14 | comment | added | François G. Dorais | @abo: If you're willing to loosen the rules that much then "A always follows from A" is not even true: in a Hilbert system with just modus ponens and no axioms, A does not follow from A. | |
| May 29, 2013 at 21:31 | comment | added | abo | Perhaps one should include "interesting" in front of "axiomatic system"? Even in an empty axiomatic system, A always follows from A, but in an empty axiomatic system, one cannot prove anything, much less A => A. By considering any set of axioms which do not allow the proof of A => A, the deduction theorem would still evidently not hold. | |
| May 29, 2013 at 20:08 | comment | added | François G. Dorais | Corollary 9.12 of Kohlenbach's Applied Proof Theory states that WE-HA$^\omega$ and hence WE-PA$^\omega$ fail to satisfy the deduction theorem. However, I must admit that I never fully understood what was going on there. | |
| May 29, 2013 at 19:51 | history | asked | Octavio | CC BY-SA 3.0 |