Timeline for Unprovable statements S where the only way to prove S is to assume S
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2012 at 21:39 | comment | added | Joel David Hamkins | Yes, I think I recall you saying that you did this. But in this case, the meet of the algebra is not conjunction; but don't most logicians want the meet of $\varphi$ and $\psi$ in the Lindenbaum algebra to be $\varphi\wedge\psi$? | |
| Nov 13, 2012 at 21:24 | comment | added | Carl Mummert | @Joel: I always arrange these algebras so that falsity is at the top, so being higher means being stronger. I realize this is not historically accurate, of course. | |
| Nov 13, 2012 at 20:53 | history | edited | François G. Dorais | CC BY-SA 3.0 |
minor edit
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| Nov 13, 2012 at 17:49 | history | edited | François G. Dorais | CC BY-SA 3.0 |
small corrections; added 3 characters in body
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| Nov 13, 2012 at 17:47 | comment | added | François G. Dorais | Absolutely, Joel! Fixing now... | |
| Nov 11, 2012 at 16:28 | comment | added | Joel David Hamkins | (And one should also say that when the theory is complete, it has no independent assertions, but the Lindenbaum algebra still has an atom.) | |
| Nov 11, 2012 at 16:27 | comment | added | Joel David Hamkins | Josh and François, isn't it instead the co-atoms in the Lindenbaum algebra that are the minimally unprovable statements? After all, being lower in the Lindenbaum algebra means being a stronger assertion (as falsity is at the bottom). The assertions min$\wedge$max and so on in François's answer are actually atoms, and their negations are co-atoms. Right? | |
| Nov 8, 2012 at 15:53 | comment | added | Joshua Grochow | François: Thanks! Not being a logician, I didn't know about the Lindenbaum-Tarski algebra. But if I've understood correctly, atoms in the Lindenbaum-Tarski algebra of $T$ (if any exist) are exactly the minimally unprovable statements over $T$. | |
| Nov 8, 2012 at 3:13 | comment | added | François G. Dorais | I forgot to exclude the trivial linear order with just one point by adding a non-triviality clause to $T$... | |
| Nov 8, 2012 at 2:39 | history | answered | François G. Dorais | CC BY-SA 3.0 |