Timeline for Boolean-Valued Models: Why is $\| x=y \| \cdot \| \phi(x) \| \leq \| \phi(y) \|$?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2015 at 1:51 | review | Close votes | |||
| Dec 21, 2015 at 10:04 | |||||
| Nov 29, 2015 at 0:54 | review | Close votes | |||
| Nov 29, 2015 at 17:16 | |||||
| Nov 29, 2015 at 0:45 | vote | accept | CommunityBot | ||
| Nov 29, 2015 at 0:42 | answer | added | Sam Roberts | timeline score: 3 | |
| Nov 29, 2015 at 0:40 | comment | added | user83393 | Thank you. I feel daft for not realizing Hamkin's approach, though I quite appreciated Roberts' approach as it was illuminating. | |
| Nov 29, 2015 at 0:38 | comment | added | Sam Roberts | @JoelDavidHamkins Sure! (I thought it might have been moved to stackexchange.) | |
| Nov 29, 2015 at 0:21 | comment | added | Joel David Hamkins | Meanwhile, it is probably better to post answers as answers, since the site works better that way. @SamRoberts, could you post your answer? | |
| Nov 29, 2015 at 0:17 | comment | added | Joel David Hamkins | Or, one can simply use contraposition. Namely, in any Boolean algebra, $a\cdot b\leq c$ just in case $a\cdot\neg c\leq \neg b$. So from $\|y=x\|\cdot\|\varphi(y)\|\leq\|\varphi(x)\|$ and the fact that $\|x=y\|=\|y=x\|$, we deduce $\|x=y\|\cdot\|\neg\varphi(x)\|\leq\|\neg\varphi(y)\|$, as desired. | |
| Nov 28, 2015 at 23:21 | history | edited | Stefan Kohl♦ |
Added top-level tag.
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| Nov 28, 2015 at 23:14 | comment | added | Peter LeFanu Lumsdaine | @SamRoberts: or to turn that around very slightly, to make it sound closer to the original approach: instead of directly showing by induction $\| x=y\| \cdot \| \varphi(x) \| \leq \|\varphi(y) \|$, strengthen the statement of the induction to $\| x = y \| \leq \left( \| \varphi(x) \| \leftrightarrow \| \varphi(y) \| \right) $. | |
| Nov 28, 2015 at 23:07 | comment | added | Sam Roberts | I think the easiest way to see this is by noting that by (i)-(iv) all atomic instances of the identity axioms $x= y \to (\phi(x) \leftrightarrow \phi(y))$ get value 1 and then note that the non-atomic instances follow in first-order logic by a simple induction. So those instances too will have value 1. | |
| Nov 28, 2015 at 22:59 | review | First posts | |||
| Nov 28, 2015 at 23:21 | |||||
| Nov 28, 2015 at 22:54 | history | asked | user83393 | CC BY-SA 3.0 |