Timeline for Modal collapse upon addition of the law of the excluded middle to an Intuitionistic modal logic
Current License: CC BY-SA 3.0
12 events
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| Mar 29, 2018 at 20:29 | comment | added | user65526 | I don't really get how the law of the excluded middle is appealed to in order to go from $\neg(\neg M\wedge\bigcirc M)$ to $M\vee \neg\bigcirc M$. Is this because you start by assuming $\neg M\wedge\bigcirc M$ and derive a contradiction, assuming that therefore $(\neg M\wedge\bigcirc M) \rightarrow \bot$ must hold? | |
| Mar 29, 2018 at 14:10 | comment | added | Bjørn Kjos-Hanssen | Probably not... you could try to make a countermodel | |
| Mar 29, 2018 at 9:01 | comment | added | user65526 | Can we get the modal collapse if we only assume the law of the excluded middle, and we don't assume $\neg \bigcirc \bot$? | |
| Mar 28, 2018 at 22:39 | vote | accept | user65526 | ||
| Mar 28, 2018 at 21:27 | comment | added | Bjørn Kjos-Hanssen | @user65526 I don't know much about structural laws... does it even matter whether you use them, given that you're using Law of Excluded Middle (and hence are not intuitionistic anymore)? | |
| Mar 28, 2018 at 20:41 | comment | added | user65526 | Are any structural laws (weakening, etc) utilised in the above proof? | |
| Mar 28, 2018 at 20:40 | comment | added | user65526 | Fantastic. So the proof wouldn't go through with Intuitionistic linear logic, I think, as in the link above, since I think Axiom S doesn't hold in it.math.stackexchange.com/questions/2395602/… | |
| Mar 28, 2018 at 20:23 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |
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| Mar 28, 2018 at 20:01 | comment | added | user65526 | If you could add proof steps it would be much appreciated! :) | |
| Mar 28, 2018 at 20:00 | comment | added | user65526 | Does the law of the excluded middle enter into this proof implicitly anywhere? Also, I don't understand why you write "so by (*) it suffices to obtain"? Couldn't the reduction work with the formula presented immediately before this remark? Also, does this proof go through in the Linear version of the logic above, discussed here: math.stackexchange.com/questions/2395602/… ? I suspect it might not, since I think (!) you don't have the converse of Axiom S, which was used in your proof. | |
| Mar 28, 2018 at 19:07 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |
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| Mar 28, 2018 at 18:58 | history | answered | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |