Equivalence of strings in modal logic
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I'm trying to solve a question which asks me to show that for any two finite strings $O_1$ and $O_2$ of $square$s and $lozenge$s, (e.g. $squarelozengelozengesquarelozengesquare)$, that
i) if $O_1equiv_S O_2$ then $OO_1 equiv_S OO_2$
and
ii) if $O_1equiv_S O_2$ then $O_1O equiv_S O_2O$
where $O$ is any other string and we say $O_1equiv_S O_2$ (they express the same modality) if $models_S (O_1phi leftrightarrow O_2phi)$, where $S$ is just some modal logic system.
I've managed to solve the first part quite easily using the fact that if $models phi_1 leftrightarrow phi_2$ then $models chi(phi_1)leftrightarrowchi(phi_2)$ where $chi(phi)$ is just the result of replacing $P$ with $phi$ in the formula $chi(P)$.
I'm not sure how to go about solving the second though, because in this case the new string is inserted between the old one and the sentence $phi$, and I can't see an obvious way to do that with this formula.
I'm also trying to find a way to show, using these, that under the modal system $B$ (where the accessibility relation is reflexive and symmetric), that there are infinitely many modalities, using some sort of informal semantic argument.
I've done the same for $S_4$ and $S_5$, but those are finite numbers of modalities so the strategy was different.
I'd really appreciate any help you could offer!
logic propositional-calculus first-order-logic modal-logic
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add a comment |
$begingroup$
I'm trying to solve a question which asks me to show that for any two finite strings $O_1$ and $O_2$ of $square$s and $lozenge$s, (e.g. $squarelozengelozengesquarelozengesquare)$, that
i) if $O_1equiv_S O_2$ then $OO_1 equiv_S OO_2$
and
ii) if $O_1equiv_S O_2$ then $O_1O equiv_S O_2O$
where $O$ is any other string and we say $O_1equiv_S O_2$ (they express the same modality) if $models_S (O_1phi leftrightarrow O_2phi)$, where $S$ is just some modal logic system.
I've managed to solve the first part quite easily using the fact that if $models phi_1 leftrightarrow phi_2$ then $models chi(phi_1)leftrightarrowchi(phi_2)$ where $chi(phi)$ is just the result of replacing $P$ with $phi$ in the formula $chi(P)$.
I'm not sure how to go about solving the second though, because in this case the new string is inserted between the old one and the sentence $phi$, and I can't see an obvious way to do that with this formula.
I'm also trying to find a way to show, using these, that under the modal system $B$ (where the accessibility relation is reflexive and symmetric), that there are infinitely many modalities, using some sort of informal semantic argument.
I've done the same for $S_4$ and $S_5$, but those are finite numbers of modalities so the strategy was different.
I'd really appreciate any help you could offer!
logic propositional-calculus first-order-logic modal-logic
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$begingroup$
You need to explain the role of $phi$ in your definition of $equiv_{S}$.
$endgroup$
– Rob Arthan
Jan 21 at 21:54
$begingroup$
I answered that here
$endgroup$
– Jishin Noben
Jan 22 at 15:14
add a comment |
$begingroup$
I'm trying to solve a question which asks me to show that for any two finite strings $O_1$ and $O_2$ of $square$s and $lozenge$s, (e.g. $squarelozengelozengesquarelozengesquare)$, that
i) if $O_1equiv_S O_2$ then $OO_1 equiv_S OO_2$
and
ii) if $O_1equiv_S O_2$ then $O_1O equiv_S O_2O$
where $O$ is any other string and we say $O_1equiv_S O_2$ (they express the same modality) if $models_S (O_1phi leftrightarrow O_2phi)$, where $S$ is just some modal logic system.
I've managed to solve the first part quite easily using the fact that if $models phi_1 leftrightarrow phi_2$ then $models chi(phi_1)leftrightarrowchi(phi_2)$ where $chi(phi)$ is just the result of replacing $P$ with $phi$ in the formula $chi(P)$.
I'm not sure how to go about solving the second though, because in this case the new string is inserted between the old one and the sentence $phi$, and I can't see an obvious way to do that with this formula.
I'm also trying to find a way to show, using these, that under the modal system $B$ (where the accessibility relation is reflexive and symmetric), that there are infinitely many modalities, using some sort of informal semantic argument.
I've done the same for $S_4$ and $S_5$, but those are finite numbers of modalities so the strategy was different.
I'd really appreciate any help you could offer!
logic propositional-calculus first-order-logic modal-logic
$endgroup$
I'm trying to solve a question which asks me to show that for any two finite strings $O_1$ and $O_2$ of $square$s and $lozenge$s, (e.g. $squarelozengelozengesquarelozengesquare)$, that
i) if $O_1equiv_S O_2$ then $OO_1 equiv_S OO_2$
and
ii) if $O_1equiv_S O_2$ then $O_1O equiv_S O_2O$
where $O$ is any other string and we say $O_1equiv_S O_2$ (they express the same modality) if $models_S (O_1phi leftrightarrow O_2phi)$, where $S$ is just some modal logic system.
I've managed to solve the first part quite easily using the fact that if $models phi_1 leftrightarrow phi_2$ then $models chi(phi_1)leftrightarrowchi(phi_2)$ where $chi(phi)$ is just the result of replacing $P$ with $phi$ in the formula $chi(P)$.
I'm not sure how to go about solving the second though, because in this case the new string is inserted between the old one and the sentence $phi$, and I can't see an obvious way to do that with this formula.
I'm also trying to find a way to show, using these, that under the modal system $B$ (where the accessibility relation is reflexive and symmetric), that there are infinitely many modalities, using some sort of informal semantic argument.
I've done the same for $S_4$ and $S_5$, but those are finite numbers of modalities so the strategy was different.
I'd really appreciate any help you could offer!
logic propositional-calculus first-order-logic modal-logic
logic propositional-calculus first-order-logic modal-logic
asked Jan 21 at 16:38
oxflopoxflop
61
61
$begingroup$
You need to explain the role of $phi$ in your definition of $equiv_{S}$.
$endgroup$
– Rob Arthan
Jan 21 at 21:54
$begingroup$
I answered that here
$endgroup$
– Jishin Noben
Jan 22 at 15:14
add a comment |
$begingroup$
You need to explain the role of $phi$ in your definition of $equiv_{S}$.
$endgroup$
– Rob Arthan
Jan 21 at 21:54
$begingroup$
I answered that here
$endgroup$
– Jishin Noben
Jan 22 at 15:14
$begingroup$
You need to explain the role of $phi$ in your definition of $equiv_{S}$.
$endgroup$
– Rob Arthan
Jan 21 at 21:54
$begingroup$
You need to explain the role of $phi$ in your definition of $equiv_{S}$.
$endgroup$
– Rob Arthan
Jan 21 at 21:54
$begingroup$
I answered that here
$endgroup$
– Jishin Noben
Jan 22 at 15:14
$begingroup$
I answered that here
$endgroup$
– Jishin Noben
Jan 22 at 15:14
add a comment |
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$begingroup$
You need to explain the role of $phi$ in your definition of $equiv_{S}$.
$endgroup$
– Rob Arthan
Jan 21 at 21:54
$begingroup$
I answered that here
$endgroup$
– Jishin Noben
Jan 22 at 15:14