prove that a frame is symmetric in modal logic
$begingroup$
Im trying to prove or refute that ⋄□A→A characterizes symmetry. I can construct a counterexample, where [W,R, V] would be an interpretation W = {w1,w2}, and the accessibility Relations w1 Rw2, w2 Rw3, Hence the Kripke Frame [W,R] is symmetric. Now Im trying to define a variable assignment that would show that symmetry doesnt hold
logic modal-logic
edited Dec 9 '18 at 14:02
David C. Ullrich
60k43994
asked Dec 9 '18 at 12:37
Martin LMartin L
61
$endgroup$
add a comment |
$begingroup$
Im trying to prove or refute that ⋄□A→A characterizes symmetry. I can construct a counterexample, where [W,R, V] would be an interpretation W = {w1,w2}, and the accessibility Relations w1 Rw2, w2 Rw3, Hence the Kripke Frame [W,R] is symmetric. Now Im trying to define a variable assignment that would show that symmetry doesnt hold
logic modal-logic
edited Dec 9 '18 at 14:02
David C. Ullrich
60k43994
asked Dec 9 '18 at 12:37
Martin LMartin L
61
$endgroup$
1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
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– José Carlos Santos
Dec 9 '18 at 12:42
$begingroup$
Let $A$ be true only in $w_3$. Then $w_1Vdash diamondBox A$.
$endgroup$
– Berci
Dec 9 '18 at 22:26
add a comment |
$begingroup$
Im trying to prove or refute that ⋄□A→A characterizes symmetry. I can construct a counterexample, where [W,R, V] would be an interpretation W = {w1,w2}, and the accessibility Relations w1 Rw2, w2 Rw3, Hence the Kripke Frame [W,R] is symmetric. Now Im trying to define a variable assignment that would show that symmetry doesnt hold
logic modal-logic
edited Dec 9 '18 at 14:02
David C. Ullrich
60k43994
asked Dec 9 '18 at 12:37
Martin LMartin L
61
$endgroup$
Im trying to prove or refute that ⋄□A→A characterizes symmetry. I can construct a counterexample, where [W,R, V] would be an interpretation W = {w1,w2}, and the accessibility Relations w1 Rw2, w2 Rw3, Hence the Kripke Frame [W,R] is symmetric. Now Im trying to define a variable assignment that would show that symmetry doesnt hold
logic modal-logic
logic modal-logic
edited Dec 9 '18 at 14:02
David C. Ullrich
60k43994
asked Dec 9 '18 at 12:37
Martin LMartin L
61
edited Dec 9 '18 at 14:02
David C. Ullrich
60k43994
asked Dec 9 '18 at 12:37
Martin LMartin L
61
edited Dec 9 '18 at 14:02
David C. Ullrich
60k43994
edited Dec 9 '18 at 14:02
David C. Ullrich
60k43994
edited Dec 9 '18 at 14:02
David C. Ullrich
60k43994
60k43994
asked Dec 9 '18 at 12:37
Martin LMartin L
61
asked Dec 9 '18 at 12:37
Martin LMartin L
61
asked Dec 9 '18 at 12:37
Martin LMartin L
61
61
1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 12:42
$begingroup$
Let $A$ be true only in $w_3$. Then $w_1Vdash diamondBox A$.
$endgroup$
– Berci
Dec 9 '18 at 22:26
add a comment |
1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 12:42
$begingroup$
Let $A$ be true only in $w_3$. Then $w_1Vdash diamondBox A$.
$endgroup$
– Berci
Dec 9 '18 at 22:26
1
1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 12:42
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 12:42
$begingroup$
Let $A$ be true only in $w_3$. Then $w_1Vdash diamondBox A$.
$endgroup$
– Berci
Dec 9 '18 at 22:26
$begingroup$
Let $A$ be true only in $w_3$. Then $w_1Vdash diamondBox A$.
$endgroup$
– Berci
Dec 9 '18 at 22:26
add a comment |
1 Answer
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$begingroup$
I think you need to show that it indeed characterizes symmetry.
Since looking at your start formula:
$diamond square A rightarrow A$
the contra positive must hold:
$neg diamond square A leftarrow neg A$
Applying the definitions of $square A leftrightarrow neg diamond neg A$ and $diamond A leftrightarrow neg square neg A$ you can derive:
$neg neg square neg neg diamond neg A leftarrow neg A$
which then can be simplefied to:
$neg A rightarrow square diamond neg A$
which looks very similar to the definiton of symmetry in modal logic:
$P rightarrow square diamond P$
Also you have a little typo in $w_2 R w_3$, which probably should be $w_2 R w_1$.
I think you can find the rest you need here https://plato.stanford.edu/entries/logic-modal/.
To show that it characterizes symmetry you have to show that $(F models diamond square A rightarrow A) Leftrightarrow (F$ is symmetric). You can do this by proving each implication direction separately. Remember that for a frame $F=<W,R>$ with $F models diamond square A rightarrow A$, it must hold that forall models of the frame $M=<W,R,V>$, $M models diamond square A rightarrow A$
answered Dec 9 '18 at 14:02
LogiCLogiC
263
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add a comment |
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$begingroup$
I think you need to show that it indeed characterizes symmetry.
Since looking at your start formula:
$diamond square A rightarrow A$
the contra positive must hold:
$neg diamond square A leftarrow neg A$
Applying the definitions of $square A leftrightarrow neg diamond neg A$ and $diamond A leftrightarrow neg square neg A$ you can derive:
$neg neg square neg neg diamond neg A leftarrow neg A$
which then can be simplefied to:
$neg A rightarrow square diamond neg A$
which looks very similar to the definiton of symmetry in modal logic:
$P rightarrow square diamond P$
Also you have a little typo in $w_2 R w_3$, which probably should be $w_2 R w_1$.
I think you can find the rest you need here https://plato.stanford.edu/entries/logic-modal/.
To show that it characterizes symmetry you have to show that $(F models diamond square A rightarrow A) Leftrightarrow (F$ is symmetric). You can do this by proving each implication direction separately. Remember that for a frame $F=<W,R>$ with $F models diamond square A rightarrow A$, it must hold that forall models of the frame $M=<W,R,V>$, $M models diamond square A rightarrow A$
answered Dec 9 '18 at 14:02
LogiCLogiC
263
$endgroup$
add a comment |
$begingroup$
I think you need to show that it indeed characterizes symmetry.
Since looking at your start formula:
$diamond square A rightarrow A$
the contra positive must hold:
$neg diamond square A leftarrow neg A$
Applying the definitions of $square A leftrightarrow neg diamond neg A$ and $diamond A leftrightarrow neg square neg A$ you can derive:
$neg neg square neg neg diamond neg A leftarrow neg A$
which then can be simplefied to:
$neg A rightarrow square diamond neg A$
which looks very similar to the definiton of symmetry in modal logic:
$P rightarrow square diamond P$
Also you have a little typo in $w_2 R w_3$, which probably should be $w_2 R w_1$.
I think you can find the rest you need here https://plato.stanford.edu/entries/logic-modal/.
To show that it characterizes symmetry you have to show that $(F models diamond square A rightarrow A) Leftrightarrow (F$ is symmetric). You can do this by proving each implication direction separately. Remember that for a frame $F=<W,R>$ with $F models diamond square A rightarrow A$, it must hold that forall models of the frame $M=<W,R,V>$, $M models diamond square A rightarrow A$
answered Dec 9 '18 at 14:02
LogiCLogiC
263
$endgroup$
add a comment |
$begingroup$
I think you need to show that it indeed characterizes symmetry.
Since looking at your start formula:
$diamond square A rightarrow A$
the contra positive must hold:
$neg diamond square A leftarrow neg A$
Applying the definitions of $square A leftrightarrow neg diamond neg A$ and $diamond A leftrightarrow neg square neg A$ you can derive:
$neg neg square neg neg diamond neg A leftarrow neg A$
which then can be simplefied to:
$neg A rightarrow square diamond neg A$
which looks very similar to the definiton of symmetry in modal logic:
$P rightarrow square diamond P$
Also you have a little typo in $w_2 R w_3$, which probably should be $w_2 R w_1$.
I think you can find the rest you need here https://plato.stanford.edu/entries/logic-modal/.
To show that it characterizes symmetry you have to show that $(F models diamond square A rightarrow A) Leftrightarrow (F$ is symmetric). You can do this by proving each implication direction separately. Remember that for a frame $F=<W,R>$ with $F models diamond square A rightarrow A$, it must hold that forall models of the frame $M=<W,R,V>$, $M models diamond square A rightarrow A$
answered Dec 9 '18 at 14:02
LogiCLogiC
263
$endgroup$
I think you need to show that it indeed characterizes symmetry.
Since looking at your start formula:
$diamond square A rightarrow A$
the contra positive must hold:
$neg diamond square A leftarrow neg A$
Applying the definitions of $square A leftrightarrow neg diamond neg A$ and $diamond A leftrightarrow neg square neg A$ you can derive:
$neg neg square neg neg diamond neg A leftarrow neg A$
which then can be simplefied to:
$neg A rightarrow square diamond neg A$
which looks very similar to the definiton of symmetry in modal logic:
$P rightarrow square diamond P$
Also you have a little typo in $w_2 R w_3$, which probably should be $w_2 R w_1$.
I think you can find the rest you need here https://plato.stanford.edu/entries/logic-modal/.
To show that it characterizes symmetry you have to show that $(F models diamond square A rightarrow A) Leftrightarrow (F$ is symmetric). You can do this by proving each implication direction separately. Remember that for a frame $F=<W,R>$ with $F models diamond square A rightarrow A$, it must hold that forall models of the frame $M=<W,R,V>$, $M models diamond square A rightarrow A$
answered Dec 9 '18 at 14:02
LogiCLogiC
263
edited Dec 9 '18 at 16:04
answered Dec 9 '18 at 14:02
LogiCLogiC
263
answered Dec 9 '18 at 14:02
LogiCLogiC
263
answered Dec 9 '18 at 14:02
LogiCLogiC
263
263
add a comment |
add a comment |
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$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 12:42
$begingroup$
Let $A$ be true only in $w_3$. Then $w_1Vdash diamondBox A$.
$endgroup$
– Berci
Dec 9 '18 at 22:26