Mixing
Validity of $(lozengephiwedgelozengepsi) rightarrow...
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I'm trying to answer a question, one part of which asks me to provide either an informal semantic argument or a counterexample to determine whether $(lozengephiwedgelozengepsi) rightarrow (lozenge(phiwedgelozengepsi)veelozenge(psiwedgelozengephi))$ is a valid formula in system K.
It then asks me, for those parts of the question I've found to be false, to determine what conditions I can impose on the accessibility relation in order to make it valid.
I haven't been able to come up with either a proof or a counterexample, so I'd really appreciate any help you could offer. If it turns out that it isn't valid, then some sort of hint for what conditions are necessary would be really helpful too. Thanks!
logic propositional-calculus first-order-logic modal-logic
asked Jan 20 at 22:35
jblogjblog
182
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add a comment |
$begingroup$
I'm trying to answer a question, one part of which asks me to provide either an informal semantic argument or a counterexample to determine whether $(lozengephiwedgelozengepsi) rightarrow (lozenge(phiwedgelozengepsi)veelozenge(psiwedgelozengephi))$ is a valid formula in system K.
It then asks me, for those parts of the question I've found to be false, to determine what conditions I can impose on the accessibility relation in order to make it valid.
I haven't been able to come up with either a proof or a counterexample, so I'd really appreciate any help you could offer. If it turns out that it isn't valid, then some sort of hint for what conditions are necessary would be really helpful too. Thanks!
logic propositional-calculus first-order-logic modal-logic
asked Jan 20 at 22:35
jblogjblog
182
$endgroup$
add a comment |
$begingroup$
I'm trying to answer a question, one part of which asks me to provide either an informal semantic argument or a counterexample to determine whether $(lozengephiwedgelozengepsi) rightarrow (lozenge(phiwedgelozengepsi)veelozenge(psiwedgelozengephi))$ is a valid formula in system K.
It then asks me, for those parts of the question I've found to be false, to determine what conditions I can impose on the accessibility relation in order to make it valid.
I haven't been able to come up with either a proof or a counterexample, so I'd really appreciate any help you could offer. If it turns out that it isn't valid, then some sort of hint for what conditions are necessary would be really helpful too. Thanks!
logic propositional-calculus first-order-logic modal-logic
asked Jan 20 at 22:35
jblogjblog
182
$endgroup$
I'm trying to answer a question, one part of which asks me to provide either an informal semantic argument or a counterexample to determine whether $(lozengephiwedgelozengepsi) rightarrow (lozenge(phiwedgelozengepsi)veelozenge(psiwedgelozengephi))$ is a valid formula in system K.
It then asks me, for those parts of the question I've found to be false, to determine what conditions I can impose on the accessibility relation in order to make it valid.
I haven't been able to come up with either a proof or a counterexample, so I'd really appreciate any help you could offer. If it turns out that it isn't valid, then some sort of hint for what conditions are necessary would be really helpful too. Thanks!
logic propositional-calculus first-order-logic modal-logic
logic propositional-calculus first-order-logic modal-logic
asked Jan 20 at 22:35
jblogjblog
182
asked Jan 20 at 22:35
jblogjblog
182
asked Jan 20 at 22:35
jblogjblog
182
asked Jan 20 at 22:35
jblogjblog
182
asked Jan 20 at 22:35
jblogjblog
182
182
add a comment |
add a comment |
1 Answer
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Consider a frame with three nodes ${a,b,c}.$ Node $a$ has $b$ and $c$ accessible from it (and $a$ is not accessible from $a).$ No nodes are accessible from $b$ and $c.$ At node $b,$ $phi$ is true and $psi$ is false, whereas at node $c,$ $phi$ is false and $psi$ is true. The statement in question is not true at node $a,$ so it is not valid.
One well-known (but strong) frame condition that guarantees the statement is valid is that the accessibility relation is an equivalence relation, i.e. system S5. In this case, if $lozengephi$ and $lozenge psi$ hold at some node $a$, then the accessible worlds $b$ and $c$ at which $phi$ and $psi$ hold respectively are accessible to each other, so $lozengephi$ and $lozenge psi$ hold at both $b$ and $c.$ Thus, even $lozenge(philand lozenge psi)land lozenge(psiland lozengephi)$ (with a $land,$ not a $lor$) holds at $a.$
A weaker condition that will get us only what we need is that if $aRb$ and $aRc$ then $bRc$ or $cRb.$ (On Wikipedia, they call this $H$ and note that it is given by the axiom $square (square Ato B)lor square(square Bto A).$)
answered Jan 21 at 1:31
spaceisdarkgreenspaceisdarkgreen
33.4k21753
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1 Answer
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1 Answer
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$begingroup$
Consider a frame with three nodes ${a,b,c}.$ Node $a$ has $b$ and $c$ accessible from it (and $a$ is not accessible from $a).$ No nodes are accessible from $b$ and $c.$ At node $b,$ $phi$ is true and $psi$ is false, whereas at node $c,$ $phi$ is false and $psi$ is true. The statement in question is not true at node $a,$ so it is not valid.
One well-known (but strong) frame condition that guarantees the statement is valid is that the accessibility relation is an equivalence relation, i.e. system S5. In this case, if $lozengephi$ and $lozenge psi$ hold at some node $a$, then the accessible worlds $b$ and $c$ at which $phi$ and $psi$ hold respectively are accessible to each other, so $lozengephi$ and $lozenge psi$ hold at both $b$ and $c.$ Thus, even $lozenge(philand lozenge psi)land lozenge(psiland lozengephi)$ (with a $land,$ not a $lor$) holds at $a.$
A weaker condition that will get us only what we need is that if $aRb$ and $aRc$ then $bRc$ or $cRb.$ (On Wikipedia, they call this $H$ and note that it is given by the axiom $square (square Ato B)lor square(square Bto A).$)
answered Jan 21 at 1:31
spaceisdarkgreenspaceisdarkgreen
33.4k21753
$endgroup$
add a comment |
$begingroup$
Consider a frame with three nodes ${a,b,c}.$ Node $a$ has $b$ and $c$ accessible from it (and $a$ is not accessible from $a).$ No nodes are accessible from $b$ and $c.$ At node $b,$ $phi$ is true and $psi$ is false, whereas at node $c,$ $phi$ is false and $psi$ is true. The statement in question is not true at node $a,$ so it is not valid.
One well-known (but strong) frame condition that guarantees the statement is valid is that the accessibility relation is an equivalence relation, i.e. system S5. In this case, if $lozengephi$ and $lozenge psi$ hold at some node $a$, then the accessible worlds $b$ and $c$ at which $phi$ and $psi$ hold respectively are accessible to each other, so $lozengephi$ and $lozenge psi$ hold at both $b$ and $c.$ Thus, even $lozenge(philand lozenge psi)land lozenge(psiland lozengephi)$ (with a $land,$ not a $lor$) holds at $a.$
A weaker condition that will get us only what we need is that if $aRb$ and $aRc$ then $bRc$ or $cRb.$ (On Wikipedia, they call this $H$ and note that it is given by the axiom $square (square Ato B)lor square(square Bto A).$)
answered Jan 21 at 1:31
spaceisdarkgreenspaceisdarkgreen
33.4k21753
$endgroup$
add a comment |
$begingroup$
Consider a frame with three nodes ${a,b,c}.$ Node $a$ has $b$ and $c$ accessible from it (and $a$ is not accessible from $a).$ No nodes are accessible from $b$ and $c.$ At node $b,$ $phi$ is true and $psi$ is false, whereas at node $c,$ $phi$ is false and $psi$ is true. The statement in question is not true at node $a,$ so it is not valid.
One well-known (but strong) frame condition that guarantees the statement is valid is that the accessibility relation is an equivalence relation, i.e. system S5. In this case, if $lozengephi$ and $lozenge psi$ hold at some node $a$, then the accessible worlds $b$ and $c$ at which $phi$ and $psi$ hold respectively are accessible to each other, so $lozengephi$ and $lozenge psi$ hold at both $b$ and $c.$ Thus, even $lozenge(philand lozenge psi)land lozenge(psiland lozengephi)$ (with a $land,$ not a $lor$) holds at $a.$
A weaker condition that will get us only what we need is that if $aRb$ and $aRc$ then $bRc$ or $cRb.$ (On Wikipedia, they call this $H$ and note that it is given by the axiom $square (square Ato B)lor square(square Bto A).$)
answered Jan 21 at 1:31
spaceisdarkgreenspaceisdarkgreen
33.4k21753
$endgroup$
Consider a frame with three nodes ${a,b,c}.$ Node $a$ has $b$ and $c$ accessible from it (and $a$ is not accessible from $a).$ No nodes are accessible from $b$ and $c.$ At node $b,$ $phi$ is true and $psi$ is false, whereas at node $c,$ $phi$ is false and $psi$ is true. The statement in question is not true at node $a,$ so it is not valid.
One well-known (but strong) frame condition that guarantees the statement is valid is that the accessibility relation is an equivalence relation, i.e. system S5. In this case, if $lozengephi$ and $lozenge psi$ hold at some node $a$, then the accessible worlds $b$ and $c$ at which $phi$ and $psi$ hold respectively are accessible to each other, so $lozengephi$ and $lozenge psi$ hold at both $b$ and $c.$ Thus, even $lozenge(philand lozenge psi)land lozenge(psiland lozengephi)$ (with a $land,$ not a $lor$) holds at $a.$
A weaker condition that will get us only what we need is that if $aRb$ and $aRc$ then $bRc$ or $cRb.$ (On Wikipedia, they call this $H$ and note that it is given by the axiom $square (square Ato B)lor square(square Bto A).$)
answered Jan 21 at 1:31
spaceisdarkgreenspaceisdarkgreen
33.4k21753
edited Jan 21 at 1:37
answered Jan 21 at 1:31
spaceisdarkgreenspaceisdarkgreen
33.4k21753
answered Jan 21 at 1:31
spaceisdarkgreenspaceisdarkgreen
33.4k21753
answered Jan 21 at 1:31
spaceisdarkgreenspaceisdarkgreen
33.4k21753
33.4k21753
add a comment |
add a comment |
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