Propositional Intuitionistic Logic QuestionIntuitionist Logic QuestionDo De Morgan's laws hold in propositional intuitionistic logic?Simple proof theory - Propositional LogicIf no interpretations satisfy a set of formulae U, is it possible for $Umodels A$?Logic - Simplifying a propositional logic expressionProgressing in Propositional LogicProofs in propositional logicProving $lnot lnot (psi lor lnot psi)$ is a theorem of intuitionistic propositional logicPropositional Logic questions (Conditionals).Intuitionistic logic and derivationIntuitionist Logic Question
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Propositional Intuitionistic Logic Question
Intuitionist Logic QuestionDo De Morgan's laws hold in propositional intuitionistic logic?Simple proof theory - Propositional LogicIf no interpretations satisfy a set of formulae U, is it possible for $Umodels A$?Logic - Simplifying a propositional logic expressionProgressing in Propositional LogicProofs in propositional logicProving $lnot lnot (psi lor lnot psi)$ is a theorem of intuitionistic propositional logicPropositional Logic questions (Conditionals).Intuitionistic logic and derivationIntuitionist Logic Question
$begingroup$
Show that it is not the case that if $models lnot (Aland B)$ then $models lnot A$ or $models lnot B$.
Show $models lnot (Aland B)$ is true and neither $models lnot A$ not $models lnot B$ is the case.
Consider the formula $lnot (pland lnot p)$.
$models lnot (pland lnot p)$ then $models lnot p$ or $models lnotlnot p$.
Show $models lnot (pland lnot p)$ is true and neither $models lnot p$ nor $models lnotlnot p$ is the case.
$models$ is for validity, the intutionist logical consequence.
$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$.
logic intuitionistic-logic
edited Mar 11 at 22:48
Alex Kruckman
28k32658
asked Mar 11 at 19:47
ContortionistContortionist
12
New contributor
Contortionist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Show that it is not the case that if $models lnot (Aland B)$ then $models lnot A$ or $models lnot B$.
Show $models lnot (Aland B)$ is true and neither $models lnot A$ not $models lnot B$ is the case.
Consider the formula $lnot (pland lnot p)$.
$models lnot (pland lnot p)$ then $models lnot p$ or $models lnotlnot p$.
Show $models lnot (pland lnot p)$ is true and neither $models lnot p$ nor $models lnotlnot p$ is the case.
$models$ is for validity, the intutionist logical consequence.
$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$.
logic intuitionistic-logic
edited Mar 11 at 22:48
Alex Kruckman
28k32658
asked Mar 11 at 19:47
ContortionistContortionist
12
New contributor
Contortionist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
It's a good idea to learn to format your questions with MathJax. Your questions will also be better received if you explain some context (like what your basic definitions are), what you've tried, and where you got stuck and why - rather than just stating the problem and a partial proof. In particular, the line "$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$" is not at all clear.
$endgroup$
– Alex Kruckman
Mar 11 at 22:41
$begingroup$
(I've added the MathJax formatting, to show you an example).
$endgroup$
– Alex Kruckman
Mar 11 at 22:48
add a comment |
$begingroup$
Show that it is not the case that if $models lnot (Aland B)$ then $models lnot A$ or $models lnot B$.
Show $models lnot (Aland B)$ is true and neither $models lnot A$ not $models lnot B$ is the case.
Consider the formula $lnot (pland lnot p)$.
$models lnot (pland lnot p)$ then $models lnot p$ or $models lnotlnot p$.
Show $models lnot (pland lnot p)$ is true and neither $models lnot p$ nor $models lnotlnot p$ is the case.
$models$ is for validity, the intutionist logical consequence.
$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$.
logic intuitionistic-logic
edited Mar 11 at 22:48
Alex Kruckman
28k32658
asked Mar 11 at 19:47
ContortionistContortionist
12
New contributor
Contortionist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Show that it is not the case that if $models lnot (Aland B)$ then $models lnot A$ or $models lnot B$.
Show $models lnot (Aland B)$ is true and neither $models lnot A$ not $models lnot B$ is the case.
Consider the formula $lnot (pland lnot p)$.
$models lnot (pland lnot p)$ then $models lnot p$ or $models lnotlnot p$.
Show $models lnot (pland lnot p)$ is true and neither $models lnot p$ nor $models lnotlnot p$ is the case.
$models$ is for validity, the intutionist logical consequence.
$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$.
logic intuitionistic-logic
logic intuitionistic-logic
edited Mar 11 at 22:48
Alex Kruckman
28k32658
asked Mar 11 at 19:47
ContortionistContortionist
12
New contributor
Contortionist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 11 at 22:48
Alex Kruckman
28k32658
asked Mar 11 at 19:47
ContortionistContortionist
12
New contributor
Contortionist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 11 at 22:48
Alex Kruckman
28k32658
edited Mar 11 at 22:48
Alex Kruckman
28k32658
edited Mar 11 at 22:48
Alex Kruckman
28k32658
28k32658
asked Mar 11 at 19:47
ContortionistContortionist
12
New contributor
Contortionist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 19:47
ContortionistContortionist
12
asked Mar 11 at 19:47
ContortionistContortionist
12
12
New contributor
Contortionist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Contortionist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Contortionist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
It's a good idea to learn to format your questions with MathJax. Your questions will also be better received if you explain some context (like what your basic definitions are), what you've tried, and where you got stuck and why - rather than just stating the problem and a partial proof. In particular, the line "$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$" is not at all clear.
$endgroup$
– Alex Kruckman
Mar 11 at 22:41
$begingroup$
(I've added the MathJax formatting, to show you an example).
$endgroup$
– Alex Kruckman
Mar 11 at 22:48
add a comment |
$begingroup$
It's a good idea to learn to format your questions with MathJax. Your questions will also be better received if you explain some context (like what your basic definitions are), what you've tried, and where you got stuck and why - rather than just stating the problem and a partial proof. In particular, the line "$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$" is not at all clear.
$endgroup$
– Alex Kruckman
Mar 11 at 22:41
$begingroup$
(I've added the MathJax formatting, to show you an example).
$endgroup$
– Alex Kruckman
Mar 11 at 22:48
$begingroup$
It's a good idea to learn to format your questions with MathJax. Your questions will also be better received if you explain some context (like what your basic definitions are), what you've tried, and where you got stuck and why - rather than just stating the problem and a partial proof. In particular, the line "$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$" is not at all clear.
$endgroup$
– Alex Kruckman
Mar 11 at 22:41
$begingroup$
It's a good idea to learn to format your questions with MathJax. Your questions will also be better received if you explain some context (like what your basic definitions are), what you've tried, and where you got stuck and why - rather than just stating the problem and a partial proof. In particular, the line "$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$" is not at all clear.
$endgroup$
– Alex Kruckman
Mar 11 at 22:41
$begingroup$
(I've added the MathJax formatting, to show you an example).
$endgroup$
– Alex Kruckman
Mar 11 at 22:48
$begingroup$
(I've added the MathJax formatting, to show you an example).
$endgroup$
– Alex Kruckman
Mar 11 at 22:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are several notions of semantics for intuitionistic logic, and hence several meanings of $models$. If you're going to talk about the $models$ relation in intuitionistic logic, you should really make sure that everyone is on the same page by specifying what it means. Based on your question, I'm going to assume you are using Kripke Semantics.
First things first: In intuitionistic logic, $lnot varphi$ can be viewed as an abbreviation for $varphirightarrow bot$. So we have $models lnot varphi$ iff for every intuitionistic Kripke model $M$ and every world $win M$, $wVdash varphirightarrow bot$. By definition, this means that for all $ugeq w$, if $uVdash varphi$, then $uVdash bot$. But we always have $unotVdash bot$. So the condition is equivalent to saying that that for all $ugeq w$, $unotVdash varphi$.
Summing up:
- We have $models lnot varphi$ if and only if for every model $M$ and every world $win M$, $wVdash lnot varphi$.
- If $M$ is a model and $win M$ is a world, then we have $wVdash lnot varphi$ if and only if for all $ugeq w$, $unotVdash varphi$.
Ok, following your question, let's show that $models lnot (pland lnot p)$, but $notmodels lnot p$ and $notmodels lnotlnot p$.
To show that $models lnot (pland lnot p)$, we have to consider an arbitrary model $M$ and an arbitrary world $win M$ and show that for all $ugeq w$, $unotVdash pland lnot p$. By definition, $uVdash pland lnot p$ if and only if $uVdash p$ and $uVdash lnot p$. But if $uVdash lnot p$, we have that for all $u'geq u$, $u'notVdash p$. In particular, taking $u' = u$, we have $unotVdash p$, which contradicts $uVdash p$. So $unotVdash pland lnot p$, as desired.
To show that $notmodels lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = w$, i.e. $wVdash p$. Then since $wgeq w$ and $wVdash p$, we have $wnotVdash lnot p$.
To show that $notmodels lnot lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnotlnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = emptyset$, i.e. $wnotVdash p$. Then for all $w'geq w$ ($w' = w$ is the only choice), we have $w'notVdash p$, so $wVdash lnot p$. And since $wgeq w$ and $wVdash lnot p$, we have $wnotVdash lnotlnot p$.
answered Mar 11 at 22:32
Alex KruckmanAlex Kruckman
28k32658
$endgroup$
$begingroup$
Thank you. Not sure if I understand. Would the model M for ⊭¬p be W = {w), wRw, vw(p) =1? And M for ⊭¬¬p be W = w, wRw, vwp(0)?
$endgroup$
– Contortionist
Mar 11 at 23:22
$begingroup$
If $vw(p)$ means the valuation of $p$ at $w$, then yes. In my notation, $v(p)$ is the set of worlds at which $p$ is true, i.e. $v(p) = win Mmid vw(p)=1$.
$endgroup$
– Alex Kruckman
Mar 12 at 0:54
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There are several notions of semantics for intuitionistic logic, and hence several meanings of $models$. If you're going to talk about the $models$ relation in intuitionistic logic, you should really make sure that everyone is on the same page by specifying what it means. Based on your question, I'm going to assume you are using Kripke Semantics.
First things first: In intuitionistic logic, $lnot varphi$ can be viewed as an abbreviation for $varphirightarrow bot$. So we have $models lnot varphi$ iff for every intuitionistic Kripke model $M$ and every world $win M$, $wVdash varphirightarrow bot$. By definition, this means that for all $ugeq w$, if $uVdash varphi$, then $uVdash bot$. But we always have $unotVdash bot$. So the condition is equivalent to saying that that for all $ugeq w$, $unotVdash varphi$.
Summing up:
- We have $models lnot varphi$ if and only if for every model $M$ and every world $win M$, $wVdash lnot varphi$.
- If $M$ is a model and $win M$ is a world, then we have $wVdash lnot varphi$ if and only if for all $ugeq w$, $unotVdash varphi$.
Ok, following your question, let's show that $models lnot (pland lnot p)$, but $notmodels lnot p$ and $notmodels lnotlnot p$.
To show that $models lnot (pland lnot p)$, we have to consider an arbitrary model $M$ and an arbitrary world $win M$ and show that for all $ugeq w$, $unotVdash pland lnot p$. By definition, $uVdash pland lnot p$ if and only if $uVdash p$ and $uVdash lnot p$. But if $uVdash lnot p$, we have that for all $u'geq u$, $u'notVdash p$. In particular, taking $u' = u$, we have $unotVdash p$, which contradicts $uVdash p$. So $unotVdash pland lnot p$, as desired.
To show that $notmodels lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = w$, i.e. $wVdash p$. Then since $wgeq w$ and $wVdash p$, we have $wnotVdash lnot p$.
To show that $notmodels lnot lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnotlnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = emptyset$, i.e. $wnotVdash p$. Then for all $w'geq w$ ($w' = w$ is the only choice), we have $w'notVdash p$, so $wVdash lnot p$. And since $wgeq w$ and $wVdash lnot p$, we have $wnotVdash lnotlnot p$.
answered Mar 11 at 22:32
Alex KruckmanAlex Kruckman
28k32658
$endgroup$
$begingroup$
Thank you. Not sure if I understand. Would the model M for ⊭¬p be W = {w), wRw, vw(p) =1? And M for ⊭¬¬p be W = w, wRw, vwp(0)?
$endgroup$
– Contortionist
Mar 11 at 23:22
$begingroup$
If $vw(p)$ means the valuation of $p$ at $w$, then yes. In my notation, $v(p)$ is the set of worlds at which $p$ is true, i.e. $v(p) = win Mmid vw(p)=1$.
$endgroup$
– Alex Kruckman
Mar 12 at 0:54
add a comment |
$begingroup$
There are several notions of semantics for intuitionistic logic, and hence several meanings of $models$. If you're going to talk about the $models$ relation in intuitionistic logic, you should really make sure that everyone is on the same page by specifying what it means. Based on your question, I'm going to assume you are using Kripke Semantics.
First things first: In intuitionistic logic, $lnot varphi$ can be viewed as an abbreviation for $varphirightarrow bot$. So we have $models lnot varphi$ iff for every intuitionistic Kripke model $M$ and every world $win M$, $wVdash varphirightarrow bot$. By definition, this means that for all $ugeq w$, if $uVdash varphi$, then $uVdash bot$. But we always have $unotVdash bot$. So the condition is equivalent to saying that that for all $ugeq w$, $unotVdash varphi$.
Summing up:
- We have $models lnot varphi$ if and only if for every model $M$ and every world $win M$, $wVdash lnot varphi$.
- If $M$ is a model and $win M$ is a world, then we have $wVdash lnot varphi$ if and only if for all $ugeq w$, $unotVdash varphi$.
Ok, following your question, let's show that $models lnot (pland lnot p)$, but $notmodels lnot p$ and $notmodels lnotlnot p$.
To show that $models lnot (pland lnot p)$, we have to consider an arbitrary model $M$ and an arbitrary world $win M$ and show that for all $ugeq w$, $unotVdash pland lnot p$. By definition, $uVdash pland lnot p$ if and only if $uVdash p$ and $uVdash lnot p$. But if $uVdash lnot p$, we have that for all $u'geq u$, $u'notVdash p$. In particular, taking $u' = u$, we have $unotVdash p$, which contradicts $uVdash p$. So $unotVdash pland lnot p$, as desired.
To show that $notmodels lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = w$, i.e. $wVdash p$. Then since $wgeq w$ and $wVdash p$, we have $wnotVdash lnot p$.
To show that $notmodels lnot lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnotlnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = emptyset$, i.e. $wnotVdash p$. Then for all $w'geq w$ ($w' = w$ is the only choice), we have $w'notVdash p$, so $wVdash lnot p$. And since $wgeq w$ and $wVdash lnot p$, we have $wnotVdash lnotlnot p$.
answered Mar 11 at 22:32
Alex KruckmanAlex Kruckman
28k32658
$endgroup$
$begingroup$
Thank you. Not sure if I understand. Would the model M for ⊭¬p be W = {w), wRw, vw(p) =1? And M for ⊭¬¬p be W = w, wRw, vwp(0)?
$endgroup$
– Contortionist
Mar 11 at 23:22
$begingroup$
If $vw(p)$ means the valuation of $p$ at $w$, then yes. In my notation, $v(p)$ is the set of worlds at which $p$ is true, i.e. $v(p) = win Mmid vw(p)=1$.
$endgroup$
– Alex Kruckman
Mar 12 at 0:54
add a comment |
$begingroup$
There are several notions of semantics for intuitionistic logic, and hence several meanings of $models$. If you're going to talk about the $models$ relation in intuitionistic logic, you should really make sure that everyone is on the same page by specifying what it means. Based on your question, I'm going to assume you are using Kripke Semantics.
First things first: In intuitionistic logic, $lnot varphi$ can be viewed as an abbreviation for $varphirightarrow bot$. So we have $models lnot varphi$ iff for every intuitionistic Kripke model $M$ and every world $win M$, $wVdash varphirightarrow bot$. By definition, this means that for all $ugeq w$, if $uVdash varphi$, then $uVdash bot$. But we always have $unotVdash bot$. So the condition is equivalent to saying that that for all $ugeq w$, $unotVdash varphi$.
Summing up:
- We have $models lnot varphi$ if and only if for every model $M$ and every world $win M$, $wVdash lnot varphi$.
- If $M$ is a model and $win M$ is a world, then we have $wVdash lnot varphi$ if and only if for all $ugeq w$, $unotVdash varphi$.
Ok, following your question, let's show that $models lnot (pland lnot p)$, but $notmodels lnot p$ and $notmodels lnotlnot p$.
To show that $models lnot (pland lnot p)$, we have to consider an arbitrary model $M$ and an arbitrary world $win M$ and show that for all $ugeq w$, $unotVdash pland lnot p$. By definition, $uVdash pland lnot p$ if and only if $uVdash p$ and $uVdash lnot p$. But if $uVdash lnot p$, we have that for all $u'geq u$, $u'notVdash p$. In particular, taking $u' = u$, we have $unotVdash p$, which contradicts $uVdash p$. So $unotVdash pland lnot p$, as desired.
To show that $notmodels lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = w$, i.e. $wVdash p$. Then since $wgeq w$ and $wVdash p$, we have $wnotVdash lnot p$.
To show that $notmodels lnot lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnotlnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = emptyset$, i.e. $wnotVdash p$. Then for all $w'geq w$ ($w' = w$ is the only choice), we have $w'notVdash p$, so $wVdash lnot p$. And since $wgeq w$ and $wVdash lnot p$, we have $wnotVdash lnotlnot p$.
answered Mar 11 at 22:32
Alex KruckmanAlex Kruckman
28k32658
$endgroup$
There are several notions of semantics for intuitionistic logic, and hence several meanings of $models$. If you're going to talk about the $models$ relation in intuitionistic logic, you should really make sure that everyone is on the same page by specifying what it means. Based on your question, I'm going to assume you are using Kripke Semantics.
First things first: In intuitionistic logic, $lnot varphi$ can be viewed as an abbreviation for $varphirightarrow bot$. So we have $models lnot varphi$ iff for every intuitionistic Kripke model $M$ and every world $win M$, $wVdash varphirightarrow bot$. By definition, this means that for all $ugeq w$, if $uVdash varphi$, then $uVdash bot$. But we always have $unotVdash bot$. So the condition is equivalent to saying that that for all $ugeq w$, $unotVdash varphi$.
Summing up:
- We have $models lnot varphi$ if and only if for every model $M$ and every world $win M$, $wVdash lnot varphi$.
- If $M$ is a model and $win M$ is a world, then we have $wVdash lnot varphi$ if and only if for all $ugeq w$, $unotVdash varphi$.
Ok, following your question, let's show that $models lnot (pland lnot p)$, but $notmodels lnot p$ and $notmodels lnotlnot p$.
To show that $models lnot (pland lnot p)$, we have to consider an arbitrary model $M$ and an arbitrary world $win M$ and show that for all $ugeq w$, $unotVdash pland lnot p$. By definition, $uVdash pland lnot p$ if and only if $uVdash p$ and $uVdash lnot p$. But if $uVdash lnot p$, we have that for all $u'geq u$, $u'notVdash p$. In particular, taking $u' = u$, we have $unotVdash p$, which contradicts $uVdash p$. So $unotVdash pland lnot p$, as desired.
To show that $notmodels lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = w$, i.e. $wVdash p$. Then since $wgeq w$ and $wVdash p$, we have $wnotVdash lnot p$.
To show that $notmodels lnot lnot p$, we just need to find a single model $M$ and a single world $w$ such that $wnotVdash lnotlnot p$. Well, consider a model $M$ with a single world $w$ such that $v(p) = emptyset$, i.e. $wnotVdash p$. Then for all $w'geq w$ ($w' = w$ is the only choice), we have $w'notVdash p$, so $wVdash lnot p$. And since $wgeq w$ and $wVdash lnot p$, we have $wnotVdash lnotlnot p$.
answered Mar 11 at 22:32
Alex KruckmanAlex Kruckman
28k32658
edited Mar 11 at 22:39
answered Mar 11 at 22:32
Alex KruckmanAlex Kruckman
28k32658
answered Mar 11 at 22:32
Alex KruckmanAlex Kruckman
28k32658
answered Mar 11 at 22:32
Alex KruckmanAlex Kruckman
28k32658
28k32658
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Thank you. Not sure if I understand. Would the model M for ⊭¬p be W = {w), wRw, vw(p) =1? And M for ⊭¬¬p be W = w, wRw, vwp(0)?
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– Contortionist
Mar 11 at 23:22
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If $vw(p)$ means the valuation of $p$ at $w$, then yes. In my notation, $v(p)$ is the set of worlds at which $p$ is true, i.e. $v(p) = win Mmid vw(p)=1$.
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– Alex Kruckman
Mar 12 at 0:54
add a comment |
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Thank you. Not sure if I understand. Would the model M for ⊭¬p be W = {w), wRw, vw(p) =1? And M for ⊭¬¬p be W = w, wRw, vwp(0)?
$endgroup$
– Contortionist
Mar 11 at 23:22
$begingroup$
If $vw(p)$ means the valuation of $p$ at $w$, then yes. In my notation, $v(p)$ is the set of worlds at which $p$ is true, i.e. $v(p) = win Mmid vw(p)=1$.
$endgroup$
– Alex Kruckman
Mar 12 at 0:54
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Thank you. Not sure if I understand. Would the model M for ⊭¬p be W = {w), wRw, vw(p) =1? And M for ⊭¬¬p be W = w, wRw, vwp(0)?
$endgroup$
– Contortionist
Mar 11 at 23:22
$begingroup$
Thank you. Not sure if I understand. Would the model M for ⊭¬p be W = {w), wRw, vw(p) =1? And M for ⊭¬¬p be W = w, wRw, vwp(0)?
$endgroup$
– Contortionist
Mar 11 at 23:22
$begingroup$
If $vw(p)$ means the valuation of $p$ at $w$, then yes. In my notation, $v(p)$ is the set of worlds at which $p$ is true, i.e. $v(p) = win Mmid vw(p)=1$.
$endgroup$
– Alex Kruckman
Mar 12 at 0:54
$begingroup$
If $vw(p)$ means the valuation of $p$ at $w$, then yes. In my notation, $v(p)$ is the set of worlds at which $p$ is true, i.e. $v(p) = win Mmid vw(p)=1$.
$endgroup$
– Alex Kruckman
Mar 12 at 0:54
add a comment |
Contortionist is a new contributor. Be nice, and check out our Code of Conduct.
Contortionist is a new contributor. Be nice, and check out our Code of Conduct.
Contortionist is a new contributor. Be nice, and check out our Code of Conduct.
Contortionist is a new contributor. Be nice, and check out our Code of Conduct.
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It's a good idea to learn to format your questions with MathJax. Your questions will also be better received if you explain some context (like what your basic definitions are), what you've tried, and where you got stuck and why - rather than just stating the problem and a partial proof. In particular, the line "$models lnot p$ if $W = w$, $wRw$, $vw(p) = 0$" is not at all clear.
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– Alex Kruckman
Mar 11 at 22:41
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(I've added the MathJax formatting, to show you an example).
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– Alex Kruckman
Mar 11 at 22:48