How can we prove soundness property if it's possible for our assumption set to contain false assumptions?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












Soundness property, to my knowledge is the property that:



$Gamma vdash varphi implies Gamma vDash varphi$



If $varphi$ is provable (a syntactic consequence) from $Gamma$ then $varphi$ is also a semantic consequence of $varphi$, which I believe is saying $varphi$ is "true".



But what if, for example, $varphi in Gamma$ and $varphi = bot$? It appears conceivable that some of the assumptions in $Gamma$ are false, and then we might be able to prove things from it, but semantically they would be false.



It's possible I just have the definition of soundness wrong but how is this accounted for? We would normally say that the Hilbert system is both complete and sound but is this still the case even if we begin with a $Gamma$ that contains some false premises? Or is it "sound only in certain cases"? How does this work?










share|cite|improve this question

















  • 2




    If $Gamma$ contains a contradiction, then $Gamma$ is unsatisfiable, so $Gamma vDash psi$ is vacuously true. The interesting and nontrivial case is when $Gamma$ is consistent.
    – Carl Mummert
    56 mins ago











  • @CarlMummert Is "contradiction" same as false, $bot$, etc? What do you mean by unsatisfiable?
    – user525966
    49 mins ago











  • I mean that no model satisfies $bot$, so a theory that contains $bot$ is not satisfiable. In that case, trivially "every model of the theory will also be a model of $phi$", because the theory has no models. For example, for any $Gamma$ we have $Gamma vdash bot Longrightarrow Gamma vDash bot$.
    – Carl Mummert
    45 mins ago











  • @CarlMummert Is a "model" a particular set of boolean inputs to the atomic variables of $Gamma$? Does "satisfies" mean "everything in $Gamma$ evaluates to true under a specific model"?
    – user525966
    41 mins ago







  • 1




    Yes. We would say $Gamma$ is satisfiable because there is at least one model/interpretation that makes every formula in $Gamma$ true.
    – Carl Mummert
    38 mins ago















up vote
1
down vote

favorite












Soundness property, to my knowledge is the property that:



$Gamma vdash varphi implies Gamma vDash varphi$



If $varphi$ is provable (a syntactic consequence) from $Gamma$ then $varphi$ is also a semantic consequence of $varphi$, which I believe is saying $varphi$ is "true".



But what if, for example, $varphi in Gamma$ and $varphi = bot$? It appears conceivable that some of the assumptions in $Gamma$ are false, and then we might be able to prove things from it, but semantically they would be false.



It's possible I just have the definition of soundness wrong but how is this accounted for? We would normally say that the Hilbert system is both complete and sound but is this still the case even if we begin with a $Gamma$ that contains some false premises? Or is it "sound only in certain cases"? How does this work?










share|cite|improve this question

















  • 2




    If $Gamma$ contains a contradiction, then $Gamma$ is unsatisfiable, so $Gamma vDash psi$ is vacuously true. The interesting and nontrivial case is when $Gamma$ is consistent.
    – Carl Mummert
    56 mins ago











  • @CarlMummert Is "contradiction" same as false, $bot$, etc? What do you mean by unsatisfiable?
    – user525966
    49 mins ago











  • I mean that no model satisfies $bot$, so a theory that contains $bot$ is not satisfiable. In that case, trivially "every model of the theory will also be a model of $phi$", because the theory has no models. For example, for any $Gamma$ we have $Gamma vdash bot Longrightarrow Gamma vDash bot$.
    – Carl Mummert
    45 mins ago











  • @CarlMummert Is a "model" a particular set of boolean inputs to the atomic variables of $Gamma$? Does "satisfies" mean "everything in $Gamma$ evaluates to true under a specific model"?
    – user525966
    41 mins ago







  • 1




    Yes. We would say $Gamma$ is satisfiable because there is at least one model/interpretation that makes every formula in $Gamma$ true.
    – Carl Mummert
    38 mins ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Soundness property, to my knowledge is the property that:



$Gamma vdash varphi implies Gamma vDash varphi$



If $varphi$ is provable (a syntactic consequence) from $Gamma$ then $varphi$ is also a semantic consequence of $varphi$, which I believe is saying $varphi$ is "true".



But what if, for example, $varphi in Gamma$ and $varphi = bot$? It appears conceivable that some of the assumptions in $Gamma$ are false, and then we might be able to prove things from it, but semantically they would be false.



It's possible I just have the definition of soundness wrong but how is this accounted for? We would normally say that the Hilbert system is both complete and sound but is this still the case even if we begin with a $Gamma$ that contains some false premises? Or is it "sound only in certain cases"? How does this work?










share|cite|improve this question













Soundness property, to my knowledge is the property that:



$Gamma vdash varphi implies Gamma vDash varphi$



If $varphi$ is provable (a syntactic consequence) from $Gamma$ then $varphi$ is also a semantic consequence of $varphi$, which I believe is saying $varphi$ is "true".



But what if, for example, $varphi in Gamma$ and $varphi = bot$? It appears conceivable that some of the assumptions in $Gamma$ are false, and then we might be able to prove things from it, but semantically they would be false.



It's possible I just have the definition of soundness wrong but how is this accounted for? We would normally say that the Hilbert system is both complete and sound but is this still the case even if we begin with a $Gamma$ that contains some false premises? Or is it "sound only in certain cases"? How does this work?







logic definition propositional-calculus hilbert-calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









user525966

1,770719




1,770719







  • 2




    If $Gamma$ contains a contradiction, then $Gamma$ is unsatisfiable, so $Gamma vDash psi$ is vacuously true. The interesting and nontrivial case is when $Gamma$ is consistent.
    – Carl Mummert
    56 mins ago











  • @CarlMummert Is "contradiction" same as false, $bot$, etc? What do you mean by unsatisfiable?
    – user525966
    49 mins ago











  • I mean that no model satisfies $bot$, so a theory that contains $bot$ is not satisfiable. In that case, trivially "every model of the theory will also be a model of $phi$", because the theory has no models. For example, for any $Gamma$ we have $Gamma vdash bot Longrightarrow Gamma vDash bot$.
    – Carl Mummert
    45 mins ago











  • @CarlMummert Is a "model" a particular set of boolean inputs to the atomic variables of $Gamma$? Does "satisfies" mean "everything in $Gamma$ evaluates to true under a specific model"?
    – user525966
    41 mins ago







  • 1




    Yes. We would say $Gamma$ is satisfiable because there is at least one model/interpretation that makes every formula in $Gamma$ true.
    – Carl Mummert
    38 mins ago













  • 2




    If $Gamma$ contains a contradiction, then $Gamma$ is unsatisfiable, so $Gamma vDash psi$ is vacuously true. The interesting and nontrivial case is when $Gamma$ is consistent.
    – Carl Mummert
    56 mins ago











  • @CarlMummert Is "contradiction" same as false, $bot$, etc? What do you mean by unsatisfiable?
    – user525966
    49 mins ago











  • I mean that no model satisfies $bot$, so a theory that contains $bot$ is not satisfiable. In that case, trivially "every model of the theory will also be a model of $phi$", because the theory has no models. For example, for any $Gamma$ we have $Gamma vdash bot Longrightarrow Gamma vDash bot$.
    – Carl Mummert
    45 mins ago











  • @CarlMummert Is a "model" a particular set of boolean inputs to the atomic variables of $Gamma$? Does "satisfies" mean "everything in $Gamma$ evaluates to true under a specific model"?
    – user525966
    41 mins ago







  • 1




    Yes. We would say $Gamma$ is satisfiable because there is at least one model/interpretation that makes every formula in $Gamma$ true.
    – Carl Mummert
    38 mins ago








2




2




If $Gamma$ contains a contradiction, then $Gamma$ is unsatisfiable, so $Gamma vDash psi$ is vacuously true. The interesting and nontrivial case is when $Gamma$ is consistent.
– Carl Mummert
56 mins ago





If $Gamma$ contains a contradiction, then $Gamma$ is unsatisfiable, so $Gamma vDash psi$ is vacuously true. The interesting and nontrivial case is when $Gamma$ is consistent.
– Carl Mummert
56 mins ago













@CarlMummert Is "contradiction" same as false, $bot$, etc? What do you mean by unsatisfiable?
– user525966
49 mins ago





@CarlMummert Is "contradiction" same as false, $bot$, etc? What do you mean by unsatisfiable?
– user525966
49 mins ago













I mean that no model satisfies $bot$, so a theory that contains $bot$ is not satisfiable. In that case, trivially "every model of the theory will also be a model of $phi$", because the theory has no models. For example, for any $Gamma$ we have $Gamma vdash bot Longrightarrow Gamma vDash bot$.
– Carl Mummert
45 mins ago





I mean that no model satisfies $bot$, so a theory that contains $bot$ is not satisfiable. In that case, trivially "every model of the theory will also be a model of $phi$", because the theory has no models. For example, for any $Gamma$ we have $Gamma vdash bot Longrightarrow Gamma vDash bot$.
– Carl Mummert
45 mins ago













@CarlMummert Is a "model" a particular set of boolean inputs to the atomic variables of $Gamma$? Does "satisfies" mean "everything in $Gamma$ evaluates to true under a specific model"?
– user525966
41 mins ago





@CarlMummert Is a "model" a particular set of boolean inputs to the atomic variables of $Gamma$? Does "satisfies" mean "everything in $Gamma$ evaluates to true under a specific model"?
– user525966
41 mins ago





1




1




Yes. We would say $Gamma$ is satisfiable because there is at least one model/interpretation that makes every formula in $Gamma$ true.
– Carl Mummert
38 mins ago





Yes. We would say $Gamma$ is satisfiable because there is at least one model/interpretation that makes every formula in $Gamma$ true.
– Carl Mummert
38 mins ago











1 Answer
1






active

oldest

votes

















up vote
5
down vote



accepted










If $botin Gamma,$ then we have $Gammavdash bot$ and $Gamma models bot,$ so this is not in conflict with the soundness theorem. It is clear that $Gamma vdash bot.$ The reason $Gamma models bot$ is that, since $botin Gamma,$ there are no interpretations in which all the sentences in $Gamma$ hold, i.e. no interpretations satisfying $Gamma$. Hence, vacuously, $bot$ holds in every interpretation satisfying $Gamma$.






share|cite|improve this answer




















  • I don't understand... I thought semantic consequence was the idea that if $A vDash varphi$ then it means under every interpretation of the contents of $A$ (i.e. all combinations of true/false among all the atomic propositional variables of all the propositions in $A$), we have $varphi = top$. Is this wrong?
    – user525966
    52 mins ago











  • @user525966 $Amodels phi$ means that under every interpretation in which all sentences in $A$ are true, $phi$ is true. (Also, I would be careful with the notation $phi=top.$ This seems to suggest $phi$ is the sentence $top,$ not that $phi$ is true in a given interpretation.)
    – spaceisdarkgreen
    49 mins ago











  • I meant it the same way as you when you say "$varphi$ is true" but I guess I have to use English to represent that and not $=$? Is there some other way to say that $varphi$ "is" true or "evaluates" to true? $v(varphi) = top$ for truth functional $v$ or however it's phrased?
    – user525966
    48 mins ago











  • Can I think of $A vDash varphi$ as $p to q$ sort of? Whenever everything in $A$ is true, then so is $varphi$? And whenever something in $A$ is false, $varphi$ is vacuously true? Only way for $A vDash varphi$ to not hold is if everything in $A$ is true but $varphi$ is somehow false?
    – user525966
    44 mins ago







  • 1




    @user525966 Yes, I figured that was what you meant. Yes, it would probably be best to just say "$phi$ is true under the interpretation" in English. (There are multiple ways to represent it in notation. Most common is if $mathcal M$ denotes the interpretation, we write $mathcal M models phi,$ in what is perhaps an annoying overloading of the "$models$" symbol.)
    – spaceisdarkgreen
    43 mins ago











Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2928386%2fhow-can-we-prove-soundness-property-if-its-possible-for-our-assumption-set-to-c%23new-answer', 'question_page');

);

Post as a guest






























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










If $botin Gamma,$ then we have $Gammavdash bot$ and $Gamma models bot,$ so this is not in conflict with the soundness theorem. It is clear that $Gamma vdash bot.$ The reason $Gamma models bot$ is that, since $botin Gamma,$ there are no interpretations in which all the sentences in $Gamma$ hold, i.e. no interpretations satisfying $Gamma$. Hence, vacuously, $bot$ holds in every interpretation satisfying $Gamma$.






share|cite|improve this answer




















  • I don't understand... I thought semantic consequence was the idea that if $A vDash varphi$ then it means under every interpretation of the contents of $A$ (i.e. all combinations of true/false among all the atomic propositional variables of all the propositions in $A$), we have $varphi = top$. Is this wrong?
    – user525966
    52 mins ago











  • @user525966 $Amodels phi$ means that under every interpretation in which all sentences in $A$ are true, $phi$ is true. (Also, I would be careful with the notation $phi=top.$ This seems to suggest $phi$ is the sentence $top,$ not that $phi$ is true in a given interpretation.)
    – spaceisdarkgreen
    49 mins ago











  • I meant it the same way as you when you say "$varphi$ is true" but I guess I have to use English to represent that and not $=$? Is there some other way to say that $varphi$ "is" true or "evaluates" to true? $v(varphi) = top$ for truth functional $v$ or however it's phrased?
    – user525966
    48 mins ago











  • Can I think of $A vDash varphi$ as $p to q$ sort of? Whenever everything in $A$ is true, then so is $varphi$? And whenever something in $A$ is false, $varphi$ is vacuously true? Only way for $A vDash varphi$ to not hold is if everything in $A$ is true but $varphi$ is somehow false?
    – user525966
    44 mins ago







  • 1




    @user525966 Yes, I figured that was what you meant. Yes, it would probably be best to just say "$phi$ is true under the interpretation" in English. (There are multiple ways to represent it in notation. Most common is if $mathcal M$ denotes the interpretation, we write $mathcal M models phi,$ in what is perhaps an annoying overloading of the "$models$" symbol.)
    – spaceisdarkgreen
    43 mins ago















up vote
5
down vote



accepted










If $botin Gamma,$ then we have $Gammavdash bot$ and $Gamma models bot,$ so this is not in conflict with the soundness theorem. It is clear that $Gamma vdash bot.$ The reason $Gamma models bot$ is that, since $botin Gamma,$ there are no interpretations in which all the sentences in $Gamma$ hold, i.e. no interpretations satisfying $Gamma$. Hence, vacuously, $bot$ holds in every interpretation satisfying $Gamma$.






share|cite|improve this answer




















  • I don't understand... I thought semantic consequence was the idea that if $A vDash varphi$ then it means under every interpretation of the contents of $A$ (i.e. all combinations of true/false among all the atomic propositional variables of all the propositions in $A$), we have $varphi = top$. Is this wrong?
    – user525966
    52 mins ago











  • @user525966 $Amodels phi$ means that under every interpretation in which all sentences in $A$ are true, $phi$ is true. (Also, I would be careful with the notation $phi=top.$ This seems to suggest $phi$ is the sentence $top,$ not that $phi$ is true in a given interpretation.)
    – spaceisdarkgreen
    49 mins ago











  • I meant it the same way as you when you say "$varphi$ is true" but I guess I have to use English to represent that and not $=$? Is there some other way to say that $varphi$ "is" true or "evaluates" to true? $v(varphi) = top$ for truth functional $v$ or however it's phrased?
    – user525966
    48 mins ago











  • Can I think of $A vDash varphi$ as $p to q$ sort of? Whenever everything in $A$ is true, then so is $varphi$? And whenever something in $A$ is false, $varphi$ is vacuously true? Only way for $A vDash varphi$ to not hold is if everything in $A$ is true but $varphi$ is somehow false?
    – user525966
    44 mins ago







  • 1




    @user525966 Yes, I figured that was what you meant. Yes, it would probably be best to just say "$phi$ is true under the interpretation" in English. (There are multiple ways to represent it in notation. Most common is if $mathcal M$ denotes the interpretation, we write $mathcal M models phi,$ in what is perhaps an annoying overloading of the "$models$" symbol.)
    – spaceisdarkgreen
    43 mins ago













up vote
5
down vote



accepted







up vote
5
down vote



accepted






If $botin Gamma,$ then we have $Gammavdash bot$ and $Gamma models bot,$ so this is not in conflict with the soundness theorem. It is clear that $Gamma vdash bot.$ The reason $Gamma models bot$ is that, since $botin Gamma,$ there are no interpretations in which all the sentences in $Gamma$ hold, i.e. no interpretations satisfying $Gamma$. Hence, vacuously, $bot$ holds in every interpretation satisfying $Gamma$.






share|cite|improve this answer












If $botin Gamma,$ then we have $Gammavdash bot$ and $Gamma models bot,$ so this is not in conflict with the soundness theorem. It is clear that $Gamma vdash bot.$ The reason $Gamma models bot$ is that, since $botin Gamma,$ there are no interpretations in which all the sentences in $Gamma$ hold, i.e. no interpretations satisfying $Gamma$. Hence, vacuously, $bot$ holds in every interpretation satisfying $Gamma$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 56 mins ago









spaceisdarkgreen

29.2k21549




29.2k21549











  • I don't understand... I thought semantic consequence was the idea that if $A vDash varphi$ then it means under every interpretation of the contents of $A$ (i.e. all combinations of true/false among all the atomic propositional variables of all the propositions in $A$), we have $varphi = top$. Is this wrong?
    – user525966
    52 mins ago











  • @user525966 $Amodels phi$ means that under every interpretation in which all sentences in $A$ are true, $phi$ is true. (Also, I would be careful with the notation $phi=top.$ This seems to suggest $phi$ is the sentence $top,$ not that $phi$ is true in a given interpretation.)
    – spaceisdarkgreen
    49 mins ago











  • I meant it the same way as you when you say "$varphi$ is true" but I guess I have to use English to represent that and not $=$? Is there some other way to say that $varphi$ "is" true or "evaluates" to true? $v(varphi) = top$ for truth functional $v$ or however it's phrased?
    – user525966
    48 mins ago











  • Can I think of $A vDash varphi$ as $p to q$ sort of? Whenever everything in $A$ is true, then so is $varphi$? And whenever something in $A$ is false, $varphi$ is vacuously true? Only way for $A vDash varphi$ to not hold is if everything in $A$ is true but $varphi$ is somehow false?
    – user525966
    44 mins ago







  • 1




    @user525966 Yes, I figured that was what you meant. Yes, it would probably be best to just say "$phi$ is true under the interpretation" in English. (There are multiple ways to represent it in notation. Most common is if $mathcal M$ denotes the interpretation, we write $mathcal M models phi,$ in what is perhaps an annoying overloading of the "$models$" symbol.)
    – spaceisdarkgreen
    43 mins ago

















  • I don't understand... I thought semantic consequence was the idea that if $A vDash varphi$ then it means under every interpretation of the contents of $A$ (i.e. all combinations of true/false among all the atomic propositional variables of all the propositions in $A$), we have $varphi = top$. Is this wrong?
    – user525966
    52 mins ago











  • @user525966 $Amodels phi$ means that under every interpretation in which all sentences in $A$ are true, $phi$ is true. (Also, I would be careful with the notation $phi=top.$ This seems to suggest $phi$ is the sentence $top,$ not that $phi$ is true in a given interpretation.)
    – spaceisdarkgreen
    49 mins ago











  • I meant it the same way as you when you say "$varphi$ is true" but I guess I have to use English to represent that and not $=$? Is there some other way to say that $varphi$ "is" true or "evaluates" to true? $v(varphi) = top$ for truth functional $v$ or however it's phrased?
    – user525966
    48 mins ago











  • Can I think of $A vDash varphi$ as $p to q$ sort of? Whenever everything in $A$ is true, then so is $varphi$? And whenever something in $A$ is false, $varphi$ is vacuously true? Only way for $A vDash varphi$ to not hold is if everything in $A$ is true but $varphi$ is somehow false?
    – user525966
    44 mins ago







  • 1




    @user525966 Yes, I figured that was what you meant. Yes, it would probably be best to just say "$phi$ is true under the interpretation" in English. (There are multiple ways to represent it in notation. Most common is if $mathcal M$ denotes the interpretation, we write $mathcal M models phi,$ in what is perhaps an annoying overloading of the "$models$" symbol.)
    – spaceisdarkgreen
    43 mins ago
















I don't understand... I thought semantic consequence was the idea that if $A vDash varphi$ then it means under every interpretation of the contents of $A$ (i.e. all combinations of true/false among all the atomic propositional variables of all the propositions in $A$), we have $varphi = top$. Is this wrong?
– user525966
52 mins ago





I don't understand... I thought semantic consequence was the idea that if $A vDash varphi$ then it means under every interpretation of the contents of $A$ (i.e. all combinations of true/false among all the atomic propositional variables of all the propositions in $A$), we have $varphi = top$. Is this wrong?
– user525966
52 mins ago













@user525966 $Amodels phi$ means that under every interpretation in which all sentences in $A$ are true, $phi$ is true. (Also, I would be careful with the notation $phi=top.$ This seems to suggest $phi$ is the sentence $top,$ not that $phi$ is true in a given interpretation.)
– spaceisdarkgreen
49 mins ago





@user525966 $Amodels phi$ means that under every interpretation in which all sentences in $A$ are true, $phi$ is true. (Also, I would be careful with the notation $phi=top.$ This seems to suggest $phi$ is the sentence $top,$ not that $phi$ is true in a given interpretation.)
– spaceisdarkgreen
49 mins ago













I meant it the same way as you when you say "$varphi$ is true" but I guess I have to use English to represent that and not $=$? Is there some other way to say that $varphi$ "is" true or "evaluates" to true? $v(varphi) = top$ for truth functional $v$ or however it's phrased?
– user525966
48 mins ago





I meant it the same way as you when you say "$varphi$ is true" but I guess I have to use English to represent that and not $=$? Is there some other way to say that $varphi$ "is" true or "evaluates" to true? $v(varphi) = top$ for truth functional $v$ or however it's phrased?
– user525966
48 mins ago













Can I think of $A vDash varphi$ as $p to q$ sort of? Whenever everything in $A$ is true, then so is $varphi$? And whenever something in $A$ is false, $varphi$ is vacuously true? Only way for $A vDash varphi$ to not hold is if everything in $A$ is true but $varphi$ is somehow false?
– user525966
44 mins ago





Can I think of $A vDash varphi$ as $p to q$ sort of? Whenever everything in $A$ is true, then so is $varphi$? And whenever something in $A$ is false, $varphi$ is vacuously true? Only way for $A vDash varphi$ to not hold is if everything in $A$ is true but $varphi$ is somehow false?
– user525966
44 mins ago





1




1




@user525966 Yes, I figured that was what you meant. Yes, it would probably be best to just say "$phi$ is true under the interpretation" in English. (There are multiple ways to represent it in notation. Most common is if $mathcal M$ denotes the interpretation, we write $mathcal M models phi,$ in what is perhaps an annoying overloading of the "$models$" symbol.)
– spaceisdarkgreen
43 mins ago





@user525966 Yes, I figured that was what you meant. Yes, it would probably be best to just say "$phi$ is true under the interpretation" in English. (There are multiple ways to represent it in notation. Most common is if $mathcal M$ denotes the interpretation, we write $mathcal M models phi,$ in what is perhaps an annoying overloading of the "$models$" symbol.)
– spaceisdarkgreen
43 mins ago


















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2928386%2fhow-can-we-prove-soundness-property-if-its-possible-for-our-assumption-set-to-c%23new-answer', 'question_page');

);

Post as a guest













































































Comments

Popular posts from this blog

What does second last employer means? [closed]

Installing NextGIS Connect into QGIS 3?

One-line joke