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How can we prove soundness property if it's possible for our assumption set to contain false assumptions?
Clash Royale CLAN TAG#URR8PPP
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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
asked 1 hour ago
user525966
1,770719
 |Â
show 2 more comments
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?
logic definition propositional-calculus hilbert-calculus
asked 1 hour ago
user525966
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
 |Â
show 2 more comments
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?
logic definition propositional-calculus hilbert-calculus
asked 1 hour ago
user525966
1,770719
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
logic definition propositional-calculus hilbert-calculus
asked 1 hour ago
user525966
1,770719
asked 1 hour ago
user525966
1,770719
asked 1 hour ago
user525966
1,770719
asked 1 hour ago
user525966
1,770719
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
 |Â
show 2 more comments
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
 |Â
show 2 more comments
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 $Gamma$.
answered 56 mins ago
spaceisdarkgreen
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
 |Â
show 9 more comments
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 $Gamma$.
answered 56 mins ago
spaceisdarkgreen
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
 |Â
show 9 more comments
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 $Gamma$.
answered 56 mins ago
spaceisdarkgreen
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
 |Â
show 9 more comments
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 $Gamma$.
answered 56 mins ago
spaceisdarkgreen
29.2k21549
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 $Gamma$.
answered 56 mins ago
spaceisdarkgreen
29.2k21549
answered 56 mins ago
spaceisdarkgreen
29.2k21549
answered 56 mins ago
spaceisdarkgreen
29.2k21549
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
 |Â
show 9 more comments
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
 |Â
show 9 more comments
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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