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Difference between consistency and satisfiability
Clarifications on proof of compactness theoremRelationship between consistency, strong completeness and soundnessQuestion about the proof of consistency iff satisfiability of a theoryFinding Satisfiability, Unsatisfiability and Valid well formed formulaDifference between validity and satisfiability?Logic and consistencyProof of SatisfiabilityUnclear why (first order) satisfiability undecidable and not semi-decidable.Undecidability of first-order satisfiability problem?What is the difference between syntactic and semantic completeness?Satisfiability and validity in first-order logicWhat is the difference between validity and satisfiability?
$begingroup$
If a set of formula is consistent, there exist a model in which every formula is true. This is only if the set is satisfiable.
But satisfiability is the fact that it can be true so what is the difference between the 2 notions ?
logic
asked May 17 at 13:14
Emma Vande WouwerEmma Vande Wouwer
332
$endgroup$
add a comment |
$begingroup$
If a set of formula is consistent, there exist a model in which every formula is true. This is only if the set is satisfiable.
But satisfiability is the fact that it can be true so what is the difference between the 2 notions ?
logic
asked May 17 at 13:14
Emma Vande WouwerEmma Vande Wouwer
332
$endgroup$
$begingroup$
Great question!
$endgroup$
– Pat Devlin
May 17 at 14:25
add a comment |
$begingroup$
If a set of formula is consistent, there exist a model in which every formula is true. This is only if the set is satisfiable.
But satisfiability is the fact that it can be true so what is the difference between the 2 notions ?
logic
asked May 17 at 13:14
Emma Vande WouwerEmma Vande Wouwer
332
$endgroup$
If a set of formula is consistent, there exist a model in which every formula is true. This is only if the set is satisfiable.
But satisfiability is the fact that it can be true so what is the difference between the 2 notions ?
logic
logic
asked May 17 at 13:14
Emma Vande WouwerEmma Vande Wouwer
332
asked May 17 at 13:14
Emma Vande WouwerEmma Vande Wouwer
332
asked May 17 at 13:14
Emma Vande WouwerEmma Vande Wouwer
332
asked May 17 at 13:14
Emma Vande WouwerEmma Vande Wouwer
332
asked May 17 at 13:14
Emma Vande WouwerEmma Vande Wouwer
332
332
$begingroup$
Great question!
$endgroup$
– Pat Devlin
May 17 at 14:25
add a comment |
$begingroup$
Great question!
$endgroup$
– Pat Devlin
May 17 at 14:25
$begingroup$
Great question!
$endgroup$
– Pat Devlin
May 17 at 14:25
$begingroup$
Great question!
$endgroup$
– Pat Devlin
May 17 at 14:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consistency is a syntactic property. It means that there is no proof of contradiction from your axioms.
Satisfiability is a semantic property. It means that there is a model of the axioms.
In first-order logic (as well as propositional logic) the two notions are equivalent because the logic is sound and complete. Meaning a satisfiable theory is consistent, and a consistent theory is satisfiable.
Other logics, however, are not so lucky to have both of these properties and the two notions separate. In fact, if we do not assume the axiom of choice, then it is consistent that there is a theory which is consistent but not satisfiable.
answered May 17 at 13:24
Asaf Karagila♦Asaf Karagila
311k33445777
$endgroup$
$begingroup$
Regarding the last point, I always feel like such issues are artifacts of allowing uncountable theories in the first place. What do you think?
$endgroup$
– user21820
May 17 at 16:44
$begingroup$
Well, it's true. If the language is countable, then choice is not necessary. But then you can argue that this requires $sf WKL_0$ over weaker theories, which are also considered "reasonable". So it's not just uncountability.
$endgroup$
– Asaf Karagila♦
May 17 at 16:45
$begingroup$
I don't find a similar issue with WKL because in some sense the notion that every arithmetic sentence has a boolean truth-value already presupposes that arithmetical properties are well-defined, and so ACA0 is well-justified once we assume PA is meaningful. What I find strange is that when we allow (in ZFC) a theory to have an uncountable language, we lose control over the theory if we lack a well-ordering of it, and a priori it's not clear that the axiom of choice is meaningful if we have full power-sets, yet it is reasonable if the intended universe is countable...
$endgroup$
– user21820
May 17 at 16:53
$begingroup$
Well, that's because uncountable things are unwieldy, which is why we need the axiom of choice to bring order to the universe of sets. If my memory serves me right, Shelah has a version of the completeness theorem in ZF and you need to require something like linear orders and that the cardinality of the language equals to its finite subsets, or finite sequences, or something like that. And then you can prove completeness without additional assumptions. This really shows you that the counterexamples I mention are very particular.
$endgroup$
– Asaf Karagila♦
May 17 at 16:58
$begingroup$
Oh I read that the completeness theorem is equivalent to BPIT, but I'd be curious if you can pull up a reference for what you just said.
$endgroup$
– user21820
May 17 at 17:02
|
show 1 more comment
$begingroup$
Consistency is defined syntactically :
a set $Gamma$ of formulas is inconsistent iff we can derive (in the proof system) a contradiction from it : $Gamma vdash bot$.
In this way, we prove that :
a set $Gamma$ is consistent iff it is satisfiable.
See also the post : Relationship between consistency, strong completeness and soundness.
answered May 17 at 13:24
Mauro ALLEGRANZAMauro ALLEGRANZA
69.1k449118
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
Consistency is a syntactic property. It means that there is no proof of contradiction from your axioms.
Satisfiability is a semantic property. It means that there is a model of the axioms.
In first-order logic (as well as propositional logic) the two notions are equivalent because the logic is sound and complete. Meaning a satisfiable theory is consistent, and a consistent theory is satisfiable.
Other logics, however, are not so lucky to have both of these properties and the two notions separate. In fact, if we do not assume the axiom of choice, then it is consistent that there is a theory which is consistent but not satisfiable.
answered May 17 at 13:24
Asaf Karagila♦Asaf Karagila
311k33445777
$endgroup$
$begingroup$
Regarding the last point, I always feel like such issues are artifacts of allowing uncountable theories in the first place. What do you think?
$endgroup$
– user21820
May 17 at 16:44
$begingroup$
Well, it's true. If the language is countable, then choice is not necessary. But then you can argue that this requires $sf WKL_0$ over weaker theories, which are also considered "reasonable". So it's not just uncountability.
$endgroup$
– Asaf Karagila♦
May 17 at 16:45
$begingroup$
I don't find a similar issue with WKL because in some sense the notion that every arithmetic sentence has a boolean truth-value already presupposes that arithmetical properties are well-defined, and so ACA0 is well-justified once we assume PA is meaningful. What I find strange is that when we allow (in ZFC) a theory to have an uncountable language, we lose control over the theory if we lack a well-ordering of it, and a priori it's not clear that the axiom of choice is meaningful if we have full power-sets, yet it is reasonable if the intended universe is countable...
$endgroup$
– user21820
May 17 at 16:53
$begingroup$
Well, that's because uncountable things are unwieldy, which is why we need the axiom of choice to bring order to the universe of sets. If my memory serves me right, Shelah has a version of the completeness theorem in ZF and you need to require something like linear orders and that the cardinality of the language equals to its finite subsets, or finite sequences, or something like that. And then you can prove completeness without additional assumptions. This really shows you that the counterexamples I mention are very particular.
$endgroup$
– Asaf Karagila♦
May 17 at 16:58
$begingroup$
Oh I read that the completeness theorem is equivalent to BPIT, but I'd be curious if you can pull up a reference for what you just said.
$endgroup$
– user21820
May 17 at 17:02
|
show 1 more comment
$begingroup$
Consistency is a syntactic property. It means that there is no proof of contradiction from your axioms.
Satisfiability is a semantic property. It means that there is a model of the axioms.
In first-order logic (as well as propositional logic) the two notions are equivalent because the logic is sound and complete. Meaning a satisfiable theory is consistent, and a consistent theory is satisfiable.
Other logics, however, are not so lucky to have both of these properties and the two notions separate. In fact, if we do not assume the axiom of choice, then it is consistent that there is a theory which is consistent but not satisfiable.
answered May 17 at 13:24
Asaf Karagila♦Asaf Karagila
311k33445777
$endgroup$
$begingroup$
Regarding the last point, I always feel like such issues are artifacts of allowing uncountable theories in the first place. What do you think?
$endgroup$
– user21820
May 17 at 16:44
$begingroup$
Well, it's true. If the language is countable, then choice is not necessary. But then you can argue that this requires $sf WKL_0$ over weaker theories, which are also considered "reasonable". So it's not just uncountability.
$endgroup$
– Asaf Karagila♦
May 17 at 16:45
$begingroup$
I don't find a similar issue with WKL because in some sense the notion that every arithmetic sentence has a boolean truth-value already presupposes that arithmetical properties are well-defined, and so ACA0 is well-justified once we assume PA is meaningful. What I find strange is that when we allow (in ZFC) a theory to have an uncountable language, we lose control over the theory if we lack a well-ordering of it, and a priori it's not clear that the axiom of choice is meaningful if we have full power-sets, yet it is reasonable if the intended universe is countable...
$endgroup$
– user21820
May 17 at 16:53
$begingroup$
Well, that's because uncountable things are unwieldy, which is why we need the axiom of choice to bring order to the universe of sets. If my memory serves me right, Shelah has a version of the completeness theorem in ZF and you need to require something like linear orders and that the cardinality of the language equals to its finite subsets, or finite sequences, or something like that. And then you can prove completeness without additional assumptions. This really shows you that the counterexamples I mention are very particular.
$endgroup$
– Asaf Karagila♦
May 17 at 16:58
$begingroup$
Oh I read that the completeness theorem is equivalent to BPIT, but I'd be curious if you can pull up a reference for what you just said.
$endgroup$
– user21820
May 17 at 17:02
|
show 1 more comment
$begingroup$
Consistency is a syntactic property. It means that there is no proof of contradiction from your axioms.
Satisfiability is a semantic property. It means that there is a model of the axioms.
In first-order logic (as well as propositional logic) the two notions are equivalent because the logic is sound and complete. Meaning a satisfiable theory is consistent, and a consistent theory is satisfiable.
Other logics, however, are not so lucky to have both of these properties and the two notions separate. In fact, if we do not assume the axiom of choice, then it is consistent that there is a theory which is consistent but not satisfiable.
answered May 17 at 13:24
Asaf Karagila♦Asaf Karagila
311k33445777
$endgroup$
Consistency is a syntactic property. It means that there is no proof of contradiction from your axioms.
Satisfiability is a semantic property. It means that there is a model of the axioms.
In first-order logic (as well as propositional logic) the two notions are equivalent because the logic is sound and complete. Meaning a satisfiable theory is consistent, and a consistent theory is satisfiable.
Other logics, however, are not so lucky to have both of these properties and the two notions separate. In fact, if we do not assume the axiom of choice, then it is consistent that there is a theory which is consistent but not satisfiable.
answered May 17 at 13:24
Asaf Karagila♦Asaf Karagila
311k33445777
answered May 17 at 13:24
Asaf Karagila♦Asaf Karagila
311k33445777
answered May 17 at 13:24
Asaf Karagila♦Asaf Karagila
311k33445777
answered May 17 at 13:24
Asaf Karagila♦Asaf Karagila
311k33445777
311k33445777
$begingroup$
Regarding the last point, I always feel like such issues are artifacts of allowing uncountable theories in the first place. What do you think?
$endgroup$
– user21820
May 17 at 16:44
$begingroup$
Well, it's true. If the language is countable, then choice is not necessary. But then you can argue that this requires $sf WKL_0$ over weaker theories, which are also considered "reasonable". So it's not just uncountability.
$endgroup$
– Asaf Karagila♦
May 17 at 16:45
$begingroup$
I don't find a similar issue with WKL because in some sense the notion that every arithmetic sentence has a boolean truth-value already presupposes that arithmetical properties are well-defined, and so ACA0 is well-justified once we assume PA is meaningful. What I find strange is that when we allow (in ZFC) a theory to have an uncountable language, we lose control over the theory if we lack a well-ordering of it, and a priori it's not clear that the axiom of choice is meaningful if we have full power-sets, yet it is reasonable if the intended universe is countable...
$endgroup$
– user21820
May 17 at 16:53
$begingroup$
Well, that's because uncountable things are unwieldy, which is why we need the axiom of choice to bring order to the universe of sets. If my memory serves me right, Shelah has a version of the completeness theorem in ZF and you need to require something like linear orders and that the cardinality of the language equals to its finite subsets, or finite sequences, or something like that. And then you can prove completeness without additional assumptions. This really shows you that the counterexamples I mention are very particular.
$endgroup$
– Asaf Karagila♦
May 17 at 16:58
$begingroup$
Oh I read that the completeness theorem is equivalent to BPIT, but I'd be curious if you can pull up a reference for what you just said.
$endgroup$
– user21820
May 17 at 17:02
|
show 1 more comment
$begingroup$
Regarding the last point, I always feel like such issues are artifacts of allowing uncountable theories in the first place. What do you think?
$endgroup$
– user21820
May 17 at 16:44
$begingroup$
Well, it's true. If the language is countable, then choice is not necessary. But then you can argue that this requires $sf WKL_0$ over weaker theories, which are also considered "reasonable". So it's not just uncountability.
$endgroup$
– Asaf Karagila♦
May 17 at 16:45
$begingroup$
I don't find a similar issue with WKL because in some sense the notion that every arithmetic sentence has a boolean truth-value already presupposes that arithmetical properties are well-defined, and so ACA0 is well-justified once we assume PA is meaningful. What I find strange is that when we allow (in ZFC) a theory to have an uncountable language, we lose control over the theory if we lack a well-ordering of it, and a priori it's not clear that the axiom of choice is meaningful if we have full power-sets, yet it is reasonable if the intended universe is countable...
$endgroup$
– user21820
May 17 at 16:53
$begingroup$
Well, that's because uncountable things are unwieldy, which is why we need the axiom of choice to bring order to the universe of sets. If my memory serves me right, Shelah has a version of the completeness theorem in ZF and you need to require something like linear orders and that the cardinality of the language equals to its finite subsets, or finite sequences, or something like that. And then you can prove completeness without additional assumptions. This really shows you that the counterexamples I mention are very particular.
$endgroup$
– Asaf Karagila♦
May 17 at 16:58
$begingroup$
Oh I read that the completeness theorem is equivalent to BPIT, but I'd be curious if you can pull up a reference for what you just said.
$endgroup$
– user21820
May 17 at 17:02
$begingroup$
Regarding the last point, I always feel like such issues are artifacts of allowing uncountable theories in the first place. What do you think?
$endgroup$
– user21820
May 17 at 16:44
$begingroup$
Regarding the last point, I always feel like such issues are artifacts of allowing uncountable theories in the first place. What do you think?
$endgroup$
– user21820
May 17 at 16:44
$begingroup$
Well, it's true. If the language is countable, then choice is not necessary. But then you can argue that this requires $sf WKL_0$ over weaker theories, which are also considered "reasonable". So it's not just uncountability.
$endgroup$
– Asaf Karagila♦
May 17 at 16:45
$begingroup$
Well, it's true. If the language is countable, then choice is not necessary. But then you can argue that this requires $sf WKL_0$ over weaker theories, which are also considered "reasonable". So it's not just uncountability.
$endgroup$
– Asaf Karagila♦
May 17 at 16:45
$begingroup$
I don't find a similar issue with WKL because in some sense the notion that every arithmetic sentence has a boolean truth-value already presupposes that arithmetical properties are well-defined, and so ACA0 is well-justified once we assume PA is meaningful. What I find strange is that when we allow (in ZFC) a theory to have an uncountable language, we lose control over the theory if we lack a well-ordering of it, and a priori it's not clear that the axiom of choice is meaningful if we have full power-sets, yet it is reasonable if the intended universe is countable...
$endgroup$
– user21820
May 17 at 16:53
$begingroup$
I don't find a similar issue with WKL because in some sense the notion that every arithmetic sentence has a boolean truth-value already presupposes that arithmetical properties are well-defined, and so ACA0 is well-justified once we assume PA is meaningful. What I find strange is that when we allow (in ZFC) a theory to have an uncountable language, we lose control over the theory if we lack a well-ordering of it, and a priori it's not clear that the axiom of choice is meaningful if we have full power-sets, yet it is reasonable if the intended universe is countable...
$endgroup$
– user21820
May 17 at 16:53
$begingroup$
Well, that's because uncountable things are unwieldy, which is why we need the axiom of choice to bring order to the universe of sets. If my memory serves me right, Shelah has a version of the completeness theorem in ZF and you need to require something like linear orders and that the cardinality of the language equals to its finite subsets, or finite sequences, or something like that. And then you can prove completeness without additional assumptions. This really shows you that the counterexamples I mention are very particular.
$endgroup$
– Asaf Karagila♦
May 17 at 16:58
$begingroup$
Well, that's because uncountable things are unwieldy, which is why we need the axiom of choice to bring order to the universe of sets. If my memory serves me right, Shelah has a version of the completeness theorem in ZF and you need to require something like linear orders and that the cardinality of the language equals to its finite subsets, or finite sequences, or something like that. And then you can prove completeness without additional assumptions. This really shows you that the counterexamples I mention are very particular.
$endgroup$
– Asaf Karagila♦
May 17 at 16:58
$begingroup$
Oh I read that the completeness theorem is equivalent to BPIT, but I'd be curious if you can pull up a reference for what you just said.
$endgroup$
– user21820
May 17 at 17:02
$begingroup$
Oh I read that the completeness theorem is equivalent to BPIT, but I'd be curious if you can pull up a reference for what you just said.
$endgroup$
– user21820
May 17 at 17:02
|
show 1 more comment
$begingroup$
Consistency is defined syntactically :
a set $Gamma$ of formulas is inconsistent iff we can derive (in the proof system) a contradiction from it : $Gamma vdash bot$.
In this way, we prove that :
a set $Gamma$ is consistent iff it is satisfiable.
See also the post : Relationship between consistency, strong completeness and soundness.
answered May 17 at 13:24
Mauro ALLEGRANZAMauro ALLEGRANZA
69.1k449118
$endgroup$
add a comment |
$begingroup$
Consistency is defined syntactically :
a set $Gamma$ of formulas is inconsistent iff we can derive (in the proof system) a contradiction from it : $Gamma vdash bot$.
In this way, we prove that :
a set $Gamma$ is consistent iff it is satisfiable.
See also the post : Relationship between consistency, strong completeness and soundness.
answered May 17 at 13:24
Mauro ALLEGRANZAMauro ALLEGRANZA
69.1k449118
$endgroup$
add a comment |
$begingroup$
Consistency is defined syntactically :
a set $Gamma$ of formulas is inconsistent iff we can derive (in the proof system) a contradiction from it : $Gamma vdash bot$.
In this way, we prove that :
a set $Gamma$ is consistent iff it is satisfiable.
See also the post : Relationship between consistency, strong completeness and soundness.
answered May 17 at 13:24
Mauro ALLEGRANZAMauro ALLEGRANZA
69.1k449118
$endgroup$
Consistency is defined syntactically :
a set $Gamma$ of formulas is inconsistent iff we can derive (in the proof system) a contradiction from it : $Gamma vdash bot$.
In this way, we prove that :
a set $Gamma$ is consistent iff it is satisfiable.
See also the post : Relationship between consistency, strong completeness and soundness.
answered May 17 at 13:24
Mauro ALLEGRANZAMauro ALLEGRANZA
69.1k449118
edited May 18 at 10:08
answered May 17 at 13:24
Mauro ALLEGRANZAMauro ALLEGRANZA
69.1k449118
answered May 17 at 13:24
Mauro ALLEGRANZAMauro ALLEGRANZA
69.1k449118
answered May 17 at 13:24
Mauro ALLEGRANZAMauro ALLEGRANZA
69.1k449118
69.1k449118
add a comment |
add a comment |
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Great question!
$endgroup$
– Pat Devlin
May 17 at 14:25