Validity of formulas The 2019 Stack Overflow Developer Survey Results Are InValidity vs. Tautology and soundnessProve tha validity of the formulaUniversal closure and t-satisfiablity / validityAre those formulas valid?Proving a formula is validProblems with using validity symbol ⊨ “vacuously”, as in “X ⊨” and “⊨ A”Interpreting a set of predicate formulas as a modelSatisfiability and validity in first-order logicFirst Order Logic - Valid FormulaValidity of trivial universally quantified formula with equality.
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Validity of formulas
The 2019 Stack Overflow Developer Survey Results Are InValidity vs. Tautology and soundnessProve tha validity of the formulaUniversal closure and t-satisfiablity / validityAre those formulas valid?Proving a formula is validProblems with using validity symbol ⊨ “vacuously”, as in “X ⊨” and “⊨ A”Interpreting a set of predicate formulas as a modelSatisfiability and validity in first-order logicFirst Order Logic - Valid FormulaValidity of trivial universally quantified formula with equality.
$begingroup$
So i know that if $(∃x ϕ)$ is valid that doesnt implie that doesnt implie that $ (∀x ϕ) $ is also valid, my question is if it is sufficient for me to get an interpretation stucture that satisfies the first and doesnt satisfy the second or i actually need to find a formula that works for every interpretation stucture and then show it doesnt satisfy the second, because it doesnt seem so easy to get that formula.
logic first-order-logic
edited Mar 30 at 11:17
blub
3,299929
asked Mar 30 at 11:05
Pedro SantosPedro Santos
16810
$endgroup$
add a comment |
$begingroup$
So i know that if $(∃x ϕ)$ is valid that doesnt implie that doesnt implie that $ (∀x ϕ) $ is also valid, my question is if it is sufficient for me to get an interpretation stucture that satisfies the first and doesnt satisfy the second or i actually need to find a formula that works for every interpretation stucture and then show it doesnt satisfy the second, because it doesnt seem so easy to get that formula.
logic first-order-logic
edited Mar 30 at 11:17
blub
3,299929
asked Mar 30 at 11:05
Pedro SantosPedro Santos
16810
$endgroup$
3
$begingroup$
You need to show that an interpretation that satisfies the first formula does not necessarily satisfy the second. In other words, you need to find an interpretation where the first formula holds but the second doesn't.
$endgroup$
– frabala
Mar 30 at 11:16
$begingroup$
Alright cool , thats what i did i just wasnt sure if that was enough , because every interpretation stucture has to satisfy the first one.
$endgroup$
– Pedro Santos
Mar 30 at 11:18
add a comment |
$begingroup$
So i know that if $(∃x ϕ)$ is valid that doesnt implie that doesnt implie that $ (∀x ϕ) $ is also valid, my question is if it is sufficient for me to get an interpretation stucture that satisfies the first and doesnt satisfy the second or i actually need to find a formula that works for every interpretation stucture and then show it doesnt satisfy the second, because it doesnt seem so easy to get that formula.
logic first-order-logic
edited Mar 30 at 11:17
blub
3,299929
asked Mar 30 at 11:05
Pedro SantosPedro Santos
16810
$endgroup$
So i know that if $(∃x ϕ)$ is valid that doesnt implie that doesnt implie that $ (∀x ϕ) $ is also valid, my question is if it is sufficient for me to get an interpretation stucture that satisfies the first and doesnt satisfy the second or i actually need to find a formula that works for every interpretation stucture and then show it doesnt satisfy the second, because it doesnt seem so easy to get that formula.
logic first-order-logic
logic first-order-logic
edited Mar 30 at 11:17
blub
3,299929
asked Mar 30 at 11:05
Pedro SantosPedro Santos
16810
edited Mar 30 at 11:17
blub
3,299929
asked Mar 30 at 11:05
Pedro SantosPedro Santos
16810
edited Mar 30 at 11:17
blub
3,299929
edited Mar 30 at 11:17
blub
3,299929
edited Mar 30 at 11:17
blub
3,299929
3,299929
asked Mar 30 at 11:05
Pedro SantosPedro Santos
16810
asked Mar 30 at 11:05
Pedro SantosPedro Santos
16810
asked Mar 30 at 11:05
Pedro SantosPedro Santos
16810
16810
3
$begingroup$
You need to show that an interpretation that satisfies the first formula does not necessarily satisfy the second. In other words, you need to find an interpretation where the first formula holds but the second doesn't.
$endgroup$
– frabala
Mar 30 at 11:16
$begingroup$
Alright cool , thats what i did i just wasnt sure if that was enough , because every interpretation stucture has to satisfy the first one.
$endgroup$
– Pedro Santos
Mar 30 at 11:18
add a comment |
3
$begingroup$
You need to show that an interpretation that satisfies the first formula does not necessarily satisfy the second. In other words, you need to find an interpretation where the first formula holds but the second doesn't.
$endgroup$
– frabala
Mar 30 at 11:16
$begingroup$
Alright cool , thats what i did i just wasnt sure if that was enough , because every interpretation stucture has to satisfy the first one.
$endgroup$
– Pedro Santos
Mar 30 at 11:18
3
3
$begingroup$
You need to show that an interpretation that satisfies the first formula does not necessarily satisfy the second. In other words, you need to find an interpretation where the first formula holds but the second doesn't.
$endgroup$
– frabala
Mar 30 at 11:16
$begingroup$
You need to show that an interpretation that satisfies the first formula does not necessarily satisfy the second. In other words, you need to find an interpretation where the first formula holds but the second doesn't.
$endgroup$
– frabala
Mar 30 at 11:16
$begingroup$
Alright cool , thats what i did i just wasnt sure if that was enough , because every interpretation stucture has to satisfy the first one.
$endgroup$
– Pedro Santos
Mar 30 at 11:18
$begingroup$
Alright cool , thats what i did i just wasnt sure if that was enough , because every interpretation stucture has to satisfy the first one.
$endgroup$
– Pedro Santos
Mar 30 at 11:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You need only show a single occurrence of the first to show that $exists...$ is valid. Then show one instance or interpretation that is false for the second to show that $forall...$ is not valid.
answered Mar 30 at 11:23
poetasispoetasis
430317
$endgroup$
$begingroup$
I thought about using an interpretation stucture where the domain are 2 elements and use the formula $∃ p(x)$ i can make the formula valid here and (∀ x p(x)) not valid. Im not sure if this does what i want. Or i need to start with a formula that is valid everywhere. Cause finding that kind of formula doesnt seem to be straight forward, maybe using an axiom but i dont know.
$endgroup$
– Pedro Santos
Mar 30 at 11:37
$begingroup$
I believe your guess in the first part of your question is the correct one. I think a formula that works $everywhere$ would show that the $forall$ case is true and that would be what you don't want to show.
$endgroup$
– poetasis
Mar 30 at 11:45
$begingroup$
Alright Thanks.
$endgroup$
– Pedro Santos
Mar 30 at 11:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need only show a single occurrence of the first to show that $exists...$ is valid. Then show one instance or interpretation that is false for the second to show that $forall...$ is not valid.
answered Mar 30 at 11:23
poetasispoetasis
430317
$endgroup$
$begingroup$
I thought about using an interpretation stucture where the domain are 2 elements and use the formula $∃ p(x)$ i can make the formula valid here and (∀ x p(x)) not valid. Im not sure if this does what i want. Or i need to start with a formula that is valid everywhere. Cause finding that kind of formula doesnt seem to be straight forward, maybe using an axiom but i dont know.
$endgroup$
– Pedro Santos
Mar 30 at 11:37
$begingroup$
I believe your guess in the first part of your question is the correct one. I think a formula that works $everywhere$ would show that the $forall$ case is true and that would be what you don't want to show.
$endgroup$
– poetasis
Mar 30 at 11:45
$begingroup$
Alright Thanks.
$endgroup$
– Pedro Santos
Mar 30 at 11:45
add a comment |
$begingroup$
You need only show a single occurrence of the first to show that $exists...$ is valid. Then show one instance or interpretation that is false for the second to show that $forall...$ is not valid.
answered Mar 30 at 11:23
poetasispoetasis
430317
$endgroup$
$begingroup$
I thought about using an interpretation stucture where the domain are 2 elements and use the formula $∃ p(x)$ i can make the formula valid here and (∀ x p(x)) not valid. Im not sure if this does what i want. Or i need to start with a formula that is valid everywhere. Cause finding that kind of formula doesnt seem to be straight forward, maybe using an axiom but i dont know.
$endgroup$
– Pedro Santos
Mar 30 at 11:37
$begingroup$
I believe your guess in the first part of your question is the correct one. I think a formula that works $everywhere$ would show that the $forall$ case is true and that would be what you don't want to show.
$endgroup$
– poetasis
Mar 30 at 11:45
$begingroup$
Alright Thanks.
$endgroup$
– Pedro Santos
Mar 30 at 11:45
add a comment |
$begingroup$
You need only show a single occurrence of the first to show that $exists...$ is valid. Then show one instance or interpretation that is false for the second to show that $forall...$ is not valid.
answered Mar 30 at 11:23
poetasispoetasis
430317
$endgroup$
You need only show a single occurrence of the first to show that $exists...$ is valid. Then show one instance or interpretation that is false for the second to show that $forall...$ is not valid.
answered Mar 30 at 11:23
poetasispoetasis
430317
answered Mar 30 at 11:23
poetasispoetasis
430317
answered Mar 30 at 11:23
poetasispoetasis
430317
answered Mar 30 at 11:23
poetasispoetasis
430317
430317
$begingroup$
I thought about using an interpretation stucture where the domain are 2 elements and use the formula $∃ p(x)$ i can make the formula valid here and (∀ x p(x)) not valid. Im not sure if this does what i want. Or i need to start with a formula that is valid everywhere. Cause finding that kind of formula doesnt seem to be straight forward, maybe using an axiom but i dont know.
$endgroup$
– Pedro Santos
Mar 30 at 11:37
$begingroup$
I believe your guess in the first part of your question is the correct one. I think a formula that works $everywhere$ would show that the $forall$ case is true and that would be what you don't want to show.
$endgroup$
– poetasis
Mar 30 at 11:45
$begingroup$
Alright Thanks.
$endgroup$
– Pedro Santos
Mar 30 at 11:45
add a comment |
$begingroup$
I thought about using an interpretation stucture where the domain are 2 elements and use the formula $∃ p(x)$ i can make the formula valid here and (∀ x p(x)) not valid. Im not sure if this does what i want. Or i need to start with a formula that is valid everywhere. Cause finding that kind of formula doesnt seem to be straight forward, maybe using an axiom but i dont know.
$endgroup$
– Pedro Santos
Mar 30 at 11:37
$begingroup$
I believe your guess in the first part of your question is the correct one. I think a formula that works $everywhere$ would show that the $forall$ case is true and that would be what you don't want to show.
$endgroup$
– poetasis
Mar 30 at 11:45
$begingroup$
Alright Thanks.
$endgroup$
– Pedro Santos
Mar 30 at 11:45
$begingroup$
I thought about using an interpretation stucture where the domain are 2 elements and use the formula $∃ p(x)$ i can make the formula valid here and (∀ x p(x)) not valid. Im not sure if this does what i want. Or i need to start with a formula that is valid everywhere. Cause finding that kind of formula doesnt seem to be straight forward, maybe using an axiom but i dont know.
$endgroup$
– Pedro Santos
Mar 30 at 11:37
$begingroup$
I thought about using an interpretation stucture where the domain are 2 elements and use the formula $∃ p(x)$ i can make the formula valid here and (∀ x p(x)) not valid. Im not sure if this does what i want. Or i need to start with a formula that is valid everywhere. Cause finding that kind of formula doesnt seem to be straight forward, maybe using an axiom but i dont know.
$endgroup$
– Pedro Santos
Mar 30 at 11:37
$begingroup$
I believe your guess in the first part of your question is the correct one. I think a formula that works $everywhere$ would show that the $forall$ case is true and that would be what you don't want to show.
$endgroup$
– poetasis
Mar 30 at 11:45
$begingroup$
I believe your guess in the first part of your question is the correct one. I think a formula that works $everywhere$ would show that the $forall$ case is true and that would be what you don't want to show.
$endgroup$
– poetasis
Mar 30 at 11:45
$begingroup$
Alright Thanks.
$endgroup$
– Pedro Santos
Mar 30 at 11:45
$begingroup$
Alright Thanks.
$endgroup$
– Pedro Santos
Mar 30 at 11:45
add a comment |
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$begingroup$
You need to show that an interpretation that satisfies the first formula does not necessarily satisfy the second. In other words, you need to find an interpretation where the first formula holds but the second doesn't.
$endgroup$
– frabala
Mar 30 at 11:16
$begingroup$
Alright cool , thats what i did i just wasnt sure if that was enough , because every interpretation stucture has to satisfy the first one.
$endgroup$
– Pedro Santos
Mar 30 at 11:18