Prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology
$begingroup$
Assume that $H,F$ are formulas. How to prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology. My intuition is that's right, but I don't know how to prove this in a clean and precise manner.
Definition $models$: $ H models F$ means that every interpretation suitable for both $H$ and $F$, which is a model for $H$ is also a model for $F$.
logic
asked Dec 1 '18 at 14:23
heaigheaig
62
$endgroup$
add a comment |
$begingroup$
Assume that $H,F$ are formulas. How to prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology. My intuition is that's right, but I don't know how to prove this in a clean and precise manner.
Definition $models$: $ H models F$ means that every interpretation suitable for both $H$ and $F$, which is a model for $H$ is also a model for $F$.
logic
asked Dec 1 '18 at 14:23
heaigheaig
62
$endgroup$
$begingroup$
Is it propositional logic? How is $models$ defined?
$endgroup$
– Berci
Dec 1 '18 at 14:28
$begingroup$
@Berci I added my definition.
$endgroup$
– heaig
Dec 1 '18 at 14:39
1
$begingroup$
Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
$endgroup$
– Mauro ALLEGRANZA
Dec 1 '18 at 14:39
add a comment |
$begingroup$
Assume that $H,F$ are formulas. How to prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology. My intuition is that's right, but I don't know how to prove this in a clean and precise manner.
Definition $models$: $ H models F$ means that every interpretation suitable for both $H$ and $F$, which is a model for $H$ is also a model for $F$.
logic
asked Dec 1 '18 at 14:23
heaigheaig
62
$endgroup$
Assume that $H,F$ are formulas. How to prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology. My intuition is that's right, but I don't know how to prove this in a clean and precise manner.
Definition $models$: $ H models F$ means that every interpretation suitable for both $H$ and $F$, which is a model for $H$ is also a model for $F$.
logic
logic
asked Dec 1 '18 at 14:23
heaigheaig
62
asked Dec 1 '18 at 14:23
heaigheaig
62
edited Dec 1 '18 at 14:39
heaig
asked Dec 1 '18 at 14:23
heaigheaig
62
asked Dec 1 '18 at 14:23
heaigheaig
62
asked Dec 1 '18 at 14:23
heaigheaig
62
62
$begingroup$
Is it propositional logic? How is $models$ defined?
$endgroup$
– Berci
Dec 1 '18 at 14:28
$begingroup$
@Berci I added my definition.
$endgroup$
– heaig
Dec 1 '18 at 14:39
1
$begingroup$
Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
$endgroup$
– Mauro ALLEGRANZA
Dec 1 '18 at 14:39
add a comment |
$begingroup$
Is it propositional logic? How is $models$ defined?
$endgroup$
– Berci
Dec 1 '18 at 14:28
$begingroup$
@Berci I added my definition.
$endgroup$
– heaig
Dec 1 '18 at 14:39
1
$begingroup$
Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
$endgroup$
– Mauro ALLEGRANZA
Dec 1 '18 at 14:39
$begingroup$
Is it propositional logic? How is $models$ defined?
$endgroup$
– Berci
Dec 1 '18 at 14:28
$begingroup$
Is it propositional logic? How is $models$ defined?
$endgroup$
– Berci
Dec 1 '18 at 14:28
$begingroup$
@Berci I added my definition.
$endgroup$
– heaig
Dec 1 '18 at 14:39
$begingroup$
@Berci I added my definition.
$endgroup$
– heaig
Dec 1 '18 at 14:39
1
1
$begingroup$
Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
$endgroup$
– Mauro ALLEGRANZA
Dec 1 '18 at 14:39
$begingroup$
Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
$endgroup$
– Mauro ALLEGRANZA
Dec 1 '18 at 14:39
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.
So we have:
$$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$
Introduce preconditions on both sides:
$${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$
Simplify the implication:
$${H_1 wedge cdots wedge H_k} models F$$
answered Dec 1 '18 at 14:48
orlporlp
7,5791433
$endgroup$
add a comment |
$begingroup$
To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.
answered Dec 1 '18 at 15:05
Andreas BlassAndreas Blass
49.8k451108
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.
So we have:
$$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$
Introduce preconditions on both sides:
$${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$
Simplify the implication:
$${H_1 wedge cdots wedge H_k} models F$$
answered Dec 1 '18 at 14:48
orlporlp
7,5791433
$endgroup$
add a comment |
$begingroup$
Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.
So we have:
$$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$
Introduce preconditions on both sides:
$${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$
Simplify the implication:
$${H_1 wedge cdots wedge H_k} models F$$
answered Dec 1 '18 at 14:48
orlporlp
7,5791433
$endgroup$
add a comment |
$begingroup$
Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.
So we have:
$$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$
Introduce preconditions on both sides:
$${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$
Simplify the implication:
$${H_1 wedge cdots wedge H_k} models F$$
answered Dec 1 '18 at 14:48
orlporlp
7,5791433
$endgroup$
Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.
So we have:
$$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$
Introduce preconditions on both sides:
$${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$
Simplify the implication:
$${H_1 wedge cdots wedge H_k} models F$$
answered Dec 1 '18 at 14:48
orlporlp
7,5791433
answered Dec 1 '18 at 14:48
orlporlp
7,5791433
answered Dec 1 '18 at 14:48
orlporlp
7,5791433
answered Dec 1 '18 at 14:48
orlporlp
7,5791433
7,5791433
add a comment |
add a comment |
$begingroup$
To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.
answered Dec 1 '18 at 15:05
Andreas BlassAndreas Blass
49.8k451108
$endgroup$
add a comment |
$begingroup$
To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.
answered Dec 1 '18 at 15:05
Andreas BlassAndreas Blass
49.8k451108
$endgroup$
add a comment |
$begingroup$
To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.
answered Dec 1 '18 at 15:05
Andreas BlassAndreas Blass
49.8k451108
$endgroup$
To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.
answered Dec 1 '18 at 15:05
Andreas BlassAndreas Blass
49.8k451108
answered Dec 1 '18 at 15:05
Andreas BlassAndreas Blass
49.8k451108
answered Dec 1 '18 at 15:05
Andreas BlassAndreas Blass
49.8k451108
answered Dec 1 '18 at 15:05
Andreas BlassAndreas Blass
49.8k451108
49.8k451108
add a comment |
add a comment |
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$begingroup$
Is it propositional logic? How is $models$ defined?
$endgroup$
– Berci
Dec 1 '18 at 14:28
$begingroup$
@Berci I added my definition.
$endgroup$
– heaig
Dec 1 '18 at 14:39
1
$begingroup$
Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
$endgroup$
– Mauro ALLEGRANZA
Dec 1 '18 at 14:39