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Tautological implication and monotony/monotonicity of Tarski condition
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I think I have a fundamental misunderstanding of the definition of an union set; the following is from the notes published by Prof David Makinson at LSE for a further logic course:
Prove that from its definition that tautological implication satisfies monotony/monotonicity of Tarski condition (Whenever $A ⊨ alpha$ then $Acup B ⊨ alpha$).
Solution: Suppose $A ⊨ alpha$; we want to show that $Acup B vDash alpha$. Let $v$ be any valuation and suppose $v$ makes all formulae in $Acup B$ true. Then it makes all formulae in $A$ true, so by the first supposition it makes $alpha$ true.
$v(forall x(xin Acup B))=T$ leads to $v(forall x (xin A))=T$ because $Acup B=forall x(xin A lor xin B) $, therefore it means everything that is either a member of $A$ or $B$ is true. From that all formulae in A are true.
But this doesn't seem to make sense if I look at the definition: IF (and this is a BIG if) I am understanding correctly, all formulae in $Acup B$ are still true under $v$ if $v(forall x (xin A))=T$ but $v(forall x (xin B))=F$, because of the inclusive disjunction in the definition of $Acup B$. (In particular $xin Acup B$ still holds if $xin A$ but $lnot xin B$)
But if so, 'All formulae in $Acup B$ are true. Then it makes all formulae in $A$ true', if I replace the 2nd sentence with 'Then it makes all formulae in $B$ true', it doesn't make sense anymore because in the current truth assignment, we are assuming all formulae in A are true but all formulae in B are false.
I am sure it is my understanding of union set that went wrong (2nd paragraph); but I can't tell how. Could anyone tell me where I went wrong please?
elementary-set-theory logic proof-explanation
asked Jul 25 at 6:21
Daniel Mak
360215
add a comment |Â
up vote
0
down vote
favorite
I think I have a fundamental misunderstanding of the definition of an union set; the following is from the notes published by Prof David Makinson at LSE for a further logic course:
Prove that from its definition that tautological implication satisfies monotony/monotonicity of Tarski condition (Whenever $A ⊨ alpha$ then $Acup B ⊨ alpha$).
Solution: Suppose $A ⊨ alpha$; we want to show that $Acup B vDash alpha$. Let $v$ be any valuation and suppose $v$ makes all formulae in $Acup B$ true. Then it makes all formulae in $A$ true, so by the first supposition it makes $alpha$ true.
$v(forall x(xin Acup B))=T$ leads to $v(forall x (xin A))=T$ because $Acup B=forall x(xin A lor xin B) $, therefore it means everything that is either a member of $A$ or $B$ is true. From that all formulae in A are true.
But this doesn't seem to make sense if I look at the definition: IF (and this is a BIG if) I am understanding correctly, all formulae in $Acup B$ are still true under $v$ if $v(forall x (xin A))=T$ but $v(forall x (xin B))=F$, because of the inclusive disjunction in the definition of $Acup B$. (In particular $xin Acup B$ still holds if $xin A$ but $lnot xin B$)
But if so, 'All formulae in $Acup B$ are true. Then it makes all formulae in $A$ true', if I replace the 2nd sentence with 'Then it makes all formulae in $B$ true', it doesn't make sense anymore because in the current truth assignment, we are assuming all formulae in A are true but all formulae in B are false.
I am sure it is my understanding of union set that went wrong (2nd paragraph); but I can't tell how. Could anyone tell me where I went wrong please?
elementary-set-theory logic proof-explanation
asked Jul 25 at 6:21
Daniel Mak
360215
It's from the notes of a course I attended earlier so I am not sure how to quote it. $v(forall x(xin Acup B))=T$ just means every formulae in $Acup B$ is true, so it's just a formulated way of saying the same thing; unless I am understanding what the solution said wrong?
– Daniel Mak
Jul 25 at 7:13
And while I do respect your concern of plagiarism, I am not sure how this is the case here, as I am certainly not taking credit for it (otherwise I would have understood it), nor am I passing it on as an answer or anything...
– Daniel Mak
Jul 25 at 7:17
Forgive my ignorance but in terms of meaning I honestly am not sure if there is a difference...both seems to mean the same thing, ie we are assigning a truth value $T$ to everything that is either in $A$ or $B$, the only difference seems to be that the latter is using $alpha$ to denote the formulae in the union while the former uses $x$
– Daniel Mak
Jul 25 at 7:24
Edited to credit Prof David Makinson at LSE
– Daniel Mak
Jul 25 at 7:26
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I think I have a fundamental misunderstanding of the definition of an union set; the following is from the notes published by Prof David Makinson at LSE for a further logic course:
Prove that from its definition that tautological implication satisfies monotony/monotonicity of Tarski condition (Whenever $A ⊨ alpha$ then $Acup B ⊨ alpha$).
Solution: Suppose $A ⊨ alpha$; we want to show that $Acup B vDash alpha$. Let $v$ be any valuation and suppose $v$ makes all formulae in $Acup B$ true. Then it makes all formulae in $A$ true, so by the first supposition it makes $alpha$ true.
$v(forall x(xin Acup B))=T$ leads to $v(forall x (xin A))=T$ because $Acup B=forall x(xin A lor xin B) $, therefore it means everything that is either a member of $A$ or $B$ is true. From that all formulae in A are true.
But this doesn't seem to make sense if I look at the definition: IF (and this is a BIG if) I am understanding correctly, all formulae in $Acup B$ are still true under $v$ if $v(forall x (xin A))=T$ but $v(forall x (xin B))=F$, because of the inclusive disjunction in the definition of $Acup B$. (In particular $xin Acup B$ still holds if $xin A$ but $lnot xin B$)
But if so, 'All formulae in $Acup B$ are true. Then it makes all formulae in $A$ true', if I replace the 2nd sentence with 'Then it makes all formulae in $B$ true', it doesn't make sense anymore because in the current truth assignment, we are assuming all formulae in A are true but all formulae in B are false.
I am sure it is my understanding of union set that went wrong (2nd paragraph); but I can't tell how. Could anyone tell me where I went wrong please?
elementary-set-theory logic proof-explanation
asked Jul 25 at 6:21
Daniel Mak
360215
I think I have a fundamental misunderstanding of the definition of an union set; the following is from the notes published by Prof David Makinson at LSE for a further logic course:
Prove that from its definition that tautological implication satisfies monotony/monotonicity of Tarski condition (Whenever $A ⊨ alpha$ then $Acup B ⊨ alpha$).
Solution: Suppose $A ⊨ alpha$; we want to show that $Acup B vDash alpha$. Let $v$ be any valuation and suppose $v$ makes all formulae in $Acup B$ true. Then it makes all formulae in $A$ true, so by the first supposition it makes $alpha$ true.
$v(forall x(xin Acup B))=T$ leads to $v(forall x (xin A))=T$ because $Acup B=forall x(xin A lor xin B) $, therefore it means everything that is either a member of $A$ or $B$ is true. From that all formulae in A are true.
But this doesn't seem to make sense if I look at the definition: IF (and this is a BIG if) I am understanding correctly, all formulae in $Acup B$ are still true under $v$ if $v(forall x (xin A))=T$ but $v(forall x (xin B))=F$, because of the inclusive disjunction in the definition of $Acup B$. (In particular $xin Acup B$ still holds if $xin A$ but $lnot xin B$)
But if so, 'All formulae in $Acup B$ are true. Then it makes all formulae in $A$ true', if I replace the 2nd sentence with 'Then it makes all formulae in $B$ true', it doesn't make sense anymore because in the current truth assignment, we are assuming all formulae in A are true but all formulae in B are false.
I am sure it is my understanding of union set that went wrong (2nd paragraph); but I can't tell how. Could anyone tell me where I went wrong please?
elementary-set-theory logic proof-explanation
asked Jul 25 at 6:21
Daniel Mak
360215
edited Jul 25 at 7:25
asked Jul 25 at 6:21
Daniel Mak
360215
asked Jul 25 at 6:21
Daniel Mak
360215
asked Jul 25 at 6:21
Daniel Mak
360215
360215
It's from the notes of a course I attended earlier so I am not sure how to quote it. $v(forall x(xin Acup B))=T$ just means every formulae in $Acup B$ is true, so it's just a formulated way of saying the same thing; unless I am understanding what the solution said wrong?
– Daniel Mak
Jul 25 at 7:13
And while I do respect your concern of plagiarism, I am not sure how this is the case here, as I am certainly not taking credit for it (otherwise I would have understood it), nor am I passing it on as an answer or anything...
– Daniel Mak
Jul 25 at 7:17
Forgive my ignorance but in terms of meaning I honestly am not sure if there is a difference...both seems to mean the same thing, ie we are assigning a truth value $T$ to everything that is either in $A$ or $B$, the only difference seems to be that the latter is using $alpha$ to denote the formulae in the union while the former uses $x$
– Daniel Mak
Jul 25 at 7:24
Edited to credit Prof David Makinson at LSE
– Daniel Mak
Jul 25 at 7:26
add a comment |Â
It's from the notes of a course I attended earlier so I am not sure how to quote it. $v(forall x(xin Acup B))=T$ just means every formulae in $Acup B$ is true, so it's just a formulated way of saying the same thing; unless I am understanding what the solution said wrong?
– Daniel Mak
Jul 25 at 7:13
And while I do respect your concern of plagiarism, I am not sure how this is the case here, as I am certainly not taking credit for it (otherwise I would have understood it), nor am I passing it on as an answer or anything...
– Daniel Mak
Jul 25 at 7:17
Forgive my ignorance but in terms of meaning I honestly am not sure if there is a difference...both seems to mean the same thing, ie we are assigning a truth value $T$ to everything that is either in $A$ or $B$, the only difference seems to be that the latter is using $alpha$ to denote the formulae in the union while the former uses $x$
– Daniel Mak
Jul 25 at 7:24
Edited to credit Prof David Makinson at LSE
– Daniel Mak
Jul 25 at 7:26
It's from the notes of a course I attended earlier so I am not sure how to quote it. $v(forall x(xin Acup B))=T$ just means every formulae in $Acup B$ is true, so it's just a formulated way of saying the same thing; unless I am understanding what the solution said wrong?
– Daniel Mak
Jul 25 at 7:13
It's from the notes of a course I attended earlier so I am not sure how to quote it. $v(forall x(xin Acup B))=T$ just means every formulae in $Acup B$ is true, so it's just a formulated way of saying the same thing; unless I am understanding what the solution said wrong?
– Daniel Mak
Jul 25 at 7:13
And while I do respect your concern of plagiarism, I am not sure how this is the case here, as I am certainly not taking credit for it (otherwise I would have understood it), nor am I passing it on as an answer or anything...
– Daniel Mak
Jul 25 at 7:17
And while I do respect your concern of plagiarism, I am not sure how this is the case here, as I am certainly not taking credit for it (otherwise I would have understood it), nor am I passing it on as an answer or anything...
– Daniel Mak
Jul 25 at 7:17
Forgive my ignorance but in terms of meaning I honestly am not sure if there is a difference...both seems to mean the same thing, ie we are assigning a truth value $T$ to everything that is either in $A$ or $B$, the only difference seems to be that the latter is using $alpha$ to denote the formulae in the union while the former uses $x$
– Daniel Mak
Jul 25 at 7:24
Forgive my ignorance but in terms of meaning I honestly am not sure if there is a difference...both seems to mean the same thing, ie we are assigning a truth value $T$ to everything that is either in $A$ or $B$, the only difference seems to be that the latter is using $alpha$ to denote the formulae in the union while the former uses $x$
– Daniel Mak
Jul 25 at 7:24
Edited to credit Prof David Makinson at LSE
– Daniel Mak
Jul 25 at 7:26
Edited to credit Prof David Makinson at LSE
– Daniel Mak
Jul 25 at 7:26
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The union $Acup B$ is defined as $x$. So, $cin Acup B$ iff it is true that "$cin A$ or $cin B$."
For example, if $A=1,2,3$ and $B=2,3,4,5$, then $Acup B = 1,2,3,4,5$. Using the definition, $2in Acup B$ because it is true that "$2in A$ or $2in B$" ... so are other elements of $Acup B$.
For now, we make the numbers "represent" propositions. Suppose a valuation $v$ makes all formulae in $A∪B$ true. That is, $v(1)=T, v(2)=T,..., v(5)=T$.
With $A=1,2,3$, do we have $v(1)=T, v(2)=T$, and $v(3)=T$ ?
Edit:
Another example: Let $A=1,2,3$ and $B=4,5$. Then, $Acup B=1,2,3,4,5$.
Now, suppose we have a valuation $v$ such that all formulae in $A$ are true but all formulae in $B$ are false. So we have $v(1)=v(2)=v(3)=T$ and $v(4)=v(5)=F$.
With $Acup B=1,2,3,4,5$, do we have $v(1)=v(2)=...=v(5)=T$ (or are all formulae in $Acup B$ true)?
answered Jul 26 at 7:07
Poypoyan
319310
Thank you so much for your reply! Yes it would seem that we do have $v(1)=v(2)=v(3)=T$. (Side question: am I right in saying that it makes no sense to say that that $v(Acup B)=T$/a valuation makes $Acup B$ true, because a set cannot be true or false, but only to say that $v$ makes ALL FORMULAE in $Acup B$ true, since only formulae can take a truth value?)
– Daniel Mak
Jul 27 at 4:55
But suppose we have a valuation st all formulae in $A$ are true but all formulae in $B$ are false (obviously your example would then not be applicable since $A$ and $B$ are mutually exclusive), would that make all formulae in $Acup B$ true? I think this is where it confuses me the most because the definition of a union utilises a disjunction, which will be true as long as one disjunct is true. Just by looking at what has been said, it would seem that the answer is no/not all formulae in $Acup B$ is true, but I am not 100% certain.
– Daniel Mak
Jul 27 at 4:57
1
(1) ...but only to say that $v$ makes ALL FORMULAE in $A∪B$ true, since only formulae can take a truth value? Yes. Take note that $v(A∪B)=T$ can be an abuse of notation...
– Poypoyan
Jul 27 at 8:57
1
(2) The disjunction utilized in the set union is just a criterion for an element to be a member of the set $Acup B$. The valuation $v$ doesn't care on set operations – it only cares on a formula $p$, which it tells whether $p$ is $True$ or $False$.
– Poypoyan
Jul 27 at 9:56
1
(3) See my edit on my answer.
– Poypoyan
Jul 27 at 10:24
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The union $Acup B$ is defined as $x$. So, $cin Acup B$ iff it is true that "$cin A$ or $cin B$."
For example, if $A=1,2,3$ and $B=2,3,4,5$, then $Acup B = 1,2,3,4,5$. Using the definition, $2in Acup B$ because it is true that "$2in A$ or $2in B$" ... so are other elements of $Acup B$.
For now, we make the numbers "represent" propositions. Suppose a valuation $v$ makes all formulae in $A∪B$ true. That is, $v(1)=T, v(2)=T,..., v(5)=T$.
With $A=1,2,3$, do we have $v(1)=T, v(2)=T$, and $v(3)=T$ ?
Edit:
Another example: Let $A=1,2,3$ and $B=4,5$. Then, $Acup B=1,2,3,4,5$.
Now, suppose we have a valuation $v$ such that all formulae in $A$ are true but all formulae in $B$ are false. So we have $v(1)=v(2)=v(3)=T$ and $v(4)=v(5)=F$.
With $Acup B=1,2,3,4,5$, do we have $v(1)=v(2)=...=v(5)=T$ (or are all formulae in $Acup B$ true)?
answered Jul 26 at 7:07
Poypoyan
319310
Thank you so much for your reply! Yes it would seem that we do have $v(1)=v(2)=v(3)=T$. (Side question: am I right in saying that it makes no sense to say that that $v(Acup B)=T$/a valuation makes $Acup B$ true, because a set cannot be true or false, but only to say that $v$ makes ALL FORMULAE in $Acup B$ true, since only formulae can take a truth value?)
– Daniel Mak
Jul 27 at 4:55
But suppose we have a valuation st all formulae in $A$ are true but all formulae in $B$ are false (obviously your example would then not be applicable since $A$ and $B$ are mutually exclusive), would that make all formulae in $Acup B$ true? I think this is where it confuses me the most because the definition of a union utilises a disjunction, which will be true as long as one disjunct is true. Just by looking at what has been said, it would seem that the answer is no/not all formulae in $Acup B$ is true, but I am not 100% certain.
– Daniel Mak
Jul 27 at 4:57
1
(1) ...but only to say that $v$ makes ALL FORMULAE in $A∪B$ true, since only formulae can take a truth value? Yes. Take note that $v(A∪B)=T$ can be an abuse of notation...
– Poypoyan
Jul 27 at 8:57
1
(2) The disjunction utilized in the set union is just a criterion for an element to be a member of the set $Acup B$. The valuation $v$ doesn't care on set operations – it only cares on a formula $p$, which it tells whether $p$ is $True$ or $False$.
– Poypoyan
Jul 27 at 9:56
1
(3) See my edit on my answer.
– Poypoyan
Jul 27 at 10:24
 |Â
show 1 more comment
up vote
2
down vote
accepted
The union $Acup B$ is defined as $x$. So, $cin Acup B$ iff it is true that "$cin A$ or $cin B$."
For example, if $A=1,2,3$ and $B=2,3,4,5$, then $Acup B = 1,2,3,4,5$. Using the definition, $2in Acup B$ because it is true that "$2in A$ or $2in B$" ... so are other elements of $Acup B$.
For now, we make the numbers "represent" propositions. Suppose a valuation $v$ makes all formulae in $A∪B$ true. That is, $v(1)=T, v(2)=T,..., v(5)=T$.
With $A=1,2,3$, do we have $v(1)=T, v(2)=T$, and $v(3)=T$ ?
Edit:
Another example: Let $A=1,2,3$ and $B=4,5$. Then, $Acup B=1,2,3,4,5$.
Now, suppose we have a valuation $v$ such that all formulae in $A$ are true but all formulae in $B$ are false. So we have $v(1)=v(2)=v(3)=T$ and $v(4)=v(5)=F$.
With $Acup B=1,2,3,4,5$, do we have $v(1)=v(2)=...=v(5)=T$ (or are all formulae in $Acup B$ true)?
answered Jul 26 at 7:07
Poypoyan
319310
Thank you so much for your reply! Yes it would seem that we do have $v(1)=v(2)=v(3)=T$. (Side question: am I right in saying that it makes no sense to say that that $v(Acup B)=T$/a valuation makes $Acup B$ true, because a set cannot be true or false, but only to say that $v$ makes ALL FORMULAE in $Acup B$ true, since only formulae can take a truth value?)
– Daniel Mak
Jul 27 at 4:55
But suppose we have a valuation st all formulae in $A$ are true but all formulae in $B$ are false (obviously your example would then not be applicable since $A$ and $B$ are mutually exclusive), would that make all formulae in $Acup B$ true? I think this is where it confuses me the most because the definition of a union utilises a disjunction, which will be true as long as one disjunct is true. Just by looking at what has been said, it would seem that the answer is no/not all formulae in $Acup B$ is true, but I am not 100% certain.
– Daniel Mak
Jul 27 at 4:57
1
(1) ...but only to say that $v$ makes ALL FORMULAE in $A∪B$ true, since only formulae can take a truth value? Yes. Take note that $v(A∪B)=T$ can be an abuse of notation...
– Poypoyan
Jul 27 at 8:57
1
(2) The disjunction utilized in the set union is just a criterion for an element to be a member of the set $Acup B$. The valuation $v$ doesn't care on set operations – it only cares on a formula $p$, which it tells whether $p$ is $True$ or $False$.
– Poypoyan
Jul 27 at 9:56
1
(3) See my edit on my answer.
– Poypoyan
Jul 27 at 10:24
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The union $Acup B$ is defined as $x$. So, $cin Acup B$ iff it is true that "$cin A$ or $cin B$."
For example, if $A=1,2,3$ and $B=2,3,4,5$, then $Acup B = 1,2,3,4,5$. Using the definition, $2in Acup B$ because it is true that "$2in A$ or $2in B$" ... so are other elements of $Acup B$.
For now, we make the numbers "represent" propositions. Suppose a valuation $v$ makes all formulae in $A∪B$ true. That is, $v(1)=T, v(2)=T,..., v(5)=T$.
With $A=1,2,3$, do we have $v(1)=T, v(2)=T$, and $v(3)=T$ ?
Edit:
Another example: Let $A=1,2,3$ and $B=4,5$. Then, $Acup B=1,2,3,4,5$.
Now, suppose we have a valuation $v$ such that all formulae in $A$ are true but all formulae in $B$ are false. So we have $v(1)=v(2)=v(3)=T$ and $v(4)=v(5)=F$.
With $Acup B=1,2,3,4,5$, do we have $v(1)=v(2)=...=v(5)=T$ (or are all formulae in $Acup B$ true)?
answered Jul 26 at 7:07
Poypoyan
319310
The union $Acup B$ is defined as $x$. So, $cin Acup B$ iff it is true that "$cin A$ or $cin B$."
For example, if $A=1,2,3$ and $B=2,3,4,5$, then $Acup B = 1,2,3,4,5$. Using the definition, $2in Acup B$ because it is true that "$2in A$ or $2in B$" ... so are other elements of $Acup B$.
For now, we make the numbers "represent" propositions. Suppose a valuation $v$ makes all formulae in $A∪B$ true. That is, $v(1)=T, v(2)=T,..., v(5)=T$.
With $A=1,2,3$, do we have $v(1)=T, v(2)=T$, and $v(3)=T$ ?
Edit:
Another example: Let $A=1,2,3$ and $B=4,5$. Then, $Acup B=1,2,3,4,5$.
Now, suppose we have a valuation $v$ such that all formulae in $A$ are true but all formulae in $B$ are false. So we have $v(1)=v(2)=v(3)=T$ and $v(4)=v(5)=F$.
With $Acup B=1,2,3,4,5$, do we have $v(1)=v(2)=...=v(5)=T$ (or are all formulae in $Acup B$ true)?
answered Jul 26 at 7:07
Poypoyan
319310
edited Jul 27 at 10:23
answered Jul 26 at 7:07
Poypoyan
319310
answered Jul 26 at 7:07
Poypoyan
319310
answered Jul 26 at 7:07
Poypoyan
319310
319310
Thank you so much for your reply! Yes it would seem that we do have $v(1)=v(2)=v(3)=T$. (Side question: am I right in saying that it makes no sense to say that that $v(Acup B)=T$/a valuation makes $Acup B$ true, because a set cannot be true or false, but only to say that $v$ makes ALL FORMULAE in $Acup B$ true, since only formulae can take a truth value?)
– Daniel Mak
Jul 27 at 4:55
But suppose we have a valuation st all formulae in $A$ are true but all formulae in $B$ are false (obviously your example would then not be applicable since $A$ and $B$ are mutually exclusive), would that make all formulae in $Acup B$ true? I think this is where it confuses me the most because the definition of a union utilises a disjunction, which will be true as long as one disjunct is true. Just by looking at what has been said, it would seem that the answer is no/not all formulae in $Acup B$ is true, but I am not 100% certain.
– Daniel Mak
Jul 27 at 4:57
1
(1) ...but only to say that $v$ makes ALL FORMULAE in $A∪B$ true, since only formulae can take a truth value? Yes. Take note that $v(A∪B)=T$ can be an abuse of notation...
– Poypoyan
Jul 27 at 8:57
1
(2) The disjunction utilized in the set union is just a criterion for an element to be a member of the set $Acup B$. The valuation $v$ doesn't care on set operations – it only cares on a formula $p$, which it tells whether $p$ is $True$ or $False$.
– Poypoyan
Jul 27 at 9:56
1
(3) See my edit on my answer.
– Poypoyan
Jul 27 at 10:24
 |Â
show 1 more comment
Thank you so much for your reply! Yes it would seem that we do have $v(1)=v(2)=v(3)=T$. (Side question: am I right in saying that it makes no sense to say that that $v(Acup B)=T$/a valuation makes $Acup B$ true, because a set cannot be true or false, but only to say that $v$ makes ALL FORMULAE in $Acup B$ true, since only formulae can take a truth value?)
– Daniel Mak
Jul 27 at 4:55
But suppose we have a valuation st all formulae in $A$ are true but all formulae in $B$ are false (obviously your example would then not be applicable since $A$ and $B$ are mutually exclusive), would that make all formulae in $Acup B$ true? I think this is where it confuses me the most because the definition of a union utilises a disjunction, which will be true as long as one disjunct is true. Just by looking at what has been said, it would seem that the answer is no/not all formulae in $Acup B$ is true, but I am not 100% certain.
– Daniel Mak
Jul 27 at 4:57
1
(1) ...but only to say that $v$ makes ALL FORMULAE in $A∪B$ true, since only formulae can take a truth value? Yes. Take note that $v(A∪B)=T$ can be an abuse of notation...
– Poypoyan
Jul 27 at 8:57
1
(2) The disjunction utilized in the set union is just a criterion for an element to be a member of the set $Acup B$. The valuation $v$ doesn't care on set operations – it only cares on a formula $p$, which it tells whether $p$ is $True$ or $False$.
– Poypoyan
Jul 27 at 9:56
1
(3) See my edit on my answer.
– Poypoyan
Jul 27 at 10:24
Thank you so much for your reply! Yes it would seem that we do have $v(1)=v(2)=v(3)=T$. (Side question: am I right in saying that it makes no sense to say that that $v(Acup B)=T$/a valuation makes $Acup B$ true, because a set cannot be true or false, but only to say that $v$ makes ALL FORMULAE in $Acup B$ true, since only formulae can take a truth value?)
– Daniel Mak
Jul 27 at 4:55
Thank you so much for your reply! Yes it would seem that we do have $v(1)=v(2)=v(3)=T$. (Side question: am I right in saying that it makes no sense to say that that $v(Acup B)=T$/a valuation makes $Acup B$ true, because a set cannot be true or false, but only to say that $v$ makes ALL FORMULAE in $Acup B$ true, since only formulae can take a truth value?)
– Daniel Mak
Jul 27 at 4:55
But suppose we have a valuation st all formulae in $A$ are true but all formulae in $B$ are false (obviously your example would then not be applicable since $A$ and $B$ are mutually exclusive), would that make all formulae in $Acup B$ true? I think this is where it confuses me the most because the definition of a union utilises a disjunction, which will be true as long as one disjunct is true. Just by looking at what has been said, it would seem that the answer is no/not all formulae in $Acup B$ is true, but I am not 100% certain.
– Daniel Mak
Jul 27 at 4:57
But suppose we have a valuation st all formulae in $A$ are true but all formulae in $B$ are false (obviously your example would then not be applicable since $A$ and $B$ are mutually exclusive), would that make all formulae in $Acup B$ true? I think this is where it confuses me the most because the definition of a union utilises a disjunction, which will be true as long as one disjunct is true. Just by looking at what has been said, it would seem that the answer is no/not all formulae in $Acup B$ is true, but I am not 100% certain.
– Daniel Mak
Jul 27 at 4:57
1
1
(1) ...but only to say that $v$ makes ALL FORMULAE in $A∪B$ true, since only formulae can take a truth value? Yes. Take note that $v(A∪B)=T$ can be an abuse of notation...
– Poypoyan
Jul 27 at 8:57
(1) ...but only to say that $v$ makes ALL FORMULAE in $A∪B$ true, since only formulae can take a truth value? Yes. Take note that $v(A∪B)=T$ can be an abuse of notation...
– Poypoyan
Jul 27 at 8:57
1
1
(2) The disjunction utilized in the set union is just a criterion for an element to be a member of the set $Acup B$. The valuation $v$ doesn't care on set operations – it only cares on a formula $p$, which it tells whether $p$ is $True$ or $False$.
– Poypoyan
Jul 27 at 9:56
(2) The disjunction utilized in the set union is just a criterion for an element to be a member of the set $Acup B$. The valuation $v$ doesn't care on set operations – it only cares on a formula $p$, which it tells whether $p$ is $True$ or $False$.
– Poypoyan
Jul 27 at 9:56
1
1
(3) See my edit on my answer.
– Poypoyan
Jul 27 at 10:24
(3) See my edit on my answer.
– Poypoyan
Jul 27 at 10:24
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It's from the notes of a course I attended earlier so I am not sure how to quote it. $v(forall x(xin Acup B))=T$ just means every formulae in $Acup B$ is true, so it's just a formulated way of saying the same thing; unless I am understanding what the solution said wrong?
– Daniel Mak
Jul 25 at 7:13
And while I do respect your concern of plagiarism, I am not sure how this is the case here, as I am certainly not taking credit for it (otherwise I would have understood it), nor am I passing it on as an answer or anything...
– Daniel Mak
Jul 25 at 7:17
Forgive my ignorance but in terms of meaning I honestly am not sure if there is a difference...both seems to mean the same thing, ie we are assigning a truth value $T$ to everything that is either in $A$ or $B$, the only difference seems to be that the latter is using $alpha$ to denote the formulae in the union while the former uses $x$
– Daniel Mak
Jul 25 at 7:24
Edited to credit Prof David Makinson at LSE
– Daniel Mak
Jul 25 at 7:26