how to prove ∃x(∃yA(y) → A(x)) is valid in classical logic
∃x(∃yA(y) → A(x))
i know how to prove it by natural deduction. but it requires to use model theoretical approach. so i'm not sure how we can prove the validity.
can anyone help me please ?
logic predicate-logic
edited Nov 21 '18 at 14:58
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 14:18
Norman
227
add a comment |
∃x(∃yA(y) → A(x))
i know how to prove it by natural deduction. but it requires to use model theoretical approach. so i'm not sure how we can prove the validity.
can anyone help me please ?
logic predicate-logic
edited Nov 21 '18 at 14:58
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 14:18
Norman
227
add a comment |
∃x(∃yA(y) → A(x))
i know how to prove it by natural deduction. but it requires to use model theoretical approach. so i'm not sure how we can prove the validity.
can anyone help me please ?
logic predicate-logic
edited Nov 21 '18 at 14:58
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 14:18
Norman
227
∃x(∃yA(y) → A(x))
i know how to prove it by natural deduction. but it requires to use model theoretical approach. so i'm not sure how we can prove the validity.
can anyone help me please ?
logic predicate-logic
logic predicate-logic
edited Nov 21 '18 at 14:58
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 14:18
Norman
227
edited Nov 21 '18 at 14:58
Mauro ALLEGRANZA
64.4k448112
asked Nov 21 '18 at 14:18
Norman
227
edited Nov 21 '18 at 14:58
Mauro ALLEGRANZA
64.4k448112
edited Nov 21 '18 at 14:58
Mauro ALLEGRANZA
64.4k448112
edited Nov 21 '18 at 14:58
Mauro ALLEGRANZA
64.4k448112
64.4k448112
asked Nov 21 '18 at 14:18
Norman
227
asked Nov 21 '18 at 14:18
Norman
227
asked Nov 21 '18 at 14:18
Norman
227
227
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃yA(y)$ is False.
Thus $∃yA(y) to A(x)$ is True.
For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.
And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.
(ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.
This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.
answered Nov 21 '18 at 14:37
Mauro ALLEGRANZA
64.4k448112
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
add a comment |
Try a proof by contradiction:
So, suppose there is some structure that makes this sentence false.
That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.
By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.
By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.
But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.
So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.
Hence, the sentence is valid.
answered Nov 21 '18 at 14:31
Bram28
60.3k44590
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
@Norman You're welcome :)
– Bram28
Nov 21 '18 at 17:37
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃yA(y)$ is False.
Thus $∃yA(y) to A(x)$ is True.
For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.
And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.
(ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.
This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.
answered Nov 21 '18 at 14:37
Mauro ALLEGRANZA
64.4k448112
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
add a comment |
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃yA(y)$ is False.
Thus $∃yA(y) to A(x)$ is True.
For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.
And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.
(ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.
This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.
answered Nov 21 '18 at 14:37
Mauro ALLEGRANZA
64.4k448112
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
add a comment |
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃yA(y)$ is False.
Thus $∃yA(y) to A(x)$ is True.
For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.
And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.
(ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.
This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.
answered Nov 21 '18 at 14:37
Mauro ALLEGRANZA
64.4k448112
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃yA(y)$ is False.
Thus $∃yA(y) to A(x)$ is True.
For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.
And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.
(ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.
This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.
answered Nov 21 '18 at 14:37
Mauro ALLEGRANZA
64.4k448112
edited Nov 21 '18 at 14:43
answered Nov 21 '18 at 14:37
Mauro ALLEGRANZA
64.4k448112
answered Nov 21 '18 at 14:37
Mauro ALLEGRANZA
64.4k448112
answered Nov 21 '18 at 14:37
Mauro ALLEGRANZA
64.4k448112
64.4k448112
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
add a comment |
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
add a comment |
Try a proof by contradiction:
So, suppose there is some structure that makes this sentence false.
That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.
By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.
By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.
But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.
So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.
Hence, the sentence is valid.
answered Nov 21 '18 at 14:31
Bram28
60.3k44590
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
@Norman You're welcome :)
– Bram28
Nov 21 '18 at 17:37
add a comment |
Try a proof by contradiction:
So, suppose there is some structure that makes this sentence false.
That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.
By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.
By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.
But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.
So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.
Hence, the sentence is valid.
answered Nov 21 '18 at 14:31
Bram28
60.3k44590
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
@Norman You're welcome :)
– Bram28
Nov 21 '18 at 17:37
add a comment |
Try a proof by contradiction:
So, suppose there is some structure that makes this sentence false.
That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.
By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.
By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.
But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.
So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.
Hence, the sentence is valid.
answered Nov 21 '18 at 14:31
Bram28
60.3k44590
Try a proof by contradiction:
So, suppose there is some structure that makes this sentence false.
That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.
By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.
By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.
But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.
So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.
Hence, the sentence is valid.
answered Nov 21 '18 at 14:31
Bram28
60.3k44590
edited Nov 21 '18 at 14:54
answered Nov 21 '18 at 14:31
Bram28
60.3k44590
answered Nov 21 '18 at 14:31
Bram28
60.3k44590
answered Nov 21 '18 at 14:31
Bram28
60.3k44590
60.3k44590
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
@Norman You're welcome :)
– Bram28
Nov 21 '18 at 17:37
add a comment |
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
@Norman You're welcome :)
– Bram28
Nov 21 '18 at 17:37
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
Thank you so much. i really appreciate it.
– Norman
Nov 21 '18 at 16:01
@Norman You're welcome :)
– Bram28
Nov 21 '18 at 17:37
@Norman You're welcome :)
– Bram28
Nov 21 '18 at 17:37
add a comment |
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