Mixing
Doubt regarding $∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x)))$ validity
I know this is valid. But from LHS we can infer only for same $x$ for both $P$ and $Q$. So, I want to know how we can directly arrive at $P(a)→Q(b)$, when only information we have is $P(a)→Q(a)$ and $P(b)→Q(b)$.
logic
edited Nov 21 '18 at 13:59
user 170039
10.4k42465
asked Nov 21 '18 at 13:11
user10143594
103
add a comment |
I know this is valid. But from LHS we can infer only for same $x$ for both $P$ and $Q$. So, I want to know how we can directly arrive at $P(a)→Q(b)$, when only information we have is $P(a)→Q(a)$ and $P(b)→Q(b)$.
logic
edited Nov 21 '18 at 13:59
user 170039
10.4k42465
asked Nov 21 '18 at 13:11
user10143594
103
We have $forall x (Px → Qx)$ and we assume $forall x Px$. Then we derive $Pa to Qa$ and $Pa$ respectively. From them : $Qa$.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:20
Consider domain of disourse consisiting of {a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
add a comment |
I know this is valid. But from LHS we can infer only for same $x$ for both $P$ and $Q$. So, I want to know how we can directly arrive at $P(a)→Q(b)$, when only information we have is $P(a)→Q(a)$ and $P(b)→Q(b)$.
logic
edited Nov 21 '18 at 13:59
user 170039
10.4k42465
asked Nov 21 '18 at 13:11
user10143594
103
I know this is valid. But from LHS we can infer only for same $x$ for both $P$ and $Q$. So, I want to know how we can directly arrive at $P(a)→Q(b)$, when only information we have is $P(a)→Q(a)$ and $P(b)→Q(b)$.
logic
logic
edited Nov 21 '18 at 13:59
user 170039
10.4k42465
asked Nov 21 '18 at 13:11
user10143594
103
edited Nov 21 '18 at 13:59
user 170039
10.4k42465
asked Nov 21 '18 at 13:11
user10143594
103
edited Nov 21 '18 at 13:59
user 170039
10.4k42465
edited Nov 21 '18 at 13:59
user 170039
10.4k42465
edited Nov 21 '18 at 13:59
user 170039
10.4k42465
10.4k42465
asked Nov 21 '18 at 13:11
user10143594
103
asked Nov 21 '18 at 13:11
user10143594
103
asked Nov 21 '18 at 13:11
user10143594
103
103
We have $forall x (Px → Qx)$ and we assume $forall x Px$. Then we derive $Pa to Qa$ and $Pa$ respectively. From them : $Qa$.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:20
Consider domain of disourse consisiting of {a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
add a comment |
We have $forall x (Px → Qx)$ and we assume $forall x Px$. Then we derive $Pa to Qa$ and $Pa$ respectively. From them : $Qa$.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:20
Consider domain of disourse consisiting of {a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
We have $forall x (Px → Qx)$ and we assume $forall x Px$. Then we derive $Pa to Qa$ and $Pa$ respectively. From them : $Qa$.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:16
We have $forall x (Px → Qx)$ and we assume $forall x Px$. Then we derive $Pa to Qa$ and $Pa$ respectively. From them : $Qa$.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:20
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:20
Consider domain of disourse consisiting of {a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
Consider domain of disourse consisiting of {a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
add a comment |
2 Answers
2
active
oldest
votes
If P holds for all x and for each x, P(x) implies Q(x), then Q holds for all x.
You never need P(a) implies Q(b), but you have guaranteed, on RHS that P(x) for all x, and by LHS, P(x) implies Q(x).
As an example, let the domain be {a, b}.
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))". Note that we are NOT proving "for all x (P(x) implies for all x Q(x))".
answered Nov 21 '18 at 13:18
Henrique Lecco
363
Consider domain of disourse as{a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
First of all consider only premise this ∀x(P(x)→Q(x)) . After that we have to prove ∀x(P(x))→∀x(Q(x))
– user10143594
Nov 21 '18 at 13:26
1
What you're saying is different from what is written. You are saying "for all x, y, P(x) implies Q(y)". This is different from "for all x P(x) implies for all x Q(x)" . Read it as "the fact that P holds for all x implies that Q also holds for all x".
– Henrique Lecco
Nov 21 '18 at 13:27
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}
– user10143594
Nov 21 '18 at 13:32
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))".
– Henrique Lecco
Nov 21 '18 at 13:37
|
show 8 more comments
Do you mean valid in a semantic or deductive way?
Deductively it can be shown as follows:
Assume $forall x( P(x) rightarrow Q(x))$
Suppose $ forall x P(x)$,
Therefore $P(x)$ and as such $Q(x)$ by the assumption. As such $forall x Q(x)$.
That completes the proof. Since natural deduction using the usual rules is semantically valid, then so is the formula. It's valid even constructively.
For an example, let the domain be $A={a,b}$. Now assume:
$$forall x( P(x) rightarrow Q(x))$$
As such:
$$ P(a) rightarrow Q(a),P(b) rightarrow Q(b)$$
Now assume:
$P(a),P(b)$
Therefore by the above: $Q(a),Q(b)$. Since $a,b$ is the whole domain we are allowed to bring in the "for all" quantifier.
And that concludes the proof. Note that model theory would be inconsistent if this was not the case. By the above proof the sentence should hold for all possible worlds.
Finally, there is small error in the question. $P(a)rightarrow Q(b)$ is not the case.
answered Nov 21 '18 at 13:49
Dole
904514
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}.
– user10143594
Nov 21 '18 at 13:50
@user10143594 Okay, so you want to show that there is some model that satisfies it? I will edit the post...
– Dole
Nov 21 '18 at 13:53
If P(a) and P(b) are true and then obviously Q(a) and Q(b) are true. So, P(a) -> Q(b)
– user10143594
Nov 21 '18 at 14:29
@user10143594 The first statement does not infer that $P(a)rightarrow Q(b)$. It's only after that, when it's assumd that $P(a)$ IS true for all a, that you can make such inference. Sure, that does imply it.
– Dole
Nov 21 '18 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 '18 at 19:00
|
show 1 more comment
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2 Answers
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2 Answers
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If P holds for all x and for each x, P(x) implies Q(x), then Q holds for all x.
You never need P(a) implies Q(b), but you have guaranteed, on RHS that P(x) for all x, and by LHS, P(x) implies Q(x).
As an example, let the domain be {a, b}.
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))". Note that we are NOT proving "for all x (P(x) implies for all x Q(x))".
answered Nov 21 '18 at 13:18
Henrique Lecco
363
Consider domain of disourse as{a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
First of all consider only premise this ∀x(P(x)→Q(x)) . After that we have to prove ∀x(P(x))→∀x(Q(x))
– user10143594
Nov 21 '18 at 13:26
1
What you're saying is different from what is written. You are saying "for all x, y, P(x) implies Q(y)". This is different from "for all x P(x) implies for all x Q(x)" . Read it as "the fact that P holds for all x implies that Q also holds for all x".
– Henrique Lecco
Nov 21 '18 at 13:27
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}
– user10143594
Nov 21 '18 at 13:32
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))".
– Henrique Lecco
Nov 21 '18 at 13:37
|
show 8 more comments
If P holds for all x and for each x, P(x) implies Q(x), then Q holds for all x.
You never need P(a) implies Q(b), but you have guaranteed, on RHS that P(x) for all x, and by LHS, P(x) implies Q(x).
As an example, let the domain be {a, b}.
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))". Note that we are NOT proving "for all x (P(x) implies for all x Q(x))".
answered Nov 21 '18 at 13:18
Henrique Lecco
363
Consider domain of disourse as{a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
First of all consider only premise this ∀x(P(x)→Q(x)) . After that we have to prove ∀x(P(x))→∀x(Q(x))
– user10143594
Nov 21 '18 at 13:26
1
What you're saying is different from what is written. You are saying "for all x, y, P(x) implies Q(y)". This is different from "for all x P(x) implies for all x Q(x)" . Read it as "the fact that P holds for all x implies that Q also holds for all x".
– Henrique Lecco
Nov 21 '18 at 13:27
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}
– user10143594
Nov 21 '18 at 13:32
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))".
– Henrique Lecco
Nov 21 '18 at 13:37
|
show 8 more comments
If P holds for all x and for each x, P(x) implies Q(x), then Q holds for all x.
You never need P(a) implies Q(b), but you have guaranteed, on RHS that P(x) for all x, and by LHS, P(x) implies Q(x).
As an example, let the domain be {a, b}.
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))". Note that we are NOT proving "for all x (P(x) implies for all x Q(x))".
answered Nov 21 '18 at 13:18
Henrique Lecco
363
If P holds for all x and for each x, P(x) implies Q(x), then Q holds for all x.
You never need P(a) implies Q(b), but you have guaranteed, on RHS that P(x) for all x, and by LHS, P(x) implies Q(x).
As an example, let the domain be {a, b}.
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))". Note that we are NOT proving "for all x (P(x) implies for all x Q(x))".
answered Nov 21 '18 at 13:18
Henrique Lecco
363
edited Nov 21 '18 at 13:44
answered Nov 21 '18 at 13:18
Henrique Lecco
363
answered Nov 21 '18 at 13:18
Henrique Lecco
363
answered Nov 21 '18 at 13:18
Henrique Lecco
363
363
Consider domain of disourse as{a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
First of all consider only premise this ∀x(P(x)→Q(x)) . After that we have to prove ∀x(P(x))→∀x(Q(x))
– user10143594
Nov 21 '18 at 13:26
1
What you're saying is different from what is written. You are saying "for all x, y, P(x) implies Q(y)". This is different from "for all x P(x) implies for all x Q(x)" . Read it as "the fact that P holds for all x implies that Q also holds for all x".
– Henrique Lecco
Nov 21 '18 at 13:27
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}
– user10143594
Nov 21 '18 at 13:32
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))".
– Henrique Lecco
Nov 21 '18 at 13:37
|
show 8 more comments
Consider domain of disourse as{a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
First of all consider only premise this ∀x(P(x)→Q(x)) . After that we have to prove ∀x(P(x))→∀x(Q(x))
– user10143594
Nov 21 '18 at 13:26
1
What you're saying is different from what is written. You are saying "for all x, y, P(x) implies Q(y)". This is different from "for all x P(x) implies for all x Q(x)" . Read it as "the fact that P holds for all x implies that Q also holds for all x".
– Henrique Lecco
Nov 21 '18 at 13:27
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}
– user10143594
Nov 21 '18 at 13:32
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))".
– Henrique Lecco
Nov 21 '18 at 13:37
Consider domain of disourse as{a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
Consider domain of disourse as{a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23
First of all consider only premise this ∀x(P(x)→Q(x)) . After that we have to prove ∀x(P(x))→∀x(Q(x))
– user10143594
Nov 21 '18 at 13:26
First of all consider only premise this ∀x(P(x)→Q(x)) . After that we have to prove ∀x(P(x))→∀x(Q(x))
– user10143594
Nov 21 '18 at 13:26
1
1
What you're saying is different from what is written. You are saying "for all x, y, P(x) implies Q(y)". This is different from "for all x P(x) implies for all x Q(x)" . Read it as "the fact that P holds for all x implies that Q also holds for all x".
– Henrique Lecco
Nov 21 '18 at 13:27
What you're saying is different from what is written. You are saying "for all x, y, P(x) implies Q(y)". This is different from "for all x P(x) implies for all x Q(x)" . Read it as "the fact that P holds for all x implies that Q also holds for all x".
– Henrique Lecco
Nov 21 '18 at 13:27
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}
– user10143594
Nov 21 '18 at 13:32
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}
– user10143594
Nov 21 '18 at 13:32
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))".
– Henrique Lecco
Nov 21 '18 at 13:37
Suppose LHS valid. Then P(a) implies Q(a) and P(b) implies Q(b). Now, on RHS, suppose for all x, P(x). That is, we have P(a) and P(b) valid. By LHS, P(a) implies Q(a), then Q(a) is valid. Also, P(b) implies Q(b), then Q(b) is valid. Then Q holds for both a and b, that is, Q holds for all x. Then we proved "(for all x P(x)) implies (for all x Q(x))".
– Henrique Lecco
Nov 21 '18 at 13:37
|
show 8 more comments
Do you mean valid in a semantic or deductive way?
Deductively it can be shown as follows:
Assume $forall x( P(x) rightarrow Q(x))$
Suppose $ forall x P(x)$,
Therefore $P(x)$ and as such $Q(x)$ by the assumption. As such $forall x Q(x)$.
That completes the proof. Since natural deduction using the usual rules is semantically valid, then so is the formula. It's valid even constructively.
For an example, let the domain be $A={a,b}$. Now assume:
$$forall x( P(x) rightarrow Q(x))$$
As such:
$$ P(a) rightarrow Q(a),P(b) rightarrow Q(b)$$
Now assume:
$P(a),P(b)$
Therefore by the above: $Q(a),Q(b)$. Since $a,b$ is the whole domain we are allowed to bring in the "for all" quantifier.
And that concludes the proof. Note that model theory would be inconsistent if this was not the case. By the above proof the sentence should hold for all possible worlds.
Finally, there is small error in the question. $P(a)rightarrow Q(b)$ is not the case.
answered Nov 21 '18 at 13:49
Dole
904514
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}.
– user10143594
Nov 21 '18 at 13:50
@user10143594 Okay, so you want to show that there is some model that satisfies it? I will edit the post...
– Dole
Nov 21 '18 at 13:53
If P(a) and P(b) are true and then obviously Q(a) and Q(b) are true. So, P(a) -> Q(b)
– user10143594
Nov 21 '18 at 14:29
@user10143594 The first statement does not infer that $P(a)rightarrow Q(b)$. It's only after that, when it's assumd that $P(a)$ IS true for all a, that you can make such inference. Sure, that does imply it.
– Dole
Nov 21 '18 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 '18 at 19:00
|
show 1 more comment
Do you mean valid in a semantic or deductive way?
Deductively it can be shown as follows:
Assume $forall x( P(x) rightarrow Q(x))$
Suppose $ forall x P(x)$,
Therefore $P(x)$ and as such $Q(x)$ by the assumption. As such $forall x Q(x)$.
That completes the proof. Since natural deduction using the usual rules is semantically valid, then so is the formula. It's valid even constructively.
For an example, let the domain be $A={a,b}$. Now assume:
$$forall x( P(x) rightarrow Q(x))$$
As such:
$$ P(a) rightarrow Q(a),P(b) rightarrow Q(b)$$
Now assume:
$P(a),P(b)$
Therefore by the above: $Q(a),Q(b)$. Since $a,b$ is the whole domain we are allowed to bring in the "for all" quantifier.
And that concludes the proof. Note that model theory would be inconsistent if this was not the case. By the above proof the sentence should hold for all possible worlds.
Finally, there is small error in the question. $P(a)rightarrow Q(b)$ is not the case.
answered Nov 21 '18 at 13:49
Dole
904514
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}.
– user10143594
Nov 21 '18 at 13:50
@user10143594 Okay, so you want to show that there is some model that satisfies it? I will edit the post...
– Dole
Nov 21 '18 at 13:53
If P(a) and P(b) are true and then obviously Q(a) and Q(b) are true. So, P(a) -> Q(b)
– user10143594
Nov 21 '18 at 14:29
@user10143594 The first statement does not infer that $P(a)rightarrow Q(b)$. It's only after that, when it's assumd that $P(a)$ IS true for all a, that you can make such inference. Sure, that does imply it.
– Dole
Nov 21 '18 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 '18 at 19:00
|
show 1 more comment
Do you mean valid in a semantic or deductive way?
Deductively it can be shown as follows:
Assume $forall x( P(x) rightarrow Q(x))$
Suppose $ forall x P(x)$,
Therefore $P(x)$ and as such $Q(x)$ by the assumption. As such $forall x Q(x)$.
That completes the proof. Since natural deduction using the usual rules is semantically valid, then so is the formula. It's valid even constructively.
For an example, let the domain be $A={a,b}$. Now assume:
$$forall x( P(x) rightarrow Q(x))$$
As such:
$$ P(a) rightarrow Q(a),P(b) rightarrow Q(b)$$
Now assume:
$P(a),P(b)$
Therefore by the above: $Q(a),Q(b)$. Since $a,b$ is the whole domain we are allowed to bring in the "for all" quantifier.
And that concludes the proof. Note that model theory would be inconsistent if this was not the case. By the above proof the sentence should hold for all possible worlds.
Finally, there is small error in the question. $P(a)rightarrow Q(b)$ is not the case.
answered Nov 21 '18 at 13:49
Dole
904514
Do you mean valid in a semantic or deductive way?
Deductively it can be shown as follows:
Assume $forall x( P(x) rightarrow Q(x))$
Suppose $ forall x P(x)$,
Therefore $P(x)$ and as such $Q(x)$ by the assumption. As such $forall x Q(x)$.
That completes the proof. Since natural deduction using the usual rules is semantically valid, then so is the formula. It's valid even constructively.
For an example, let the domain be $A={a,b}$. Now assume:
$$forall x( P(x) rightarrow Q(x))$$
As such:
$$ P(a) rightarrow Q(a),P(b) rightarrow Q(b)$$
Now assume:
$P(a),P(b)$
Therefore by the above: $Q(a),Q(b)$. Since $a,b$ is the whole domain we are allowed to bring in the "for all" quantifier.
And that concludes the proof. Note that model theory would be inconsistent if this was not the case. By the above proof the sentence should hold for all possible worlds.
Finally, there is small error in the question. $P(a)rightarrow Q(b)$ is not the case.
answered Nov 21 '18 at 13:49
Dole
904514
edited Nov 21 '18 at 14:05
answered Nov 21 '18 at 13:49
Dole
904514
answered Nov 21 '18 at 13:49
Dole
904514
answered Nov 21 '18 at 13:49
Dole
904514
904514
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}.
– user10143594
Nov 21 '18 at 13:50
@user10143594 Okay, so you want to show that there is some model that satisfies it? I will edit the post...
– Dole
Nov 21 '18 at 13:53
If P(a) and P(b) are true and then obviously Q(a) and Q(b) are true. So, P(a) -> Q(b)
– user10143594
Nov 21 '18 at 14:29
@user10143594 The first statement does not infer that $P(a)rightarrow Q(b)$. It's only after that, when it's assumd that $P(a)$ IS true for all a, that you can make such inference. Sure, that does imply it.
– Dole
Nov 21 '18 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 '18 at 19:00
|
show 1 more comment
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}.
– user10143594
Nov 21 '18 at 13:50
@user10143594 Okay, so you want to show that there is some model that satisfies it? I will edit the post...
– Dole
Nov 21 '18 at 13:53
If P(a) and P(b) are true and then obviously Q(a) and Q(b) are true. So, P(a) -> Q(b)
– user10143594
Nov 21 '18 at 14:29
@user10143594 The first statement does not infer that $P(a)rightarrow Q(b)$. It's only after that, when it's assumd that $P(a)$ IS true for all a, that you can make such inference. Sure, that does imply it.
– Dole
Nov 21 '18 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 '18 at 19:00
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}.
– user10143594
Nov 21 '18 at 13:50
I am saying all we have is this ∀x(P(x)→Q(x)) and the domain of discourse. Now, I am asking how we can prove it to ∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x))) . I do not want to take this formula as given and then proving it correct. I want to know how to derive it using a example. And that example is using domain of discourse {a,b}.
– user10143594
Nov 21 '18 at 13:50
@user10143594 Okay, so you want to show that there is some model that satisfies it? I will edit the post...
– Dole
Nov 21 '18 at 13:53
@user10143594 Okay, so you want to show that there is some model that satisfies it? I will edit the post...
– Dole
Nov 21 '18 at 13:53
If P(a) and P(b) are true and then obviously Q(a) and Q(b) are true. So, P(a) -> Q(b)
– user10143594
Nov 21 '18 at 14:29
If P(a) and P(b) are true and then obviously Q(a) and Q(b) are true. So, P(a) -> Q(b)
– user10143594
Nov 21 '18 at 14:29
@user10143594 The first statement does not infer that $P(a)rightarrow Q(b)$. It's only after that, when it's assumd that $P(a)$ IS true for all a, that you can make such inference. Sure, that does imply it.
– Dole
Nov 21 '18 at 18:55
@user10143594 The first statement does not infer that $P(a)rightarrow Q(b)$. It's only after that, when it's assumd that $P(a)$ IS true for all a, that you can make such inference. Sure, that does imply it.
– Dole
Nov 21 '18 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 '18 at 19:00
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 '18 at 19:00
|
show 1 more comment
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We have $forall x (Px → Qx)$ and we assume $forall x Px$. Then we derive $Pa to Qa$ and $Pa$ respectively. From them : $Qa$.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 '18 at 13:20
Consider domain of disourse consisiting of {a,b}. From premises we have P(a)→Q(a) and P(b)→Q(b). Now logically how can we say about P(a)→Q(b) ?
– user10143594
Nov 21 '18 at 13:23