Doubt regarding $∀x(P(x)→Q(x))→(∀x(P(x))→∀x(Q(x)))$ validity
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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 at 13:59
user 170039
10.4k42465
asked Nov 21 at 13:11
user10143594
103
add a comment |
up vote
-1
down vote
favorite
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 at 13:59
user 170039
10.4k42465
asked Nov 21 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 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 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 at 13:23
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
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 at 13:59
user 170039
10.4k42465
asked Nov 21 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 at 13:59
user 170039
10.4k42465
asked Nov 21 at 13:11
user10143594
103
edited Nov 21 at 13:59
user 170039
10.4k42465
asked Nov 21 at 13:11
user10143594
103
edited Nov 21 at 13:59
user 170039
10.4k42465
edited Nov 21 at 13:59
user 170039
10.4k42465
edited Nov 21 at 13:59
user 170039
10.4k42465
10.4k42465
asked Nov 21 at 13:11
user10143594
103
asked Nov 21 at 13:11
user10143594
103
asked Nov 21 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 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 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 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 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 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 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 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 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 at 13:20
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 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 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 at 13:23
add a comment |
2 Answers
2
active
oldest
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up vote
0
down vote
accepted
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 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 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 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 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 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 at 13:37
|
show 8 more comments
up vote
0
down vote
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 at 13:49
Dole
898514
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 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 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 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 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 at 19:00
|
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
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 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 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 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 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 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 at 13:37
|
show 8 more comments
up vote
0
down vote
accepted
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 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 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 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 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 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 at 13:37
|
show 8 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
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 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 at 13:18
Henrique Lecco
363
edited Nov 21 at 13:44
answered Nov 21 at 13:18
Henrique Lecco
363
answered Nov 21 at 13:18
Henrique Lecco
363
answered Nov 21 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 at 13:37
|
show 8 more comments
up vote
0
down vote
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 at 13:49
Dole
898514
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 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 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 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 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 at 19:00
|
show 1 more comment
up vote
0
down vote
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 at 13:49
Dole
898514
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 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 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 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 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 at 19:00
|
show 1 more comment
up vote
0
down vote
up vote
0
down vote
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 at 13:49
Dole
898514
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 at 13:49
Dole
898514
edited Nov 21 at 14:05
answered Nov 21 at 13:49
Dole
898514
answered Nov 21 at 13:49
Dole
898514
answered Nov 21 at 13:49
Dole
898514
898514
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 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 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 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 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 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 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 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 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 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 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 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 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 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 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 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 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 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 at 18:55
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 at 19:00
In my above comment, all I wrote was based on your answer only
– user10143594
Nov 21 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 at 13:16
But $a$ is whatever; thus, we can "generalize" on it.
– Mauro ALLEGRANZA
Nov 21 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 at 13:23