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Quantifiers (logic)
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I have a question about quantifiers in logic.
I know that we can switch the quantifiers "$forall$" between them (e.g., $forall x in X, forall y in Y, p(x, y) Leftrightarrow forall y in Y, forall x in X, p(x, y)$), the quantifiers "$exists$" between them and that we cannot do this for two quantifiers "$exists$" and "$forall$" (e.g., we just have : $exists x in X, forall y in Y, p(x, y) Rightarrow forall y in Y, exists x in X, p(x, y)$)
My problem is the following, if I take for example the proposition :
$forall x in mathbbR, forall y in z in mathbbR , p(x, y)$,
the set to which $y$ belongs depends of $x$ right ?
In this case, it does not make any sense for me to switch the two quantifiers "$forall$" here (because $x$ has to be defined "first").
You could say that it's not a problem because we can switch the two "$forall"$ in the previous proposition, but then, if we take :
$exists y in z in mathbbR , forall x in mathbbR, p(x, y)$,
We cannot switch "$forall$" and "$exists$" and again, it has no sense for me...
(I have the feeling that it is simply not a good way to write it and that obviously, we have to define $x$ first, but I'm not 100% sure...)
Can you enlight me ? Thank you !
logic predicate-logic quantifiers
edited 29 mins ago
Mauro ALLEGRANZA
62.2k447106
asked 2 hours ago
deeppinkwater
265
add a comment |Â
up vote
2
down vote
favorite
I have a question about quantifiers in logic.
I know that we can switch the quantifiers "$forall$" between them (e.g., $forall x in X, forall y in Y, p(x, y) Leftrightarrow forall y in Y, forall x in X, p(x, y)$), the quantifiers "$exists$" between them and that we cannot do this for two quantifiers "$exists$" and "$forall$" (e.g., we just have : $exists x in X, forall y in Y, p(x, y) Rightarrow forall y in Y, exists x in X, p(x, y)$)
My problem is the following, if I take for example the proposition :
$forall x in mathbbR, forall y in z in mathbbR , p(x, y)$,
the set to which $y$ belongs depends of $x$ right ?
In this case, it does not make any sense for me to switch the two quantifiers "$forall$" here (because $x$ has to be defined "first").
You could say that it's not a problem because we can switch the two "$forall"$ in the previous proposition, but then, if we take :
$exists y in z in mathbbR , forall x in mathbbR, p(x, y)$,
We cannot switch "$forall$" and "$exists$" and again, it has no sense for me...
(I have the feeling that it is simply not a good way to write it and that obviously, we have to define $x$ first, but I'm not 100% sure...)
Can you enlight me ? Thank you !
logic predicate-logic quantifiers
edited 29 mins ago
Mauro ALLEGRANZA
62.2k447106
asked 2 hours ago
deeppinkwater
265
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a question about quantifiers in logic.
I know that we can switch the quantifiers "$forall$" between them (e.g., $forall x in X, forall y in Y, p(x, y) Leftrightarrow forall y in Y, forall x in X, p(x, y)$), the quantifiers "$exists$" between them and that we cannot do this for two quantifiers "$exists$" and "$forall$" (e.g., we just have : $exists x in X, forall y in Y, p(x, y) Rightarrow forall y in Y, exists x in X, p(x, y)$)
My problem is the following, if I take for example the proposition :
$forall x in mathbbR, forall y in z in mathbbR , p(x, y)$,
the set to which $y$ belongs depends of $x$ right ?
In this case, it does not make any sense for me to switch the two quantifiers "$forall$" here (because $x$ has to be defined "first").
You could say that it's not a problem because we can switch the two "$forall"$ in the previous proposition, but then, if we take :
$exists y in z in mathbbR , forall x in mathbbR, p(x, y)$,
We cannot switch "$forall$" and "$exists$" and again, it has no sense for me...
(I have the feeling that it is simply not a good way to write it and that obviously, we have to define $x$ first, but I'm not 100% sure...)
Can you enlight me ? Thank you !
logic predicate-logic quantifiers
edited 29 mins ago
Mauro ALLEGRANZA
62.2k447106
asked 2 hours ago
deeppinkwater
265
I have a question about quantifiers in logic.
I know that we can switch the quantifiers "$forall$" between them (e.g., $forall x in X, forall y in Y, p(x, y) Leftrightarrow forall y in Y, forall x in X, p(x, y)$), the quantifiers "$exists$" between them and that we cannot do this for two quantifiers "$exists$" and "$forall$" (e.g., we just have : $exists x in X, forall y in Y, p(x, y) Rightarrow forall y in Y, exists x in X, p(x, y)$)
My problem is the following, if I take for example the proposition :
$forall x in mathbbR, forall y in z in mathbbR , p(x, y)$,
the set to which $y$ belongs depends of $x$ right ?
In this case, it does not make any sense for me to switch the two quantifiers "$forall$" here (because $x$ has to be defined "first").
You could say that it's not a problem because we can switch the two "$forall"$ in the previous proposition, but then, if we take :
$exists y in z in mathbbR , forall x in mathbbR, p(x, y)$,
We cannot switch "$forall$" and "$exists$" and again, it has no sense for me...
(I have the feeling that it is simply not a good way to write it and that obviously, we have to define $x$ first, but I'm not 100% sure...)
Can you enlight me ? Thank you !
logic predicate-logic quantifiers
logic predicate-logic quantifiers
edited 29 mins ago
Mauro ALLEGRANZA
62.2k447106
asked 2 hours ago
deeppinkwater
265
edited 29 mins ago
Mauro ALLEGRANZA
62.2k447106
asked 2 hours ago
deeppinkwater
265
edited 29 mins ago
Mauro ALLEGRANZA
62.2k447106
edited 29 mins ago
Mauro ALLEGRANZA
62.2k447106
edited 29 mins ago
Mauro ALLEGRANZA
62.2k447106
62.2k447106
asked 2 hours ago
deeppinkwater
265
asked 2 hours ago
deeppinkwater
265
asked 2 hours ago
deeppinkwater
265
265
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
Long comment
In this case, the use of set theory symbols seems to complicate things.
Assume for simplicity $mathbb R$ as domain; in this way, we can simply write $forall x$.
What does it mean : $∀y ∈ z mid z=x+3 $ ? It simply means : $y=x+3$.
Thus, the formula will be : $forall x forall y (y=x+3 to p(x,y))$.
In this case, we have no issue with the swap of the two quantifiers.
answered 2 hours ago
Mauro ALLEGRANZA
62.2k447106
We can even write it without $y$: $forall x in mathbb R, p(x,x+3)$. There is no need of $forall y$ because $y=x+3$ only have one solution in $mathbb R$. $yequiv x+3 pmod n$ would have been a better exemple. (With $x,y in mathbb N or mathbb Z$, of course)
– F.Carette
2 hours ago
add a comment |Â
up vote
1
down vote
Long story short, you are right!
the set to which $y$ belongs depends of $x$ right ? In this case, it does not make any sense for me to switch the two quantifiers $forall$ here (because $x$ has to be defined "first").
That's it, if $x$ and $y$ are independant, you can switch the $forall$ quantifiers, if they're not, you can't.
The reason behind this, as explained with your exemple in Mauro ALLEGRANZA's answer, is that it can be written:
$$forall x in mathbb X, forall y in mathbb Y_x, p(x,y)$$
Which is the same as writting:
$$forall x in mathbb X, forall y in mathbb Y, (yin mathbb Y_x implies p(x,y))$$
Now you can see that $x$ and $y$ don't play symetrical roles.
answered 1 hour ago
F.Carette
96610
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Long comment
In this case, the use of set theory symbols seems to complicate things.
Assume for simplicity $mathbb R$ as domain; in this way, we can simply write $forall x$.
What does it mean : $∀y ∈ z mid z=x+3 $ ? It simply means : $y=x+3$.
Thus, the formula will be : $forall x forall y (y=x+3 to p(x,y))$.
In this case, we have no issue with the swap of the two quantifiers.
answered 2 hours ago
Mauro ALLEGRANZA
62.2k447106
We can even write it without $y$: $forall x in mathbb R, p(x,x+3)$. There is no need of $forall y$ because $y=x+3$ only have one solution in $mathbb R$. $yequiv x+3 pmod n$ would have been a better exemple. (With $x,y in mathbb N or mathbb Z$, of course)
– F.Carette
2 hours ago
add a comment |Â
up vote
4
down vote
Long comment
In this case, the use of set theory symbols seems to complicate things.
Assume for simplicity $mathbb R$ as domain; in this way, we can simply write $forall x$.
What does it mean : $∀y ∈ z mid z=x+3 $ ? It simply means : $y=x+3$.
Thus, the formula will be : $forall x forall y (y=x+3 to p(x,y))$.
In this case, we have no issue with the swap of the two quantifiers.
answered 2 hours ago
Mauro ALLEGRANZA
62.2k447106
We can even write it without $y$: $forall x in mathbb R, p(x,x+3)$. There is no need of $forall y$ because $y=x+3$ only have one solution in $mathbb R$. $yequiv x+3 pmod n$ would have been a better exemple. (With $x,y in mathbb N or mathbb Z$, of course)
– F.Carette
2 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Long comment
In this case, the use of set theory symbols seems to complicate things.
Assume for simplicity $mathbb R$ as domain; in this way, we can simply write $forall x$.
What does it mean : $∀y ∈ z mid z=x+3 $ ? It simply means : $y=x+3$.
Thus, the formula will be : $forall x forall y (y=x+3 to p(x,y))$.
In this case, we have no issue with the swap of the two quantifiers.
answered 2 hours ago
Mauro ALLEGRANZA
62.2k447106
Long comment
In this case, the use of set theory symbols seems to complicate things.
Assume for simplicity $mathbb R$ as domain; in this way, we can simply write $forall x$.
What does it mean : $∀y ∈ z mid z=x+3 $ ? It simply means : $y=x+3$.
Thus, the formula will be : $forall x forall y (y=x+3 to p(x,y))$.
In this case, we have no issue with the swap of the two quantifiers.
answered 2 hours ago
Mauro ALLEGRANZA
62.2k447106
answered 2 hours ago
Mauro ALLEGRANZA
62.2k447106
answered 2 hours ago
Mauro ALLEGRANZA
62.2k447106
answered 2 hours ago
Mauro ALLEGRANZA
62.2k447106
62.2k447106
We can even write it without $y$: $forall x in mathbb R, p(x,x+3)$. There is no need of $forall y$ because $y=x+3$ only have one solution in $mathbb R$. $yequiv x+3 pmod n$ would have been a better exemple. (With $x,y in mathbb N or mathbb Z$, of course)
– F.Carette
2 hours ago
add a comment |Â
We can even write it without $y$: $forall x in mathbb R, p(x,x+3)$. There is no need of $forall y$ because $y=x+3$ only have one solution in $mathbb R$. $yequiv x+3 pmod n$ would have been a better exemple. (With $x,y in mathbb N or mathbb Z$, of course)
– F.Carette
2 hours ago
We can even write it without $y$: $forall x in mathbb R, p(x,x+3)$. There is no need of $forall y$ because $y=x+3$ only have one solution in $mathbb R$. $yequiv x+3 pmod n$ would have been a better exemple. (With $x,y in mathbb N or mathbb Z$, of course)
– F.Carette
2 hours ago
We can even write it without $y$: $forall x in mathbb R, p(x,x+3)$. There is no need of $forall y$ because $y=x+3$ only have one solution in $mathbb R$. $yequiv x+3 pmod n$ would have been a better exemple. (With $x,y in mathbb N or mathbb Z$, of course)
– F.Carette
2 hours ago
add a comment |Â
up vote
1
down vote
Long story short, you are right!
the set to which $y$ belongs depends of $x$ right ? In this case, it does not make any sense for me to switch the two quantifiers $forall$ here (because $x$ has to be defined "first").
That's it, if $x$ and $y$ are independant, you can switch the $forall$ quantifiers, if they're not, you can't.
The reason behind this, as explained with your exemple in Mauro ALLEGRANZA's answer, is that it can be written:
$$forall x in mathbb X, forall y in mathbb Y_x, p(x,y)$$
Which is the same as writting:
$$forall x in mathbb X, forall y in mathbb Y, (yin mathbb Y_x implies p(x,y))$$
Now you can see that $x$ and $y$ don't play symetrical roles.
answered 1 hour ago
F.Carette
96610
add a comment |Â
up vote
1
down vote
Long story short, you are right!
the set to which $y$ belongs depends of $x$ right ? In this case, it does not make any sense for me to switch the two quantifiers $forall$ here (because $x$ has to be defined "first").
That's it, if $x$ and $y$ are independant, you can switch the $forall$ quantifiers, if they're not, you can't.
The reason behind this, as explained with your exemple in Mauro ALLEGRANZA's answer, is that it can be written:
$$forall x in mathbb X, forall y in mathbb Y_x, p(x,y)$$
Which is the same as writting:
$$forall x in mathbb X, forall y in mathbb Y, (yin mathbb Y_x implies p(x,y))$$
Now you can see that $x$ and $y$ don't play symetrical roles.
answered 1 hour ago
F.Carette
96610
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Long story short, you are right!
the set to which $y$ belongs depends of $x$ right ? In this case, it does not make any sense for me to switch the two quantifiers $forall$ here (because $x$ has to be defined "first").
That's it, if $x$ and $y$ are independant, you can switch the $forall$ quantifiers, if they're not, you can't.
The reason behind this, as explained with your exemple in Mauro ALLEGRANZA's answer, is that it can be written:
$$forall x in mathbb X, forall y in mathbb Y_x, p(x,y)$$
Which is the same as writting:
$$forall x in mathbb X, forall y in mathbb Y, (yin mathbb Y_x implies p(x,y))$$
Now you can see that $x$ and $y$ don't play symetrical roles.
answered 1 hour ago
F.Carette
96610
Long story short, you are right!
the set to which $y$ belongs depends of $x$ right ? In this case, it does not make any sense for me to switch the two quantifiers $forall$ here (because $x$ has to be defined "first").
That's it, if $x$ and $y$ are independant, you can switch the $forall$ quantifiers, if they're not, you can't.
The reason behind this, as explained with your exemple in Mauro ALLEGRANZA's answer, is that it can be written:
$$forall x in mathbb X, forall y in mathbb Y_x, p(x,y)$$
Which is the same as writting:
$$forall x in mathbb X, forall y in mathbb Y, (yin mathbb Y_x implies p(x,y))$$
Now you can see that $x$ and $y$ don't play symetrical roles.
answered 1 hour ago
F.Carette
96610
answered 1 hour ago
F.Carette
96610
answered 1 hour ago
F.Carette
96610
answered 1 hour ago
F.Carette
96610
96610
add a comment |Â
add a comment |Â
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