Why $∀x(P(x)→Q)$ is equivalent to $∃xP(x)→Q$
$begingroup$
Given $x$ occurs free in $P$. If $x$ does not occur free in $Q$, then $∀x(P(x)→Q)$ is semantically equivalent with $∃xP(x)→Q$. How to understand this statement. And also, need an example to show that the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
I guess what struggles me is I'm not quite understand what does "free" mean in this case. But it would be better to have answer for the above question.
discrete-mathematics logic predicate-logic quantifiers
edited Dec 9 '18 at 11:29
Mauro ALLEGRANZA
65.5k448112
asked Dec 9 '18 at 6:31
keqiao likeqiao li
374
$endgroup$
add a comment |
$begingroup$
Given $x$ occurs free in $P$. If $x$ does not occur free in $Q$, then $∀x(P(x)→Q)$ is semantically equivalent with $∃xP(x)→Q$. How to understand this statement. And also, need an example to show that the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
I guess what struggles me is I'm not quite understand what does "free" mean in this case. But it would be better to have answer for the above question.
discrete-mathematics logic predicate-logic quantifiers
edited Dec 9 '18 at 11:29
Mauro ALLEGRANZA
65.5k448112
asked Dec 9 '18 at 6:31
keqiao likeqiao li
374
$endgroup$
$begingroup$
Do you have a formal language to work with? Semantical equivalences hints towards models, which in turn require a formal language.
$endgroup$
– MacRance
Dec 9 '18 at 7:18
1
$begingroup$
An occurance of a variable x is free when it is not in the scope of some "exists x" or "for all x".
$endgroup$
– William Elliot
Dec 9 '18 at 8:02
add a comment |
$begingroup$
Given $x$ occurs free in $P$. If $x$ does not occur free in $Q$, then $∀x(P(x)→Q)$ is semantically equivalent with $∃xP(x)→Q$. How to understand this statement. And also, need an example to show that the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
I guess what struggles me is I'm not quite understand what does "free" mean in this case. But it would be better to have answer for the above question.
discrete-mathematics logic predicate-logic quantifiers
edited Dec 9 '18 at 11:29
Mauro ALLEGRANZA
65.5k448112
asked Dec 9 '18 at 6:31
keqiao likeqiao li
374
$endgroup$
Given $x$ occurs free in $P$. If $x$ does not occur free in $Q$, then $∀x(P(x)→Q)$ is semantically equivalent with $∃xP(x)→Q$. How to understand this statement. And also, need an example to show that the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
I guess what struggles me is I'm not quite understand what does "free" mean in this case. But it would be better to have answer for the above question.
discrete-mathematics logic predicate-logic quantifiers
discrete-mathematics logic predicate-logic quantifiers
edited Dec 9 '18 at 11:29
Mauro ALLEGRANZA
65.5k448112
asked Dec 9 '18 at 6:31
keqiao likeqiao li
374
edited Dec 9 '18 at 11:29
Mauro ALLEGRANZA
65.5k448112
asked Dec 9 '18 at 6:31
keqiao likeqiao li
374
edited Dec 9 '18 at 11:29
Mauro ALLEGRANZA
65.5k448112
edited Dec 9 '18 at 11:29
Mauro ALLEGRANZA
65.5k448112
edited Dec 9 '18 at 11:29
Mauro ALLEGRANZA
65.5k448112
65.5k448112
asked Dec 9 '18 at 6:31
keqiao likeqiao li
374
asked Dec 9 '18 at 6:31
keqiao likeqiao li
374
asked Dec 9 '18 at 6:31
keqiao likeqiao li
374
374
$begingroup$
Do you have a formal language to work with? Semantical equivalences hints towards models, which in turn require a formal language.
$endgroup$
– MacRance
Dec 9 '18 at 7:18
1
$begingroup$
An occurance of a variable x is free when it is not in the scope of some "exists x" or "for all x".
$endgroup$
– William Elliot
Dec 9 '18 at 8:02
add a comment |
$begingroup$
Do you have a formal language to work with? Semantical equivalences hints towards models, which in turn require a formal language.
$endgroup$
– MacRance
Dec 9 '18 at 7:18
1
$begingroup$
An occurance of a variable x is free when it is not in the scope of some "exists x" or "for all x".
$endgroup$
– William Elliot
Dec 9 '18 at 8:02
$begingroup$
Do you have a formal language to work with? Semantical equivalences hints towards models, which in turn require a formal language.
$endgroup$
– MacRance
Dec 9 '18 at 7:18
$begingroup$
Do you have a formal language to work with? Semantical equivalences hints towards models, which in turn require a formal language.
$endgroup$
– MacRance
Dec 9 '18 at 7:18
1
1
$begingroup$
An occurance of a variable x is free when it is not in the scope of some "exists x" or "for all x".
$endgroup$
– William Elliot
Dec 9 '18 at 8:02
$begingroup$
An occurance of a variable x is free when it is not in the scope of some "exists x" or "for all x".
$endgroup$
– William Elliot
Dec 9 '18 at 8:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃xP(x)$ is False.
Thus $∃xP(x) to Q$ is True.
But to say that $∃xP(x)$ is False in $mathcal M$ means that for $a in D$ whatever, we have that $P(x)[a/x]$ is False, and thus $(P(x) to Q)[a/x]$ is True in $mathcal M$.
And this means in turn that $mathcal M vDash ∀x(P(x) to Q)$.
(ii) $∃xP(x)$ is True, i.e. there is an object $a in D$ such that $P(x)[a/x]$ is True.
Two sub-cases :
(a) $Q$ is False. Thus $∃xP(x) to Q$ is False.
But $(P(x) to Q)[a/x]$ is False for the $a in D$ above, and thus $∀x(P(x) to Q)$ is False in $mathcal M$.
(b) $Q$ is True. Thus $∃xP(x) to Q$ is True.
But if $Q$ is True, also $(P(x) to Q)[a/x]$ is True, for $a in D$ whatever.
And this means that $∀x(P(x) to Q)$ is True in $mathcal M$.
Why the condition "$x$ not free in $Q$" is necessary ?
First of all we have to agree on the way to define what does it mean : $mathcal M vDash varphi(x)$ with $x$ free.
One convention is to consider the $text {Closure}$ $forall x varphi(x)$ of the formula, i.e. :
$mathcal M vDash varphi(x) text { iff } mathcal M vDash text {Cl}(varphi(x))$.
Consider now a simple interpretation with domain $D = { 0,1 }$ and interpret both $P(x)$ and $Q(x)$ as $(x=0)$.
The LH formula is : $forall x ((x=0) to (x=0))$ while the RH formula is : $exists x (x=0) to (x=0)$.
answered Dec 9 '18 at 8:41
Mauro ALLEGRANZAMauro ALLEGRANZA
65.5k448112
$endgroup$
$begingroup$
Thank you for the response. I think I'm getting understand the logic behind these question. So what would be a proper example for the second part of the statement: the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
$endgroup$
– keqiao li
Dec 9 '18 at 20:46
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃xP(x)$ is False.
Thus $∃xP(x) to Q$ is True.
But to say that $∃xP(x)$ is False in $mathcal M$ means that for $a in D$ whatever, we have that $P(x)[a/x]$ is False, and thus $(P(x) to Q)[a/x]$ is True in $mathcal M$.
And this means in turn that $mathcal M vDash ∀x(P(x) to Q)$.
(ii) $∃xP(x)$ is True, i.e. there is an object $a in D$ such that $P(x)[a/x]$ is True.
Two sub-cases :
(a) $Q$ is False. Thus $∃xP(x) to Q$ is False.
But $(P(x) to Q)[a/x]$ is False for the $a in D$ above, and thus $∀x(P(x) to Q)$ is False in $mathcal M$.
(b) $Q$ is True. Thus $∃xP(x) to Q$ is True.
But if $Q$ is True, also $(P(x) to Q)[a/x]$ is True, for $a in D$ whatever.
And this means that $∀x(P(x) to Q)$ is True in $mathcal M$.
Why the condition "$x$ not free in $Q$" is necessary ?
First of all we have to agree on the way to define what does it mean : $mathcal M vDash varphi(x)$ with $x$ free.
One convention is to consider the $text {Closure}$ $forall x varphi(x)$ of the formula, i.e. :
$mathcal M vDash varphi(x) text { iff } mathcal M vDash text {Cl}(varphi(x))$.
Consider now a simple interpretation with domain $D = { 0,1 }$ and interpret both $P(x)$ and $Q(x)$ as $(x=0)$.
The LH formula is : $forall x ((x=0) to (x=0))$ while the RH formula is : $exists x (x=0) to (x=0)$.
answered Dec 9 '18 at 8:41
Mauro ALLEGRANZAMauro ALLEGRANZA
65.5k448112
$endgroup$
$begingroup$
Thank you for the response. I think I'm getting understand the logic behind these question. So what would be a proper example for the second part of the statement: the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
$endgroup$
– keqiao li
Dec 9 '18 at 20:46
add a comment |
$begingroup$
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃xP(x)$ is False.
Thus $∃xP(x) to Q$ is True.
But to say that $∃xP(x)$ is False in $mathcal M$ means that for $a in D$ whatever, we have that $P(x)[a/x]$ is False, and thus $(P(x) to Q)[a/x]$ is True in $mathcal M$.
And this means in turn that $mathcal M vDash ∀x(P(x) to Q)$.
(ii) $∃xP(x)$ is True, i.e. there is an object $a in D$ such that $P(x)[a/x]$ is True.
Two sub-cases :
(a) $Q$ is False. Thus $∃xP(x) to Q$ is False.
But $(P(x) to Q)[a/x]$ is False for the $a in D$ above, and thus $∀x(P(x) to Q)$ is False in $mathcal M$.
(b) $Q$ is True. Thus $∃xP(x) to Q$ is True.
But if $Q$ is True, also $(P(x) to Q)[a/x]$ is True, for $a in D$ whatever.
And this means that $∀x(P(x) to Q)$ is True in $mathcal M$.
Why the condition "$x$ not free in $Q$" is necessary ?
First of all we have to agree on the way to define what does it mean : $mathcal M vDash varphi(x)$ with $x$ free.
One convention is to consider the $text {Closure}$ $forall x varphi(x)$ of the formula, i.e. :
$mathcal M vDash varphi(x) text { iff } mathcal M vDash text {Cl}(varphi(x))$.
Consider now a simple interpretation with domain $D = { 0,1 }$ and interpret both $P(x)$ and $Q(x)$ as $(x=0)$.
The LH formula is : $forall x ((x=0) to (x=0))$ while the RH formula is : $exists x (x=0) to (x=0)$.
answered Dec 9 '18 at 8:41
Mauro ALLEGRANZAMauro ALLEGRANZA
65.5k448112
$endgroup$
$begingroup$
Thank you for the response. I think I'm getting understand the logic behind these question. So what would be a proper example for the second part of the statement: the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
$endgroup$
– keqiao li
Dec 9 '18 at 20:46
add a comment |
$begingroup$
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃xP(x)$ is False.
Thus $∃xP(x) to Q$ is True.
But to say that $∃xP(x)$ is False in $mathcal M$ means that for $a in D$ whatever, we have that $P(x)[a/x]$ is False, and thus $(P(x) to Q)[a/x]$ is True in $mathcal M$.
And this means in turn that $mathcal M vDash ∀x(P(x) to Q)$.
(ii) $∃xP(x)$ is True, i.e. there is an object $a in D$ such that $P(x)[a/x]$ is True.
Two sub-cases :
(a) $Q$ is False. Thus $∃xP(x) to Q$ is False.
But $(P(x) to Q)[a/x]$ is False for the $a in D$ above, and thus $∀x(P(x) to Q)$ is False in $mathcal M$.
(b) $Q$ is True. Thus $∃xP(x) to Q$ is True.
But if $Q$ is True, also $(P(x) to Q)[a/x]$ is True, for $a in D$ whatever.
And this means that $∀x(P(x) to Q)$ is True in $mathcal M$.
Why the condition "$x$ not free in $Q$" is necessary ?
First of all we have to agree on the way to define what does it mean : $mathcal M vDash varphi(x)$ with $x$ free.
One convention is to consider the $text {Closure}$ $forall x varphi(x)$ of the formula, i.e. :
$mathcal M vDash varphi(x) text { iff } mathcal M vDash text {Cl}(varphi(x))$.
Consider now a simple interpretation with domain $D = { 0,1 }$ and interpret both $P(x)$ and $Q(x)$ as $(x=0)$.
The LH formula is : $forall x ((x=0) to (x=0))$ while the RH formula is : $exists x (x=0) to (x=0)$.
answered Dec 9 '18 at 8:41
Mauro ALLEGRANZAMauro ALLEGRANZA
65.5k448112
$endgroup$
Let $mathcal M$ a structure whatever, with domain $D$.
Two cases :
(i) $∃xP(x)$ is False.
Thus $∃xP(x) to Q$ is True.
But to say that $∃xP(x)$ is False in $mathcal M$ means that for $a in D$ whatever, we have that $P(x)[a/x]$ is False, and thus $(P(x) to Q)[a/x]$ is True in $mathcal M$.
And this means in turn that $mathcal M vDash ∀x(P(x) to Q)$.
(ii) $∃xP(x)$ is True, i.e. there is an object $a in D$ such that $P(x)[a/x]$ is True.
Two sub-cases :
(a) $Q$ is False. Thus $∃xP(x) to Q$ is False.
But $(P(x) to Q)[a/x]$ is False for the $a in D$ above, and thus $∀x(P(x) to Q)$ is False in $mathcal M$.
(b) $Q$ is True. Thus $∃xP(x) to Q$ is True.
But if $Q$ is True, also $(P(x) to Q)[a/x]$ is True, for $a in D$ whatever.
And this means that $∀x(P(x) to Q)$ is True in $mathcal M$.
Why the condition "$x$ not free in $Q$" is necessary ?
First of all we have to agree on the way to define what does it mean : $mathcal M vDash varphi(x)$ with $x$ free.
One convention is to consider the $text {Closure}$ $forall x varphi(x)$ of the formula, i.e. :
$mathcal M vDash varphi(x) text { iff } mathcal M vDash text {Cl}(varphi(x))$.
Consider now a simple interpretation with domain $D = { 0,1 }$ and interpret both $P(x)$ and $Q(x)$ as $(x=0)$.
The LH formula is : $forall x ((x=0) to (x=0))$ while the RH formula is : $exists x (x=0) to (x=0)$.
answered Dec 9 '18 at 8:41
Mauro ALLEGRANZAMauro ALLEGRANZA
65.5k448112
edited Dec 10 '18 at 8:12
answered Dec 9 '18 at 8:41
Mauro ALLEGRANZAMauro ALLEGRANZA
65.5k448112
answered Dec 9 '18 at 8:41
Mauro ALLEGRANZAMauro ALLEGRANZA
65.5k448112
answered Dec 9 '18 at 8:41
Mauro ALLEGRANZAMauro ALLEGRANZA
65.5k448112
65.5k448112
$begingroup$
Thank you for the response. I think I'm getting understand the logic behind these question. So what would be a proper example for the second part of the statement: the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
$endgroup$
– keqiao li
Dec 9 '18 at 20:46
add a comment |
$begingroup$
Thank you for the response. I think I'm getting understand the logic behind these question. So what would be a proper example for the second part of the statement: the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
$endgroup$
– keqiao li
Dec 9 '18 at 20:46
$begingroup$
Thank you for the response. I think I'm getting understand the logic behind these question. So what would be a proper example for the second part of the statement: the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
$endgroup$
– keqiao li
Dec 9 '18 at 20:46
$begingroup$
Thank you for the response. I think I'm getting understand the logic behind these question. So what would be a proper example for the second part of the statement: the formula will not necessarily be equivalent if $x$ does occur free in $Q$.
$endgroup$
– keqiao li
Dec 9 '18 at 20:46
add a comment |
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$begingroup$
Do you have a formal language to work with? Semantical equivalences hints towards models, which in turn require a formal language.
$endgroup$
– MacRance
Dec 9 '18 at 7:18
1
$begingroup$
An occurance of a variable x is free when it is not in the scope of some "exists x" or "for all x".
$endgroup$
– William Elliot
Dec 9 '18 at 8:02