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Does an implication have to be true in all interpretations to be true?
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I am wondering about this example:
$P: x = 3$
$Q: x^2 = 9$
To me it seems clear that $P implies Q$ is true. If $x=3$ it is true that $x^2=9$.
But what about the converse here?
The statement $Q implies P$ "feels" false to me at first, but is it? If $x^2=9$ is it true that $x=3$? It could be $-3$, but it's possible that $x=3$ as well, so it is possibly true, which means it is not always false. This statement is sometimes true, sometimes false.
So how do we evaluate it, and what rule in logic tells us how to interpret it?
logic definition boolean-algebra
asked Jul 21 at 3:41
user51819
5411314
add a comment |Â
up vote
0
down vote
favorite
I am wondering about this example:
$P: x = 3$
$Q: x^2 = 9$
To me it seems clear that $P implies Q$ is true. If $x=3$ it is true that $x^2=9$.
But what about the converse here?
The statement $Q implies P$ "feels" false to me at first, but is it? If $x^2=9$ is it true that $x=3$? It could be $-3$, but it's possible that $x=3$ as well, so it is possibly true, which means it is not always false. This statement is sometimes true, sometimes false.
So how do we evaluate it, and what rule in logic tells us how to interpret it?
logic definition boolean-algebra
asked Jul 21 at 3:41
user51819
5411314
1
See Why is statement and its converse not equivalent? for the general answer. That said, your example is not the best chosen, since for example $Q implies P$ is true if $x in Bbb N$ but false if $x in Bbb R$.
– dxiv
Jul 21 at 3:50
@dxiv Re: That link, I am asking something a little different here. Your point about it being true in some cases and false in others is what I am asking about.
– user51819
Jul 21 at 3:53
Even it $Q implies P$ happens to be truein some casesthat's just because you can independently prove it true in those cases. It is not because you proved $P implies Q$ first, and it has no direct relationship with the converse.
– dxiv
Jul 21 at 3:55
@dxiv I understand that -- I am asking if this specific $Q implies P$ is considered true or false in traditional true/false Boolean logic, or if it's an example of an implication that doesn't have such an answer since it depends on what we look at.
– user51819
Jul 21 at 4:15
That is thoroughly answered in the question I linked in my first comment. $big(P implies Qbig) vdash big(Q implies Pbig)$ is false. As a consequence, the truth value of $Q implies P$ is undetermined regardless of whether $P implies Q$ is true or false.
– dxiv
Jul 21 at 4:26
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am wondering about this example:
$P: x = 3$
$Q: x^2 = 9$
To me it seems clear that $P implies Q$ is true. If $x=3$ it is true that $x^2=9$.
But what about the converse here?
The statement $Q implies P$ "feels" false to me at first, but is it? If $x^2=9$ is it true that $x=3$? It could be $-3$, but it's possible that $x=3$ as well, so it is possibly true, which means it is not always false. This statement is sometimes true, sometimes false.
So how do we evaluate it, and what rule in logic tells us how to interpret it?
logic definition boolean-algebra
asked Jul 21 at 3:41
user51819
5411314
I am wondering about this example:
$P: x = 3$
$Q: x^2 = 9$
To me it seems clear that $P implies Q$ is true. If $x=3$ it is true that $x^2=9$.
But what about the converse here?
The statement $Q implies P$ "feels" false to me at first, but is it? If $x^2=9$ is it true that $x=3$? It could be $-3$, but it's possible that $x=3$ as well, so it is possibly true, which means it is not always false. This statement is sometimes true, sometimes false.
So how do we evaluate it, and what rule in logic tells us how to interpret it?
logic definition boolean-algebra
asked Jul 21 at 3:41
user51819
5411314
asked Jul 21 at 3:41
user51819
5411314
asked Jul 21 at 3:41
user51819
5411314
asked Jul 21 at 3:41
user51819
5411314
5411314
1
See Why is statement and its converse not equivalent? for the general answer. That said, your example is not the best chosen, since for example $Q implies P$ is true if $x in Bbb N$ but false if $x in Bbb R$.
– dxiv
Jul 21 at 3:50
@dxiv Re: That link, I am asking something a little different here. Your point about it being true in some cases and false in others is what I am asking about.
– user51819
Jul 21 at 3:53
Even it $Q implies P$ happens to be truein some casesthat's just because you can independently prove it true in those cases. It is not because you proved $P implies Q$ first, and it has no direct relationship with the converse.
– dxiv
Jul 21 at 3:55
@dxiv I understand that -- I am asking if this specific $Q implies P$ is considered true or false in traditional true/false Boolean logic, or if it's an example of an implication that doesn't have such an answer since it depends on what we look at.
– user51819
Jul 21 at 4:15
That is thoroughly answered in the question I linked in my first comment. $big(P implies Qbig) vdash big(Q implies Pbig)$ is false. As a consequence, the truth value of $Q implies P$ is undetermined regardless of whether $P implies Q$ is true or false.
– dxiv
Jul 21 at 4:26
add a comment |Â
1
See Why is statement and its converse not equivalent? for the general answer. That said, your example is not the best chosen, since for example $Q implies P$ is true if $x in Bbb N$ but false if $x in Bbb R$.
– dxiv
Jul 21 at 3:50
@dxiv Re: That link, I am asking something a little different here. Your point about it being true in some cases and false in others is what I am asking about.
– user51819
Jul 21 at 3:53
Even it $Q implies P$ happens to be truein some casesthat's just because you can independently prove it true in those cases. It is not because you proved $P implies Q$ first, and it has no direct relationship with the converse.
– dxiv
Jul 21 at 3:55
@dxiv I understand that -- I am asking if this specific $Q implies P$ is considered true or false in traditional true/false Boolean logic, or if it's an example of an implication that doesn't have such an answer since it depends on what we look at.
– user51819
Jul 21 at 4:15
That is thoroughly answered in the question I linked in my first comment. $big(P implies Qbig) vdash big(Q implies Pbig)$ is false. As a consequence, the truth value of $Q implies P$ is undetermined regardless of whether $P implies Q$ is true or false.
– dxiv
Jul 21 at 4:26
1
1
See Why is statement and its converse not equivalent? for the general answer. That said, your example is not the best chosen, since for example $Q implies P$ is true if $x in Bbb N$ but false if $x in Bbb R$.
– dxiv
Jul 21 at 3:50
See Why is statement and its converse not equivalent? for the general answer. That said, your example is not the best chosen, since for example $Q implies P$ is true if $x in Bbb N$ but false if $x in Bbb R$.
– dxiv
Jul 21 at 3:50
@dxiv Re: That link, I am asking something a little different here. Your point about it being true in some cases and false in others is what I am asking about.
– user51819
Jul 21 at 3:53
@dxiv Re: That link, I am asking something a little different here. Your point about it being true in some cases and false in others is what I am asking about.
– user51819
Jul 21 at 3:53
Even it $Q implies P$ happens to be true
in some cases that's just because you can independently prove it true in those cases. It is not because you proved $P implies Q$ first, and it has no direct relationship with the converse.– dxiv
Jul 21 at 3:55
Even it $Q implies P$ happens to be true
in some cases that's just because you can independently prove it true in those cases. It is not because you proved $P implies Q$ first, and it has no direct relationship with the converse.– dxiv
Jul 21 at 3:55
@dxiv I understand that -- I am asking if this specific $Q implies P$ is considered true or false in traditional true/false Boolean logic, or if it's an example of an implication that doesn't have such an answer since it depends on what we look at.
– user51819
Jul 21 at 4:15
@dxiv I understand that -- I am asking if this specific $Q implies P$ is considered true or false in traditional true/false Boolean logic, or if it's an example of an implication that doesn't have such an answer since it depends on what we look at.
– user51819
Jul 21 at 4:15
That is thoroughly answered in the question I linked in my first comment. $big(P implies Qbig) vdash big(Q implies Pbig)$ is false. As a consequence, the truth value of $Q implies P$ is undetermined regardless of whether $P implies Q$ is true or false.
– dxiv
Jul 21 at 4:26
That is thoroughly answered in the question I linked in my first comment. $big(P implies Qbig) vdash big(Q implies Pbig)$ is false. As a consequence, the truth value of $Q implies P$ is undetermined regardless of whether $P implies Q$ is true or false.
– dxiv
Jul 21 at 4:26
add a comment |Â
1 Answer
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First, there is a distinction between true/false and valid/invalid. True/false are usually used with respect to a given model. Valid(which corresponds to provable for typical logics) means "true in all models".
We don't usually talk about open formulas, i.e. formulas containing free variables like $x$ in your example, as being true/false or valid/invalid. Usually, only closed formulas with no free variables are allowed to be considered to be true/false or valid/invalid.
One common convention is to treat all free variables as implicitly universally quantified when asking about their truth/validity. In your example, this would mean considering $forall x.Qimplies P$ which is false in e.g. a model of the reals where the operations are interpreted as usual, but true in a model of the naturals as dxiv mentions. This assumes those are models which depends on the full theory you are using which you haven't specified. As such, whether the statement is valid or not depends on the full theory, though in $mathbb R$ with the usual interpretation of the operations is a model, then it is not valid.
Alternatively, you can consider an open formula as specifying a relation on a given domain set. In your example, taking the domain set to be $mathbb R$ (assuming it's a model), the formula $Qimplies P$ would be $xinmathbb Rmid x^2=9 implies x=3=mathbb R --3$ because the formula is true for all $xinmathbb R$ except for $x=-3$. Connecting to the previous paragraph, $forall x.varphi(x)$ is true (for a given domain, $D$) when $xin Dmid varphi(x)=D$. Thus $mathbb R--3neqmathbb R$ is another way of saying the universally quantified formula is false (in this model, but if it is false in any model then it is invalid). If we use $mathbb N$ as the domain set (again, assuming this is a model of your theory), then $xinmathbb Nmid x^2=9implies x=3=mathbb N$ corresponding to the universally quantified formula being true in this model.
To reiterate though, the interpretation of a formula with $n$ free variables is a $n$-ary relation on the domain set. A closed formula corresponds to $n=0$, and a $0$-ary relation is just a subset of a singleton set and is thus either the singleton set itself or empty, corresponding to true and false respectively.
answered Jul 21 at 4:26
Derek Elkins
14.9k11035
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First, there is a distinction between true/false and valid/invalid. True/false are usually used with respect to a given model. Valid(which corresponds to provable for typical logics) means "true in all models".
We don't usually talk about open formulas, i.e. formulas containing free variables like $x$ in your example, as being true/false or valid/invalid. Usually, only closed formulas with no free variables are allowed to be considered to be true/false or valid/invalid.
One common convention is to treat all free variables as implicitly universally quantified when asking about their truth/validity. In your example, this would mean considering $forall x.Qimplies P$ which is false in e.g. a model of the reals where the operations are interpreted as usual, but true in a model of the naturals as dxiv mentions. This assumes those are models which depends on the full theory you are using which you haven't specified. As such, whether the statement is valid or not depends on the full theory, though in $mathbb R$ with the usual interpretation of the operations is a model, then it is not valid.
Alternatively, you can consider an open formula as specifying a relation on a given domain set. In your example, taking the domain set to be $mathbb R$ (assuming it's a model), the formula $Qimplies P$ would be $xinmathbb Rmid x^2=9 implies x=3=mathbb R --3$ because the formula is true for all $xinmathbb R$ except for $x=-3$. Connecting to the previous paragraph, $forall x.varphi(x)$ is true (for a given domain, $D$) when $xin Dmid varphi(x)=D$. Thus $mathbb R--3neqmathbb R$ is another way of saying the universally quantified formula is false (in this model, but if it is false in any model then it is invalid). If we use $mathbb N$ as the domain set (again, assuming this is a model of your theory), then $xinmathbb Nmid x^2=9implies x=3=mathbb N$ corresponding to the universally quantified formula being true in this model.
To reiterate though, the interpretation of a formula with $n$ free variables is a $n$-ary relation on the domain set. A closed formula corresponds to $n=0$, and a $0$-ary relation is just a subset of a singleton set and is thus either the singleton set itself or empty, corresponding to true and false respectively.
answered Jul 21 at 4:26
Derek Elkins
14.9k11035
add a comment |Â
up vote
2
down vote
accepted
First, there is a distinction between true/false and valid/invalid. True/false are usually used with respect to a given model. Valid(which corresponds to provable for typical logics) means "true in all models".
We don't usually talk about open formulas, i.e. formulas containing free variables like $x$ in your example, as being true/false or valid/invalid. Usually, only closed formulas with no free variables are allowed to be considered to be true/false or valid/invalid.
One common convention is to treat all free variables as implicitly universally quantified when asking about their truth/validity. In your example, this would mean considering $forall x.Qimplies P$ which is false in e.g. a model of the reals where the operations are interpreted as usual, but true in a model of the naturals as dxiv mentions. This assumes those are models which depends on the full theory you are using which you haven't specified. As such, whether the statement is valid or not depends on the full theory, though in $mathbb R$ with the usual interpretation of the operations is a model, then it is not valid.
Alternatively, you can consider an open formula as specifying a relation on a given domain set. In your example, taking the domain set to be $mathbb R$ (assuming it's a model), the formula $Qimplies P$ would be $xinmathbb Rmid x^2=9 implies x=3=mathbb R --3$ because the formula is true for all $xinmathbb R$ except for $x=-3$. Connecting to the previous paragraph, $forall x.varphi(x)$ is true (for a given domain, $D$) when $xin Dmid varphi(x)=D$. Thus $mathbb R--3neqmathbb R$ is another way of saying the universally quantified formula is false (in this model, but if it is false in any model then it is invalid). If we use $mathbb N$ as the domain set (again, assuming this is a model of your theory), then $xinmathbb Nmid x^2=9implies x=3=mathbb N$ corresponding to the universally quantified formula being true in this model.
To reiterate though, the interpretation of a formula with $n$ free variables is a $n$-ary relation on the domain set. A closed formula corresponds to $n=0$, and a $0$-ary relation is just a subset of a singleton set and is thus either the singleton set itself or empty, corresponding to true and false respectively.
answered Jul 21 at 4:26
Derek Elkins
14.9k11035
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First, there is a distinction between true/false and valid/invalid. True/false are usually used with respect to a given model. Valid(which corresponds to provable for typical logics) means "true in all models".
We don't usually talk about open formulas, i.e. formulas containing free variables like $x$ in your example, as being true/false or valid/invalid. Usually, only closed formulas with no free variables are allowed to be considered to be true/false or valid/invalid.
One common convention is to treat all free variables as implicitly universally quantified when asking about their truth/validity. In your example, this would mean considering $forall x.Qimplies P$ which is false in e.g. a model of the reals where the operations are interpreted as usual, but true in a model of the naturals as dxiv mentions. This assumes those are models which depends on the full theory you are using which you haven't specified. As such, whether the statement is valid or not depends on the full theory, though in $mathbb R$ with the usual interpretation of the operations is a model, then it is not valid.
Alternatively, you can consider an open formula as specifying a relation on a given domain set. In your example, taking the domain set to be $mathbb R$ (assuming it's a model), the formula $Qimplies P$ would be $xinmathbb Rmid x^2=9 implies x=3=mathbb R --3$ because the formula is true for all $xinmathbb R$ except for $x=-3$. Connecting to the previous paragraph, $forall x.varphi(x)$ is true (for a given domain, $D$) when $xin Dmid varphi(x)=D$. Thus $mathbb R--3neqmathbb R$ is another way of saying the universally quantified formula is false (in this model, but if it is false in any model then it is invalid). If we use $mathbb N$ as the domain set (again, assuming this is a model of your theory), then $xinmathbb Nmid x^2=9implies x=3=mathbb N$ corresponding to the universally quantified formula being true in this model.
To reiterate though, the interpretation of a formula with $n$ free variables is a $n$-ary relation on the domain set. A closed formula corresponds to $n=0$, and a $0$-ary relation is just a subset of a singleton set and is thus either the singleton set itself or empty, corresponding to true and false respectively.
answered Jul 21 at 4:26
Derek Elkins
14.9k11035
First, there is a distinction between true/false and valid/invalid. True/false are usually used with respect to a given model. Valid(which corresponds to provable for typical logics) means "true in all models".
We don't usually talk about open formulas, i.e. formulas containing free variables like $x$ in your example, as being true/false or valid/invalid. Usually, only closed formulas with no free variables are allowed to be considered to be true/false or valid/invalid.
One common convention is to treat all free variables as implicitly universally quantified when asking about their truth/validity. In your example, this would mean considering $forall x.Qimplies P$ which is false in e.g. a model of the reals where the operations are interpreted as usual, but true in a model of the naturals as dxiv mentions. This assumes those are models which depends on the full theory you are using which you haven't specified. As such, whether the statement is valid or not depends on the full theory, though in $mathbb R$ with the usual interpretation of the operations is a model, then it is not valid.
Alternatively, you can consider an open formula as specifying a relation on a given domain set. In your example, taking the domain set to be $mathbb R$ (assuming it's a model), the formula $Qimplies P$ would be $xinmathbb Rmid x^2=9 implies x=3=mathbb R --3$ because the formula is true for all $xinmathbb R$ except for $x=-3$. Connecting to the previous paragraph, $forall x.varphi(x)$ is true (for a given domain, $D$) when $xin Dmid varphi(x)=D$. Thus $mathbb R--3neqmathbb R$ is another way of saying the universally quantified formula is false (in this model, but if it is false in any model then it is invalid). If we use $mathbb N$ as the domain set (again, assuming this is a model of your theory), then $xinmathbb Nmid x^2=9implies x=3=mathbb N$ corresponding to the universally quantified formula being true in this model.
To reiterate though, the interpretation of a formula with $n$ free variables is a $n$-ary relation on the domain set. A closed formula corresponds to $n=0$, and a $0$-ary relation is just a subset of a singleton set and is thus either the singleton set itself or empty, corresponding to true and false respectively.
answered Jul 21 at 4:26
Derek Elkins
14.9k11035
answered Jul 21 at 4:26
Derek Elkins
14.9k11035
answered Jul 21 at 4:26
Derek Elkins
14.9k11035
answered Jul 21 at 4:26
Derek Elkins
14.9k11035
14.9k11035
add a comment |Â
add a comment |Â
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1
See Why is statement and its converse not equivalent? for the general answer. That said, your example is not the best chosen, since for example $Q implies P$ is true if $x in Bbb N$ but false if $x in Bbb R$.
– dxiv
Jul 21 at 3:50
@dxiv Re: That link, I am asking something a little different here. Your point about it being true in some cases and false in others is what I am asking about.
– user51819
Jul 21 at 3:53
Even it $Q implies P$ happens to be true
in some casesthat's just because you can independently prove it true in those cases. It is not because you proved $P implies Q$ first, and it has no direct relationship with the converse.– dxiv
Jul 21 at 3:55
@dxiv I understand that -- I am asking if this specific $Q implies P$ is considered true or false in traditional true/false Boolean logic, or if it's an example of an implication that doesn't have such an answer since it depends on what we look at.
– user51819
Jul 21 at 4:15
That is thoroughly answered in the question I linked in my first comment. $big(P implies Qbig) vdash big(Q implies Pbig)$ is false. As a consequence, the truth value of $Q implies P$ is undetermined regardless of whether $P implies Q$ is true or false.
– dxiv
Jul 21 at 4:26