The barbers paradox first order logic formalizationfirst order logic resolution unificationWhat is the relation between First Order Logic and First Order Theory?Quantified Boolean Formula vs First-order logicAbout the first order logic (valid, Unsatisfiable, Syntactically wrong)Resolution in First Order LogicExercise about First-order logicFirst Order Logic : PredicatesFirst Order Logic, First Order Logic + Recurrence and SQLReal world applications of first order logicTerminology First-Order Logic
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The barbers paradox first order logic formalization
first order logic resolution unificationWhat is the relation between First Order Logic and First Order Theory?Quantified Boolean Formula vs First-order logicAbout the first order logic (valid, Unsatisfiable, Syntactically wrong)Resolution in First Order LogicExercise about First-order logicFirst Order Logic : PredicatesFirst Order Logic, First Order Logic + Recurrence and SQLReal world applications of first order logicTerminology First-Order Logic
$begingroup$
I tried to look on the site and while I found some similar questions, I did not find the first order logic formalization of the following sentence (the basic barber's paradox), so I wanted to ask if I got the right first order logic formalization of it, and I am sure others will benefit from this thread as well.
Sentence: there exists a barber who shaves all the people that don't shave themselves.
My attempt: this complicated sentence can be made simpler by inferring that "for all of those who does not shave themselves, are shaved by the barber"
And now it seems to be easier to substitute with variables so:
$$exists x(lnot S(x,x) rightarrow S(b,x))$$
where $S(x,x)$ is shaving(verb), $x$ are the persons (who don't shave themselves), and $b$ is a shortcut for barber, so if a person does not shave himself, it can be inferred that he is shaved by the barber.
logic artificial-intelligence first-order-logic
edited Apr 29 at 21:47
David Richerby
71.3k16109199
asked Apr 29 at 21:37
hps13hps13
377
$endgroup$
add a comment |
$begingroup$
I tried to look on the site and while I found some similar questions, I did not find the first order logic formalization of the following sentence (the basic barber's paradox), so I wanted to ask if I got the right first order logic formalization of it, and I am sure others will benefit from this thread as well.
Sentence: there exists a barber who shaves all the people that don't shave themselves.
My attempt: this complicated sentence can be made simpler by inferring that "for all of those who does not shave themselves, are shaved by the barber"
And now it seems to be easier to substitute with variables so:
$$exists x(lnot S(x,x) rightarrow S(b,x))$$
where $S(x,x)$ is shaving(verb), $x$ are the persons (who don't shave themselves), and $b$ is a shortcut for barber, so if a person does not shave himself, it can be inferred that he is shaved by the barber.
logic artificial-intelligence first-order-logic
edited Apr 29 at 21:47
David Richerby
71.3k16109199
asked Apr 29 at 21:37
hps13hps13
377
$endgroup$
add a comment |
$begingroup$
I tried to look on the site and while I found some similar questions, I did not find the first order logic formalization of the following sentence (the basic barber's paradox), so I wanted to ask if I got the right first order logic formalization of it, and I am sure others will benefit from this thread as well.
Sentence: there exists a barber who shaves all the people that don't shave themselves.
My attempt: this complicated sentence can be made simpler by inferring that "for all of those who does not shave themselves, are shaved by the barber"
And now it seems to be easier to substitute with variables so:
$$exists x(lnot S(x,x) rightarrow S(b,x))$$
where $S(x,x)$ is shaving(verb), $x$ are the persons (who don't shave themselves), and $b$ is a shortcut for barber, so if a person does not shave himself, it can be inferred that he is shaved by the barber.
logic artificial-intelligence first-order-logic
edited Apr 29 at 21:47
David Richerby
71.3k16109199
asked Apr 29 at 21:37
hps13hps13
377
$endgroup$
I tried to look on the site and while I found some similar questions, I did not find the first order logic formalization of the following sentence (the basic barber's paradox), so I wanted to ask if I got the right first order logic formalization of it, and I am sure others will benefit from this thread as well.
Sentence: there exists a barber who shaves all the people that don't shave themselves.
My attempt: this complicated sentence can be made simpler by inferring that "for all of those who does not shave themselves, are shaved by the barber"
And now it seems to be easier to substitute with variables so:
$$exists x(lnot S(x,x) rightarrow S(b,x))$$
where $S(x,x)$ is shaving(verb), $x$ are the persons (who don't shave themselves), and $b$ is a shortcut for barber, so if a person does not shave himself, it can be inferred that he is shaved by the barber.
logic artificial-intelligence first-order-logic
logic artificial-intelligence first-order-logic
edited Apr 29 at 21:47
David Richerby
71.3k16109199
asked Apr 29 at 21:37
hps13hps13
377
edited Apr 29 at 21:47
David Richerby
71.3k16109199
asked Apr 29 at 21:37
hps13hps13
377
edited Apr 29 at 21:47
David Richerby
71.3k16109199
edited Apr 29 at 21:47
David Richerby
71.3k16109199
edited Apr 29 at 21:47
David Richerby
71.3k16109199
71.3k16109199
asked Apr 29 at 21:37
hps13hps13
377
asked Apr 29 at 21:37
hps13hps13
377
asked Apr 29 at 21:37
hps13hps13
377
377
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're not being asked to do any inference; just to express something.
there exists a barber who shaves all the people that don't shave themselves
translates directly as
$$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$
(There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)
Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtextanything$ is true.
answered Apr 29 at 21:55
David RicherbyDavid Richerby
71.3k16109199
$endgroup$
add a comment |
$begingroup$
There are two issues about your formula:
First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.
Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.
Whith this, the sentence becomes
There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.
which translates as
$$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$
answered Apr 29 at 22:02
lemontreelemontree
1664
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
You're not being asked to do any inference; just to express something.
there exists a barber who shaves all the people that don't shave themselves
translates directly as
$$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$
(There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)
Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtextanything$ is true.
answered Apr 29 at 21:55
David RicherbyDavid Richerby
71.3k16109199
$endgroup$
add a comment |
$begingroup$
You're not being asked to do any inference; just to express something.
there exists a barber who shaves all the people that don't shave themselves
translates directly as
$$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$
(There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)
Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtextanything$ is true.
answered Apr 29 at 21:55
David RicherbyDavid Richerby
71.3k16109199
$endgroup$
add a comment |
$begingroup$
You're not being asked to do any inference; just to express something.
there exists a barber who shaves all the people that don't shave themselves
translates directly as
$$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$
(There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)
Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtextanything$ is true.
answered Apr 29 at 21:55
David RicherbyDavid Richerby
71.3k16109199
$endgroup$
You're not being asked to do any inference; just to express something.
there exists a barber who shaves all the people that don't shave themselves
translates directly as
$$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$
(There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)
Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtextanything$ is true.
answered Apr 29 at 21:55
David RicherbyDavid Richerby
71.3k16109199
answered Apr 29 at 21:55
David RicherbyDavid Richerby
71.3k16109199
answered Apr 29 at 21:55
David RicherbyDavid Richerby
71.3k16109199
answered Apr 29 at 21:55
David RicherbyDavid Richerby
71.3k16109199
71.3k16109199
add a comment |
add a comment |
$begingroup$
There are two issues about your formula:
First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.
Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.
Whith this, the sentence becomes
There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.
which translates as
$$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$
answered Apr 29 at 22:02
lemontreelemontree
1664
$endgroup$
add a comment |
$begingroup$
There are two issues about your formula:
First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.
Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.
Whith this, the sentence becomes
There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.
which translates as
$$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$
answered Apr 29 at 22:02
lemontreelemontree
1664
$endgroup$
add a comment |
$begingroup$
There are two issues about your formula:
First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.
Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.
Whith this, the sentence becomes
There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.
which translates as
$$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$
answered Apr 29 at 22:02
lemontreelemontree
1664
$endgroup$
There are two issues about your formula:
First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.
Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.
Whith this, the sentence becomes
There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.
which translates as
$$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$
answered Apr 29 at 22:02
lemontreelemontree
1664
edited Apr 29 at 22:08
answered Apr 29 at 22:02
lemontreelemontree
1664
answered Apr 29 at 22:02
lemontreelemontree
1664
answered Apr 29 at 22:02
lemontreelemontree
1664
1664
add a comment |
add a comment |
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