Translation into predicate calculus
$begingroup$
I want to translate the following sentence into predicate calculus:
"Anything taller than something Alice is taller than is taller than Alice."
Let $a$ denote Alice and $T(x,y)$ the predicate asserting that $x$ is taller than $y$. I believe the tranlation is $$(forall x)(forall y)[(T(x,y)land T(a,y))implies T(x,a)].$$
But I also think it might be $$(forall x)[(exists y)(T(x,y)land T(a,y))implies T(x,a)].$$
Which is the correct one, and why?
logic predicate-logic logic-translation
edited Dec 18 '18 at 1:45
Bram28
63.1k44793
asked Oct 15 '16 at 22:46
satokunsatokun
403412
$endgroup$
add a comment |
$begingroup$
I want to translate the following sentence into predicate calculus:
"Anything taller than something Alice is taller than is taller than Alice."
Let $a$ denote Alice and $T(x,y)$ the predicate asserting that $x$ is taller than $y$. I believe the tranlation is $$(forall x)(forall y)[(T(x,y)land T(a,y))implies T(x,a)].$$
But I also think it might be $$(forall x)[(exists y)(T(x,y)land T(a,y))implies T(x,a)].$$
Which is the correct one, and why?
logic predicate-logic logic-translation
edited Dec 18 '18 at 1:45
Bram28
63.1k44793
asked Oct 15 '16 at 22:46
satokunsatokun
403412
$endgroup$
add a comment |
$begingroup$
I want to translate the following sentence into predicate calculus:
"Anything taller than something Alice is taller than is taller than Alice."
Let $a$ denote Alice and $T(x,y)$ the predicate asserting that $x$ is taller than $y$. I believe the tranlation is $$(forall x)(forall y)[(T(x,y)land T(a,y))implies T(x,a)].$$
But I also think it might be $$(forall x)[(exists y)(T(x,y)land T(a,y))implies T(x,a)].$$
Which is the correct one, and why?
logic predicate-logic logic-translation
edited Dec 18 '18 at 1:45
Bram28
63.1k44793
asked Oct 15 '16 at 22:46
satokunsatokun
403412
$endgroup$
I want to translate the following sentence into predicate calculus:
"Anything taller than something Alice is taller than is taller than Alice."
Let $a$ denote Alice and $T(x,y)$ the predicate asserting that $x$ is taller than $y$. I believe the tranlation is $$(forall x)(forall y)[(T(x,y)land T(a,y))implies T(x,a)].$$
But I also think it might be $$(forall x)[(exists y)(T(x,y)land T(a,y))implies T(x,a)].$$
Which is the correct one, and why?
logic predicate-logic logic-translation
logic predicate-logic logic-translation
edited Dec 18 '18 at 1:45
Bram28
63.1k44793
asked Oct 15 '16 at 22:46
satokunsatokun
403412
edited Dec 18 '18 at 1:45
Bram28
63.1k44793
asked Oct 15 '16 at 22:46
satokunsatokun
403412
edited Dec 18 '18 at 1:45
Bram28
63.1k44793
edited Dec 18 '18 at 1:45
Bram28
63.1k44793
edited Dec 18 '18 at 1:45
Bram28
63.1k44793
63.1k44793
asked Oct 15 '16 at 22:46
satokunsatokun
403412
asked Oct 15 '16 at 22:46
satokunsatokun
403412
asked Oct 15 '16 at 22:46
satokunsatokun
403412
403412
add a comment |
add a comment |
2 Answers
2
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votes
$begingroup$
In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.3k42966
$endgroup$
add a comment |
$begingroup$
The direct translation would be (with $>$ instead of $T$):
$$ forall x( exists y(x > y land a > y) to x > a ) $$
where the $exists y$ is inside the premise of the implication.
This is equivalent to
$$ forall x forall y( (x>y land a>y) to x > a ) $$
where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".
Note, though, that
$$ forall x exists y( (x>y land a>y) to x > a ) $$
is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
240k17306544
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.3k42966
$endgroup$
add a comment |
$begingroup$
In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.3k42966
$endgroup$
add a comment |
$begingroup$
In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.3k42966
$endgroup$
In classical logic, $(forall y)[phi(y, x) Rightarrow psi(x)]$ is equivalent to $[(exists y)phi(y, x)] Rightarrow psi(x)$, so both of your proposed answers are correct, if classical logic is the right logic to use in Wonderland.
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.3k42966
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.3k42966
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.3k42966
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.3k42966
29.3k42966
add a comment |
add a comment |
$begingroup$
The direct translation would be (with $>$ instead of $T$):
$$ forall x( exists y(x > y land a > y) to x > a ) $$
where the $exists y$ is inside the premise of the implication.
This is equivalent to
$$ forall x forall y( (x>y land a>y) to x > a ) $$
where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".
Note, though, that
$$ forall x exists y( (x>y land a>y) to x > a ) $$
is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
240k17306544
$endgroup$
add a comment |
$begingroup$
The direct translation would be (with $>$ instead of $T$):
$$ forall x( exists y(x > y land a > y) to x > a ) $$
where the $exists y$ is inside the premise of the implication.
This is equivalent to
$$ forall x forall y( (x>y land a>y) to x > a ) $$
where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".
Note, though, that
$$ forall x exists y( (x>y land a>y) to x > a ) $$
is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
240k17306544
$endgroup$
add a comment |
$begingroup$
The direct translation would be (with $>$ instead of $T$):
$$ forall x( exists y(x > y land a > y) to x > a ) $$
where the $exists y$ is inside the premise of the implication.
This is equivalent to
$$ forall x forall y( (x>y land a>y) to x > a ) $$
where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".
Note, though, that
$$ forall x exists y( (x>y land a>y) to x > a ) $$
is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
240k17306544
$endgroup$
The direct translation would be (with $>$ instead of $T$):
$$ forall x( exists y(x > y land a > y) to x > a ) $$
where the $exists y$ is inside the premise of the implication.
This is equivalent to
$$ forall x forall y( (x>y land a>y) to x > a ) $$
where $forall y$ ranges over the entire implication. So arguably both of your proposals are right, but the one with an $exists$ can be said to be "more verbatim".
Note, though, that
$$ forall x exists y( (x>y land a>y) to x > a ) $$
is something different. This is not even particularly suited to be rendered in English; the meaning of $exists x(cdotstocdots)$ is not very intuitive.
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
240k17306544
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
240k17306544
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
240k17306544
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
240k17306544
240k17306544
add a comment |
add a comment |
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