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
63.7k44793
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
63.7k44793
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
63.7k44793
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
63.7k44793
asked Oct 15 '16 at 22:46
satokunsatokun
403412
edited Dec 18 '18 at 1:45
Bram28
63.7k44793
asked Oct 15 '16 at 22:46
satokunsatokun
403412
edited Dec 18 '18 at 1:45
Bram28
63.7k44793
edited Dec 18 '18 at 1:45
Bram28
63.7k44793
edited Dec 18 '18 at 1:45
Bram28
63.7k44793
63.7k44793
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
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
29.5k42967
$endgroup$
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
242k17308550
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1970197%2ftranslation-into-predicate-calculus%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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
29.5k42967
$endgroup$
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
29.5k42967
$endgroup$
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
29.5k42967
$endgroup$
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
29.5k42967
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.5k42967
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.5k42967
answered Oct 15 '16 at 22:59
Rob ArthanRob Arthan
29.5k42967
29.5k42967
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
242k17308550
$endgroup$
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
242k17308550
$endgroup$
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
242k17308550
$endgroup$
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
242k17308550
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
242k17308550
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
242k17308550
answered Oct 15 '16 at 22:58
Henning MakholmHenning Makholm
242k17308550
242k17308550
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1970197%2ftranslation-into-predicate-calculus%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
