Rewriting a logical statement
$begingroup$
Only lakers are irrational people.
I believe it technically should be translated as:
All irrational people are lakers.
Is there is any way at all to rewrite the above statement to mean the following and be logically correct:
All lakers are irrational people.
How would you justify it? (If it is possible)
logic logic-translation
$endgroup$
add a comment |
$begingroup$
Only lakers are irrational people.
I believe it technically should be translated as:
All irrational people are lakers.
Is there is any way at all to rewrite the above statement to mean the following and be logically correct:
All lakers are irrational people.
How would you justify it? (If it is possible)
logic logic-translation
$endgroup$
add a comment |
$begingroup$
Only lakers are irrational people.
I believe it technically should be translated as:
All irrational people are lakers.
Is there is any way at all to rewrite the above statement to mean the following and be logically correct:
All lakers are irrational people.
How would you justify it? (If it is possible)
logic logic-translation
$endgroup$
Only lakers are irrational people.
I believe it technically should be translated as:
All irrational people are lakers.
Is there is any way at all to rewrite the above statement to mean the following and be logically correct:
All lakers are irrational people.
How would you justify it? (If it is possible)
logic logic-translation
logic logic-translation
asked Nov 26 '18 at 17:42
PhillipPhillip
112
112
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Indicating with $L$ the set of lakers $l$ and with $Pi$ the set of irrational people $pi$, the first statement is equivalent to
$$forall piin Pi quad piin L$$
the second one is
$$forall lin Lquad lin Pi $$
which is not equivalent to the first one, indeed from this last one we could also have $pi not in L$ for aome $pi$.
$endgroup$
add a comment |
$begingroup$
All rational people are not lakers.
The opposite converse is equally valid.
The converse that you posit is not equally valid.
$endgroup$
add a comment |
$begingroup$
Your initial translation is correct, though in standard form I would write
All non-rational people are lakers.
This is an Aristotelian A type statement, and A type statements have valid obverses and valid contrapositives. They do not have valid converses.
Obverse: No non-rational people are non-lakers.
Contrapositive: All non-lakers are rational people.
So the ultimate answer to your question is no, because you are trying to get the converse of an A statement to be true, which does not happen in general.
If you had an E or I statement, the converse would be valid.
$endgroup$
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%2f3014649%2frewriting-a-logical-statement%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Indicating with $L$ the set of lakers $l$ and with $Pi$ the set of irrational people $pi$, the first statement is equivalent to
$$forall piin Pi quad piin L$$
the second one is
$$forall lin Lquad lin Pi $$
which is not equivalent to the first one, indeed from this last one we could also have $pi not in L$ for aome $pi$.
$endgroup$
add a comment |
$begingroup$
Indicating with $L$ the set of lakers $l$ and with $Pi$ the set of irrational people $pi$, the first statement is equivalent to
$$forall piin Pi quad piin L$$
the second one is
$$forall lin Lquad lin Pi $$
which is not equivalent to the first one, indeed from this last one we could also have $pi not in L$ for aome $pi$.
$endgroup$
add a comment |
$begingroup$
Indicating with $L$ the set of lakers $l$ and with $Pi$ the set of irrational people $pi$, the first statement is equivalent to
$$forall piin Pi quad piin L$$
the second one is
$$forall lin Lquad lin Pi $$
which is not equivalent to the first one, indeed from this last one we could also have $pi not in L$ for aome $pi$.
$endgroup$
Indicating with $L$ the set of lakers $l$ and with $Pi$ the set of irrational people $pi$, the first statement is equivalent to
$$forall piin Pi quad piin L$$
the second one is
$$forall lin Lquad lin Pi $$
which is not equivalent to the first one, indeed from this last one we could also have $pi not in L$ for aome $pi$.
answered Nov 26 '18 at 17:49
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
All rational people are not lakers.
The opposite converse is equally valid.
The converse that you posit is not equally valid.
$endgroup$
add a comment |
$begingroup$
All rational people are not lakers.
The opposite converse is equally valid.
The converse that you posit is not equally valid.
$endgroup$
add a comment |
$begingroup$
All rational people are not lakers.
The opposite converse is equally valid.
The converse that you posit is not equally valid.
$endgroup$
All rational people are not lakers.
The opposite converse is equally valid.
The converse that you posit is not equally valid.
answered Nov 26 '18 at 17:46
John L WintersJohn L Winters
829
829
add a comment |
add a comment |
$begingroup$
Your initial translation is correct, though in standard form I would write
All non-rational people are lakers.
This is an Aristotelian A type statement, and A type statements have valid obverses and valid contrapositives. They do not have valid converses.
Obverse: No non-rational people are non-lakers.
Contrapositive: All non-lakers are rational people.
So the ultimate answer to your question is no, because you are trying to get the converse of an A statement to be true, which does not happen in general.
If you had an E or I statement, the converse would be valid.
$endgroup$
add a comment |
$begingroup$
Your initial translation is correct, though in standard form I would write
All non-rational people are lakers.
This is an Aristotelian A type statement, and A type statements have valid obverses and valid contrapositives. They do not have valid converses.
Obverse: No non-rational people are non-lakers.
Contrapositive: All non-lakers are rational people.
So the ultimate answer to your question is no, because you are trying to get the converse of an A statement to be true, which does not happen in general.
If you had an E or I statement, the converse would be valid.
$endgroup$
add a comment |
$begingroup$
Your initial translation is correct, though in standard form I would write
All non-rational people are lakers.
This is an Aristotelian A type statement, and A type statements have valid obverses and valid contrapositives. They do not have valid converses.
Obverse: No non-rational people are non-lakers.
Contrapositive: All non-lakers are rational people.
So the ultimate answer to your question is no, because you are trying to get the converse of an A statement to be true, which does not happen in general.
If you had an E or I statement, the converse would be valid.
$endgroup$
Your initial translation is correct, though in standard form I would write
All non-rational people are lakers.
This is an Aristotelian A type statement, and A type statements have valid obverses and valid contrapositives. They do not have valid converses.
Obverse: No non-rational people are non-lakers.
Contrapositive: All non-lakers are rational people.
So the ultimate answer to your question is no, because you are trying to get the converse of an A statement to be true, which does not happen in general.
If you had an E or I statement, the converse would be valid.
answered Nov 26 '18 at 17:51
Adrian KeisterAdrian Keister
4,90261933
4,90261933
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%2f3014649%2frewriting-a-logical-statement%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