logic contradiction












1












$begingroup$


Im learning about predicate logic from this textbook and I stumbled upon something that really confused me (both in phrase and in the contradiction I think I found).



On page 64 there is a sentence "There is someone who likes everyone who likes everyone that he likes". The purpose is to break it down to be put into predicate logic. Ok, confusing, but paraphrasing "There is a person who likes all individuals, such that the individual likes all the same people that he also likes". But what if the individual didn't like themselves? Everyone else the individual likes the same as the person but if they didn't like themselves this would be a contradiction. The person likes the individual, but the individual didn't like themselves, then they don't like all the same people. But if the person now doesn't like the individual, they now like all the same people so the person likes the individual. Am I not understanding something, or is this really a very confusing paradox?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It might be paradoxical, but even so, you should still be able to translate it. In fact, once you have that translation, you might be able to formally show that it is indeed paradoxical! So, don't worry about any paradoxes for now, just translate it.
    $endgroup$
    – Bram28
    Jan 13 at 3:49
















1












$begingroup$


Im learning about predicate logic from this textbook and I stumbled upon something that really confused me (both in phrase and in the contradiction I think I found).



On page 64 there is a sentence "There is someone who likes everyone who likes everyone that he likes". The purpose is to break it down to be put into predicate logic. Ok, confusing, but paraphrasing "There is a person who likes all individuals, such that the individual likes all the same people that he also likes". But what if the individual didn't like themselves? Everyone else the individual likes the same as the person but if they didn't like themselves this would be a contradiction. The person likes the individual, but the individual didn't like themselves, then they don't like all the same people. But if the person now doesn't like the individual, they now like all the same people so the person likes the individual. Am I not understanding something, or is this really a very confusing paradox?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It might be paradoxical, but even so, you should still be able to translate it. In fact, once you have that translation, you might be able to formally show that it is indeed paradoxical! So, don't worry about any paradoxes for now, just translate it.
    $endgroup$
    – Bram28
    Jan 13 at 3:49














1












1








1





$begingroup$


Im learning about predicate logic from this textbook and I stumbled upon something that really confused me (both in phrase and in the contradiction I think I found).



On page 64 there is a sentence "There is someone who likes everyone who likes everyone that he likes". The purpose is to break it down to be put into predicate logic. Ok, confusing, but paraphrasing "There is a person who likes all individuals, such that the individual likes all the same people that he also likes". But what if the individual didn't like themselves? Everyone else the individual likes the same as the person but if they didn't like themselves this would be a contradiction. The person likes the individual, but the individual didn't like themselves, then they don't like all the same people. But if the person now doesn't like the individual, they now like all the same people so the person likes the individual. Am I not understanding something, or is this really a very confusing paradox?










share|cite|improve this question











$endgroup$




Im learning about predicate logic from this textbook and I stumbled upon something that really confused me (both in phrase and in the contradiction I think I found).



On page 64 there is a sentence "There is someone who likes everyone who likes everyone that he likes". The purpose is to break it down to be put into predicate logic. Ok, confusing, but paraphrasing "There is a person who likes all individuals, such that the individual likes all the same people that he also likes". But what if the individual didn't like themselves? Everyone else the individual likes the same as the person but if they didn't like themselves this would be a contradiction. The person likes the individual, but the individual didn't like themselves, then they don't like all the same people. But if the person now doesn't like the individual, they now like all the same people so the person likes the individual. Am I not understanding something, or is this really a very confusing paradox?







logic propositional-calculus predicate-logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 3:58







Colin Hicks

















asked Jan 13 at 3:45









Colin HicksColin Hicks

38129




38129








  • 1




    $begingroup$
    It might be paradoxical, but even so, you should still be able to translate it. In fact, once you have that translation, you might be able to formally show that it is indeed paradoxical! So, don't worry about any paradoxes for now, just translate it.
    $endgroup$
    – Bram28
    Jan 13 at 3:49














  • 1




    $begingroup$
    It might be paradoxical, but even so, you should still be able to translate it. In fact, once you have that translation, you might be able to formally show that it is indeed paradoxical! So, don't worry about any paradoxes for now, just translate it.
    $endgroup$
    – Bram28
    Jan 13 at 3:49








1




1




$begingroup$
It might be paradoxical, but even so, you should still be able to translate it. In fact, once you have that translation, you might be able to formally show that it is indeed paradoxical! So, don't worry about any paradoxes for now, just translate it.
$endgroup$
– Bram28
Jan 13 at 3:49




$begingroup$
It might be paradoxical, but even so, you should still be able to translate it. In fact, once you have that translation, you might be able to formally show that it is indeed paradoxical! So, don't worry about any paradoxes for now, just translate it.
$endgroup$
– Bram28
Jan 13 at 3:49










1 Answer
1






active

oldest

votes


















1












$begingroup$

What you show is that it is not possible for this person not to like themselves. That is not a paradox, however ... it just means that apparently this person must like themselves.



That is, there is indeed a contradiction if this person does not like themselves. But there is no contradiction if this person does like themselves.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So If I Included as a premise that this person didn't like themselves, it would indeed lead to a contradiction?
    $endgroup$
    – Colin Hicks
    Jan 13 at 3:57








  • 1




    $begingroup$
    @ColinHicks Yes. It would be a good exercise to formally show that ... but that requires you do symbolize the statement correctly.
    $endgroup$
    – Bram28
    Jan 13 at 3:58











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071681%2flogic-contradiction%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

What you show is that it is not possible for this person not to like themselves. That is not a paradox, however ... it just means that apparently this person must like themselves.



That is, there is indeed a contradiction if this person does not like themselves. But there is no contradiction if this person does like themselves.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So If I Included as a premise that this person didn't like themselves, it would indeed lead to a contradiction?
    $endgroup$
    – Colin Hicks
    Jan 13 at 3:57








  • 1




    $begingroup$
    @ColinHicks Yes. It would be a good exercise to formally show that ... but that requires you do symbolize the statement correctly.
    $endgroup$
    – Bram28
    Jan 13 at 3:58
















1












$begingroup$

What you show is that it is not possible for this person not to like themselves. That is not a paradox, however ... it just means that apparently this person must like themselves.



That is, there is indeed a contradiction if this person does not like themselves. But there is no contradiction if this person does like themselves.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So If I Included as a premise that this person didn't like themselves, it would indeed lead to a contradiction?
    $endgroup$
    – Colin Hicks
    Jan 13 at 3:57








  • 1




    $begingroup$
    @ColinHicks Yes. It would be a good exercise to formally show that ... but that requires you do symbolize the statement correctly.
    $endgroup$
    – Bram28
    Jan 13 at 3:58














1












1








1





$begingroup$

What you show is that it is not possible for this person not to like themselves. That is not a paradox, however ... it just means that apparently this person must like themselves.



That is, there is indeed a contradiction if this person does not like themselves. But there is no contradiction if this person does like themselves.






share|cite|improve this answer











$endgroup$



What you show is that it is not possible for this person not to like themselves. That is not a paradox, however ... it just means that apparently this person must like themselves.



That is, there is indeed a contradiction if this person does not like themselves. But there is no contradiction if this person does like themselves.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 3:57

























answered Jan 13 at 3:55









Bram28Bram28

61.5k44792




61.5k44792












  • $begingroup$
    So If I Included as a premise that this person didn't like themselves, it would indeed lead to a contradiction?
    $endgroup$
    – Colin Hicks
    Jan 13 at 3:57








  • 1




    $begingroup$
    @ColinHicks Yes. It would be a good exercise to formally show that ... but that requires you do symbolize the statement correctly.
    $endgroup$
    – Bram28
    Jan 13 at 3:58


















  • $begingroup$
    So If I Included as a premise that this person didn't like themselves, it would indeed lead to a contradiction?
    $endgroup$
    – Colin Hicks
    Jan 13 at 3:57








  • 1




    $begingroup$
    @ColinHicks Yes. It would be a good exercise to formally show that ... but that requires you do symbolize the statement correctly.
    $endgroup$
    – Bram28
    Jan 13 at 3:58
















$begingroup$
So If I Included as a premise that this person didn't like themselves, it would indeed lead to a contradiction?
$endgroup$
– Colin Hicks
Jan 13 at 3:57






$begingroup$
So If I Included as a premise that this person didn't like themselves, it would indeed lead to a contradiction?
$endgroup$
– Colin Hicks
Jan 13 at 3:57






1




1




$begingroup$
@ColinHicks Yes. It would be a good exercise to formally show that ... but that requires you do symbolize the statement correctly.
$endgroup$
– Bram28
Jan 13 at 3:58




$begingroup$
@ColinHicks Yes. It would be a good exercise to formally show that ... but that requires you do symbolize the statement correctly.
$endgroup$
– Bram28
Jan 13 at 3:58


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071681%2flogic-contradiction%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Mario Kart Wii

The Binding of Isaac: Rebirth/Afterbirth

What does “Dominus providebit” mean?