Universal Quantified Statement being equivalent when variables are swapped
I was given this statement and asked to express this with universal quantifiers.
Likes(x,y) is a Binary Relation that means that person x likes person y.
The statement given was:
"Everybody is liked by somebody*". Easy Enough, or so I thought? I put:
∀x∃y, (Likes(y,x)). Apparently this is wrong and the correct solution (that a TA displayed as the answer) is ∀y∃x (Likes(x,y)).
For clarity my question is generally :
is ∀x∃y, (P(y,x)) ≡ ∀y∃x, (P(x,y)) ?
I understand the variables are swapped, but i believe that it is logically equivalent, as i just made the "everbody" labled as x and the "somebody" labeled as "y".
Thank you!
discrete-mathematics logic quantifiers logic-translation
edited Dec 11 '18 at 20:48
Bram28
60.3k44590
asked Dec 11 '18 at 5:33
Stephen Isaac
11
add a comment |
I was given this statement and asked to express this with universal quantifiers.
Likes(x,y) is a Binary Relation that means that person x likes person y.
The statement given was:
"Everybody is liked by somebody*". Easy Enough, or so I thought? I put:
∀x∃y, (Likes(y,x)). Apparently this is wrong and the correct solution (that a TA displayed as the answer) is ∀y∃x (Likes(x,y)).
For clarity my question is generally :
is ∀x∃y, (P(y,x)) ≡ ∀y∃x, (P(x,y)) ?
I understand the variables are swapped, but i believe that it is logically equivalent, as i just made the "everbody" labled as x and the "somebody" labeled as "y".
Thank you!
discrete-mathematics logic quantifiers logic-translation
edited Dec 11 '18 at 20:48
Bram28
60.3k44590
asked Dec 11 '18 at 5:33
Stephen Isaac
11
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
– Lord Shark the Unknown
Dec 11 '18 at 5:39
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
– Stephen Isaac
Dec 11 '18 at 5:42
add a comment |
I was given this statement and asked to express this with universal quantifiers.
Likes(x,y) is a Binary Relation that means that person x likes person y.
The statement given was:
"Everybody is liked by somebody*". Easy Enough, or so I thought? I put:
∀x∃y, (Likes(y,x)). Apparently this is wrong and the correct solution (that a TA displayed as the answer) is ∀y∃x (Likes(x,y)).
For clarity my question is generally :
is ∀x∃y, (P(y,x)) ≡ ∀y∃x, (P(x,y)) ?
I understand the variables are swapped, but i believe that it is logically equivalent, as i just made the "everbody" labled as x and the "somebody" labeled as "y".
Thank you!
discrete-mathematics logic quantifiers logic-translation
edited Dec 11 '18 at 20:48
Bram28
60.3k44590
asked Dec 11 '18 at 5:33
Stephen Isaac
11
I was given this statement and asked to express this with universal quantifiers.
Likes(x,y) is a Binary Relation that means that person x likes person y.
The statement given was:
"Everybody is liked by somebody*". Easy Enough, or so I thought? I put:
∀x∃y, (Likes(y,x)). Apparently this is wrong and the correct solution (that a TA displayed as the answer) is ∀y∃x (Likes(x,y)).
For clarity my question is generally :
is ∀x∃y, (P(y,x)) ≡ ∀y∃x, (P(x,y)) ?
I understand the variables are swapped, but i believe that it is logically equivalent, as i just made the "everbody" labled as x and the "somebody" labeled as "y".
Thank you!
discrete-mathematics logic quantifiers logic-translation
discrete-mathematics logic quantifiers logic-translation
edited Dec 11 '18 at 20:48
Bram28
60.3k44590
asked Dec 11 '18 at 5:33
Stephen Isaac
11
edited Dec 11 '18 at 20:48
Bram28
60.3k44590
asked Dec 11 '18 at 5:33
Stephen Isaac
11
edited Dec 11 '18 at 20:48
Bram28
60.3k44590
edited Dec 11 '18 at 20:48
Bram28
60.3k44590
edited Dec 11 '18 at 20:48
Bram28
60.3k44590
60.3k44590
asked Dec 11 '18 at 5:33
Stephen Isaac
11
asked Dec 11 '18 at 5:33
Stephen Isaac
11
asked Dec 11 '18 at 5:33
Stephen Isaac
11
11
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
– Lord Shark the Unknown
Dec 11 '18 at 5:39
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
– Stephen Isaac
Dec 11 '18 at 5:42
add a comment |
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
– Lord Shark the Unknown
Dec 11 '18 at 5:39
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
– Stephen Isaac
Dec 11 '18 at 5:42
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
– Lord Shark the Unknown
Dec 11 '18 at 5:39
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
– Lord Shark the Unknown
Dec 11 '18 at 5:39
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
– Stephen Isaac
Dec 11 '18 at 5:42
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
– Stephen Isaac
Dec 11 '18 at 5:42
add a comment |
1 Answer
1
active
oldest
votes
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28
60.3k44590
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%2f3034934%2funiversal-quantified-statement-being-equivalent-when-variables-are-swapped%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
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28
60.3k44590
add a comment |
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28
60.3k44590
add a comment |
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28
60.3k44590
Yes, they are equivalent.
To actually formally show this, use the following two equivalences:
$$forall x varphi(x) Leftrightarrow forall y varphi(y)$$
$$exists x varphi(x) Leftrightarrow exists y varphi(y)$$
where $varphi$ is any formula, and where $varphi(x)$ is the same as $varphi(y)$, but with all free variables $x$ and $y$ swapped.
With these, we can do:
$$forall x exists y P(y,x) Leftrightarrow forall z exists y P(y,z) Leftrightarrow forall z exists x P(x,z) Leftrightarrow forall y exists x P(x,y)$$
answered Dec 11 '18 at 20:48
Bram28
60.3k44590
answered Dec 11 '18 at 20:48
Bram28
60.3k44590
answered Dec 11 '18 at 20:48
Bram28
60.3k44590
answered Dec 11 '18 at 20:48
Bram28
60.3k44590
60.3k44590
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3034934%2funiversal-quantified-statement-being-equivalent-when-variables-are-swapped%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
Those statements are logically equivalent, but surely the correct answer to the original problem is $forall xexists y text{Likes}(x,y)$.
– Lord Shark the Unknown
Dec 11 '18 at 5:39
Ah yes. My bad. I meant the statement to be the following: "Everybody is liked by somebody"
– Stephen Isaac
Dec 11 '18 at 5:42