Proving adsorption law
$begingroup$
Prove the absorption law
$Acap(Acup B)=A$
My Try:
Attempt 1:
$xin Acap(Acup B)$
$xin A$ and $xin Acup Brightarrow xin A$
Thus $Acap(Acup B)subset A$
$xin Arightarrow xin A$ and $xin Acup Brightarrow xin Acap(Acup B)$
Thus $Asubset Acap(Acup B)$
So, $Acap(Acup B)=A$
Attempt 2:
$xin Acap(Acup B)iff xin A$ and $xin Acup B$
$iff xin A$ and $[xin A$ or $xin B]$
$iff xin A$
$Acap(Acup B)=A$
My Question: Which of my above attempts is most precise?
discrete-mathematics proof-verification elementary-set-theory logic
asked Dec 2 '18 at 19:42
user982787user982787
1117
$endgroup$
add a comment |
$begingroup$
Prove the absorption law
$Acap(Acup B)=A$
My Try:
Attempt 1:
$xin Acap(Acup B)$
$xin A$ and $xin Acup Brightarrow xin A$
Thus $Acap(Acup B)subset A$
$xin Arightarrow xin A$ and $xin Acup Brightarrow xin Acap(Acup B)$
Thus $Asubset Acap(Acup B)$
So, $Acap(Acup B)=A$
Attempt 2:
$xin Acap(Acup B)iff xin A$ and $xin Acup B$
$iff xin A$ and $[xin A$ or $xin B]$
$iff xin A$
$Acap(Acup B)=A$
My Question: Which of my above attempts is most precise?
discrete-mathematics proof-verification elementary-set-theory logic
asked Dec 2 '18 at 19:42
user982787user982787
1117
$endgroup$
$begingroup$
The former is more readable. Both are missing a whole bunch of words, though.
$endgroup$
– user3482749
Dec 2 '18 at 19:46
add a comment |
$begingroup$
Prove the absorption law
$Acap(Acup B)=A$
My Try:
Attempt 1:
$xin Acap(Acup B)$
$xin A$ and $xin Acup Brightarrow xin A$
Thus $Acap(Acup B)subset A$
$xin Arightarrow xin A$ and $xin Acup Brightarrow xin Acap(Acup B)$
Thus $Asubset Acap(Acup B)$
So, $Acap(Acup B)=A$
Attempt 2:
$xin Acap(Acup B)iff xin A$ and $xin Acup B$
$iff xin A$ and $[xin A$ or $xin B]$
$iff xin A$
$Acap(Acup B)=A$
My Question: Which of my above attempts is most precise?
discrete-mathematics proof-verification elementary-set-theory logic
asked Dec 2 '18 at 19:42
user982787user982787
1117
$endgroup$
Prove the absorption law
$Acap(Acup B)=A$
My Try:
Attempt 1:
$xin Acap(Acup B)$
$xin A$ and $xin Acup Brightarrow xin A$
Thus $Acap(Acup B)subset A$
$xin Arightarrow xin A$ and $xin Acup Brightarrow xin Acap(Acup B)$
Thus $Asubset Acap(Acup B)$
So, $Acap(Acup B)=A$
Attempt 2:
$xin Acap(Acup B)iff xin A$ and $xin Acup B$
$iff xin A$ and $[xin A$ or $xin B]$
$iff xin A$
$Acap(Acup B)=A$
My Question: Which of my above attempts is most precise?
discrete-mathematics proof-verification elementary-set-theory logic
discrete-mathematics proof-verification elementary-set-theory logic
asked Dec 2 '18 at 19:42
user982787user982787
1117
asked Dec 2 '18 at 19:42
user982787user982787
1117
asked Dec 2 '18 at 19:42
user982787user982787
1117
asked Dec 2 '18 at 19:42
user982787user982787
1117
asked Dec 2 '18 at 19:42
user982787user982787
1117
1117
$begingroup$
The former is more readable. Both are missing a whole bunch of words, though.
$endgroup$
– user3482749
Dec 2 '18 at 19:46
add a comment |
$begingroup$
The former is more readable. Both are missing a whole bunch of words, though.
$endgroup$
– user3482749
Dec 2 '18 at 19:46
$begingroup$
The former is more readable. Both are missing a whole bunch of words, though.
$endgroup$
– user3482749
Dec 2 '18 at 19:46
$begingroup$
The former is more readable. Both are missing a whole bunch of words, though.
$endgroup$
– user3482749
Dec 2 '18 at 19:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 '18 at 23:51
William ElliotWilliam Elliot
7,4742720
$endgroup$
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%2f3023093%2fproving-adsorption-law%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
$begingroup$
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 '18 at 23:51
William ElliotWilliam Elliot
7,4742720
$endgroup$
add a comment |
$begingroup$
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 '18 at 23:51
William ElliotWilliam Elliot
7,4742720
$endgroup$
add a comment |
$begingroup$
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 '18 at 23:51
William ElliotWilliam Elliot
7,4742720
$endgroup$
Instead of a point by point analysis
here is a set alegbra proof.
As U $cap$ V $subseteq$ U $subseteq$ U $cup$ V,
A $cap$ (A $cup$ B) $subseteq$ A = A $cap$ A $subseteq$ A $cap$ (A $cup$ B).
Thus A = A $cap$ (A $cup$ B).
The dual theorem A = A $cup$ (A $cap$ B)
is mostly an application of DeMorgans rules or one
could adapt the above proof to prove the dual theorem.
answered Dec 2 '18 at 23:51
William ElliotWilliam Elliot
7,4742720
answered Dec 2 '18 at 23:51
William ElliotWilliam Elliot
7,4742720
answered Dec 2 '18 at 23:51
William ElliotWilliam Elliot
7,4742720
answered Dec 2 '18 at 23:51
William ElliotWilliam Elliot
7,4742720
7,4742720
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%2f3023093%2fproving-adsorption-law%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
$begingroup$
The former is more readable. Both are missing a whole bunch of words, though.
$endgroup$
– user3482749
Dec 2 '18 at 19:46