How to prove $Acap (Bsetminus C) = (Acap B)setminus(Acap C) = (Acap B)setminus C$? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to prove $forall Y subset X forall U subset X (Y setminus U = Y cap (X setminus U))$?Prove that $X cap (Y - Z) = (X cap Y) - (X cap Z)$Prove that $Acap (Bsetminus C)=(A cap B)setminus(A cap C)$.How do I show a set $A = (Asetminus B)cup (Acap B)$ for discrete math?Proving equalities with sets: $A cap (B cup C) = (A cap B) cup (A cap C)$ and $X setminus (A cap B) = (X setminus A) cup (X setminus B)$Prove that $A cap B = B setminus (barA cap B)$Proofs involving $ (Asetminus B) cup (A cap B)$$(X setminus F) cap Y = Y setminus A Leftrightarrow A = F cap Y$Prove the set identity $(A cap B) setminus (Bcap C) = A cap (B setminus C)$$ Asetminus(Bsetminus C) = (Asetminus B)cup (Acap C) $ how to prove?
What did Darwin mean by 'squib' here?
How did the aliens keep their waters separated?
Was credit for the black hole image misattributed?
Why don't the Weasley twins use magic outside of school if the Trace can only find the location of spells cast?
How to say that you spent the night with someone, you were only sleeping and nothing else?
New Order #5: where Fibonacci and Beatty meet at Wythoff
Simulating Exploding Dice
Why does tar appear to skip file contents when output file is /dev/null?
Interesting examples of non-locally compact topological groups
Need a suitable toxic chemical for a murder plot in my novel
3 doors, three guards, one stone
Am I ethically obligated to go into work on an off day if the reason is sudden?
Estimate capacitor parameters
Is it possible to ask for a hotel room without minibar/extra services?
When is phishing education going too far?
Stop battery usage [Ubuntu 18]
Single author papers against my advisor's will?
How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time
Can a zero nonce be safely used with AES-GCM if the key is random and never used again?
I'm thinking of a number
How do I automatically answer y in bash script?
Strange behaviour of Check
Using "nakedly" instead of "with nothing on"
What's the difference between (size_t)-1 and ~0?
How to prove $Acap (Bsetminus C) = (Acap B)setminus(Acap C) = (Acap B)setminus C$?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to prove $forall Y subset X forall U subset X (Y setminus U = Y cap (X setminus U))$?Prove that $X cap (Y - Z) = (X cap Y) - (X cap Z)$Prove that $Acap (Bsetminus C)=(A cap B)setminus(A cap C)$.How do I show a set $A = (Asetminus B)cup (Acap B)$ for discrete math?Proving equalities with sets: $A cap (B cup C) = (A cap B) cup (A cap C)$ and $X setminus (A cap B) = (X setminus A) cup (X setminus B)$Prove that $A cap B = B setminus (barA cap B)$Proofs involving $ (Asetminus B) cup (A cap B)$$(X setminus F) cap Y = Y setminus A Leftrightarrow A = F cap Y$Prove the set identity $(A cap B) setminus (Bcap C) = A cap (B setminus C)$$ Asetminus(Bsetminus C) = (Asetminus B)cup (Acap C) $ how to prove?
$begingroup$
I do understand how to prove something like
$$Acap (Bsetminus C)= (Acap B)setminus C$$ but I don't really understand how do I do it with something like this $$Acap (Bsetminus C) = (Acap B)setminus(Acap C) = (Acap B)setminus C$$
discrete-mathematics elementary-set-theory
asked Mar 31 at 18:42
KatinelisKatinelis
61
$endgroup$
add a comment |
$begingroup$
I do understand how to prove something like
$$Acap (Bsetminus C)= (Acap B)setminus C$$ but I don't really understand how do I do it with something like this $$Acap (Bsetminus C) = (Acap B)setminus(Acap C) = (Acap B)setminus C$$
discrete-mathematics elementary-set-theory
asked Mar 31 at 18:42
KatinelisKatinelis
61
$endgroup$
$begingroup$
To prove $A = B$ you have to prove $x in A Leftrightarrow x in B$. So prove $x in A cap (B setminus C) Leftrightarrow x in (A cap B) setminus C)$
$endgroup$
– Andrew Li
Mar 31 at 18:48
add a comment |
$begingroup$
I do understand how to prove something like
$$Acap (Bsetminus C)= (Acap B)setminus C$$ but I don't really understand how do I do it with something like this $$Acap (Bsetminus C) = (Acap B)setminus(Acap C) = (Acap B)setminus C$$
discrete-mathematics elementary-set-theory
asked Mar 31 at 18:42
KatinelisKatinelis
61
$endgroup$
I do understand how to prove something like
$$Acap (Bsetminus C)= (Acap B)setminus C$$ but I don't really understand how do I do it with something like this $$Acap (Bsetminus C) = (Acap B)setminus(Acap C) = (Acap B)setminus C$$
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
asked Mar 31 at 18:42
KatinelisKatinelis
61
asked Mar 31 at 18:42
KatinelisKatinelis
61
asked Mar 31 at 18:42
KatinelisKatinelis
61
asked Mar 31 at 18:42
KatinelisKatinelis
61
asked Mar 31 at 18:42
KatinelisKatinelis
61
61
$begingroup$
To prove $A = B$ you have to prove $x in A Leftrightarrow x in B$. So prove $x in A cap (B setminus C) Leftrightarrow x in (A cap B) setminus C)$
$endgroup$
– Andrew Li
Mar 31 at 18:48
add a comment |
$begingroup$
To prove $A = B$ you have to prove $x in A Leftrightarrow x in B$. So prove $x in A cap (B setminus C) Leftrightarrow x in (A cap B) setminus C)$
$endgroup$
– Andrew Li
Mar 31 at 18:48
$begingroup$
To prove $A = B$ you have to prove $x in A Leftrightarrow x in B$. So prove $x in A cap (B setminus C) Leftrightarrow x in (A cap B) setminus C)$
$endgroup$
– Andrew Li
Mar 31 at 18:48
$begingroup$
To prove $A = B$ you have to prove $x in A Leftrightarrow x in B$. So prove $x in A cap (B setminus C) Leftrightarrow x in (A cap B) setminus C)$
$endgroup$
– Andrew Li
Mar 31 at 18:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In fact $Xsetminus Y=Xcap Y^complement$
So the above equality is just the expression of intersection commutativity and associativity.
$Acap(Bsetminus C)=Acap(B cap C^complement)=(Acap B)cap C^complement=(Acap B)setminus C$
The middle statement is less immediate, it is easier to develop it.
We will use the rule : $(Xcap Y)^complement=X^complement cup Y^complement$
You get $(Acap B)setminus(Acap C)=(Acap B)cap(A^complementcup C^complement)$
And now we distribute $cup$ over $cap$
Same as you do with addition an,d multiplication. eg. $(atimes b)times(a'+c')=(aba')+(abc')$
$cdots=underbrace(Acap Bcap A^complement)_varnothingcup(Acap Bcap C^complement)=(Acap B)setminus C$
The first term being empty set since $(Acap A^complement=varnothing)$
Here is a table summarizing De Morgan's laws
answered Mar 31 at 20:08
zwimzwim
12.7k832
$endgroup$
add a comment |
$begingroup$
The way to solve questions about equality of subsets is to appeal to the definition, and show that each is a subset of the other (as Andrew Li has pointed out). In your case, you need to prove that three things are equal to each other. One way to do this is to show that the first two things are equal to each other, and then that the second two things are equal to each other, which by the transitivity of equality will imply that the first and third things are equal (since they are both equal to the second thing.
In your case, you've already proved that $A cap (B setminus C) = (A cap B) setminus C$ (or at least you know how to prove this), so that the first and third sets are equal. Now you only need to show that the first and second sets are equal, or that the second and third sets are equal, which will imply that all three sets are equal.
Of those two options, I think it is easier to show that the second and third sets are equal, and you can do this (again, as Andrew Li suggested), by showing that each set is a subset of the other.
answered Mar 31 at 18:51
Caleb MillerCaleb Miller
111
$endgroup$
add a comment |
Your Answer
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%2f3169753%2fhow-to-prove-a-cap-b-setminus-c-a-cap-b-setminusa-cap-c-a-cap-b-set%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 fact $Xsetminus Y=Xcap Y^complement$
So the above equality is just the expression of intersection commutativity and associativity.
$Acap(Bsetminus C)=Acap(B cap C^complement)=(Acap B)cap C^complement=(Acap B)setminus C$
The middle statement is less immediate, it is easier to develop it.
We will use the rule : $(Xcap Y)^complement=X^complement cup Y^complement$
You get $(Acap B)setminus(Acap C)=(Acap B)cap(A^complementcup C^complement)$
And now we distribute $cup$ over $cap$
Same as you do with addition an,d multiplication. eg. $(atimes b)times(a'+c')=(aba')+(abc')$
$cdots=underbrace(Acap Bcap A^complement)_varnothingcup(Acap Bcap C^complement)=(Acap B)setminus C$
The first term being empty set since $(Acap A^complement=varnothing)$
Here is a table summarizing De Morgan's laws
answered Mar 31 at 20:08
zwimzwim
12.7k832
$endgroup$
add a comment |
$begingroup$
In fact $Xsetminus Y=Xcap Y^complement$
So the above equality is just the expression of intersection commutativity and associativity.
$Acap(Bsetminus C)=Acap(B cap C^complement)=(Acap B)cap C^complement=(Acap B)setminus C$
The middle statement is less immediate, it is easier to develop it.
We will use the rule : $(Xcap Y)^complement=X^complement cup Y^complement$
You get $(Acap B)setminus(Acap C)=(Acap B)cap(A^complementcup C^complement)$
And now we distribute $cup$ over $cap$
Same as you do with addition an,d multiplication. eg. $(atimes b)times(a'+c')=(aba')+(abc')$
$cdots=underbrace(Acap Bcap A^complement)_varnothingcup(Acap Bcap C^complement)=(Acap B)setminus C$
The first term being empty set since $(Acap A^complement=varnothing)$
Here is a table summarizing De Morgan's laws
answered Mar 31 at 20:08
zwimzwim
12.7k832
$endgroup$
add a comment |
$begingroup$
In fact $Xsetminus Y=Xcap Y^complement$
So the above equality is just the expression of intersection commutativity and associativity.
$Acap(Bsetminus C)=Acap(B cap C^complement)=(Acap B)cap C^complement=(Acap B)setminus C$
The middle statement is less immediate, it is easier to develop it.
We will use the rule : $(Xcap Y)^complement=X^complement cup Y^complement$
You get $(Acap B)setminus(Acap C)=(Acap B)cap(A^complementcup C^complement)$
And now we distribute $cup$ over $cap$
Same as you do with addition an,d multiplication. eg. $(atimes b)times(a'+c')=(aba')+(abc')$
$cdots=underbrace(Acap Bcap A^complement)_varnothingcup(Acap Bcap C^complement)=(Acap B)setminus C$
The first term being empty set since $(Acap A^complement=varnothing)$
Here is a table summarizing De Morgan's laws
answered Mar 31 at 20:08
zwimzwim
12.7k832
$endgroup$
In fact $Xsetminus Y=Xcap Y^complement$
So the above equality is just the expression of intersection commutativity and associativity.
$Acap(Bsetminus C)=Acap(B cap C^complement)=(Acap B)cap C^complement=(Acap B)setminus C$
The middle statement is less immediate, it is easier to develop it.
We will use the rule : $(Xcap Y)^complement=X^complement cup Y^complement$
You get $(Acap B)setminus(Acap C)=(Acap B)cap(A^complementcup C^complement)$
And now we distribute $cup$ over $cap$
Same as you do with addition an,d multiplication. eg. $(atimes b)times(a'+c')=(aba')+(abc')$
$cdots=underbrace(Acap Bcap A^complement)_varnothingcup(Acap Bcap C^complement)=(Acap B)setminus C$
The first term being empty set since $(Acap A^complement=varnothing)$
Here is a table summarizing De Morgan's laws
answered Mar 31 at 20:08
zwimzwim
12.7k832
edited Mar 31 at 20:20
answered Mar 31 at 20:08
zwimzwim
12.7k832
answered Mar 31 at 20:08
zwimzwim
12.7k832
answered Mar 31 at 20:08
zwimzwim
12.7k832
12.7k832
add a comment |
add a comment |
$begingroup$
The way to solve questions about equality of subsets is to appeal to the definition, and show that each is a subset of the other (as Andrew Li has pointed out). In your case, you need to prove that three things are equal to each other. One way to do this is to show that the first two things are equal to each other, and then that the second two things are equal to each other, which by the transitivity of equality will imply that the first and third things are equal (since they are both equal to the second thing.
In your case, you've already proved that $A cap (B setminus C) = (A cap B) setminus C$ (or at least you know how to prove this), so that the first and third sets are equal. Now you only need to show that the first and second sets are equal, or that the second and third sets are equal, which will imply that all three sets are equal.
Of those two options, I think it is easier to show that the second and third sets are equal, and you can do this (again, as Andrew Li suggested), by showing that each set is a subset of the other.
answered Mar 31 at 18:51
Caleb MillerCaleb Miller
111
$endgroup$
add a comment |
$begingroup$
The way to solve questions about equality of subsets is to appeal to the definition, and show that each is a subset of the other (as Andrew Li has pointed out). In your case, you need to prove that three things are equal to each other. One way to do this is to show that the first two things are equal to each other, and then that the second two things are equal to each other, which by the transitivity of equality will imply that the first and third things are equal (since they are both equal to the second thing.
In your case, you've already proved that $A cap (B setminus C) = (A cap B) setminus C$ (or at least you know how to prove this), so that the first and third sets are equal. Now you only need to show that the first and second sets are equal, or that the second and third sets are equal, which will imply that all three sets are equal.
Of those two options, I think it is easier to show that the second and third sets are equal, and you can do this (again, as Andrew Li suggested), by showing that each set is a subset of the other.
answered Mar 31 at 18:51
Caleb MillerCaleb Miller
111
$endgroup$
add a comment |
$begingroup$
The way to solve questions about equality of subsets is to appeal to the definition, and show that each is a subset of the other (as Andrew Li has pointed out). In your case, you need to prove that three things are equal to each other. One way to do this is to show that the first two things are equal to each other, and then that the second two things are equal to each other, which by the transitivity of equality will imply that the first and third things are equal (since they are both equal to the second thing.
In your case, you've already proved that $A cap (B setminus C) = (A cap B) setminus C$ (or at least you know how to prove this), so that the first and third sets are equal. Now you only need to show that the first and second sets are equal, or that the second and third sets are equal, which will imply that all three sets are equal.
Of those two options, I think it is easier to show that the second and third sets are equal, and you can do this (again, as Andrew Li suggested), by showing that each set is a subset of the other.
answered Mar 31 at 18:51
Caleb MillerCaleb Miller
111
$endgroup$
The way to solve questions about equality of subsets is to appeal to the definition, and show that each is a subset of the other (as Andrew Li has pointed out). In your case, you need to prove that three things are equal to each other. One way to do this is to show that the first two things are equal to each other, and then that the second two things are equal to each other, which by the transitivity of equality will imply that the first and third things are equal (since they are both equal to the second thing.
In your case, you've already proved that $A cap (B setminus C) = (A cap B) setminus C$ (or at least you know how to prove this), so that the first and third sets are equal. Now you only need to show that the first and second sets are equal, or that the second and third sets are equal, which will imply that all three sets are equal.
Of those two options, I think it is easier to show that the second and third sets are equal, and you can do this (again, as Andrew Li suggested), by showing that each set is a subset of the other.
answered Mar 31 at 18:51
Caleb MillerCaleb Miller
111
answered Mar 31 at 18:51
Caleb MillerCaleb Miller
111
answered Mar 31 at 18:51
Caleb MillerCaleb Miller
111
answered Mar 31 at 18:51
Caleb MillerCaleb Miller
111
111
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%2f3169753%2fhow-to-prove-a-cap-b-setminus-c-a-cap-b-setminusa-cap-c-a-cap-b-set%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
$begingroup$
To prove $A = B$ you have to prove $x in A Leftrightarrow x in B$. So prove $x in A cap (B setminus C) Leftrightarrow x in (A cap B) setminus C)$
$endgroup$
– Andrew Li
Mar 31 at 18:48