$ Asetminus(Bsetminus C) = (Asetminus B)cup (Acap C) $ how to prove? [duplicate] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove the following $A setminus (Bsetminus C) = (Asetminus B)cap (A cup C)$Proving statement - $(A setminus B) cup (A setminus C) = BLeftrightarrow A=B , Ccap B=varnothing$Prove the following $A setminus (Bsetminus C) = (Asetminus B)cap (A cup C)$How to prove $A=(Asetminus B)cup (Acap B)$Show that $Asetminus (Bsetminus C)=(Asetminus B)cup(Acap Bcap C)$Proving equalities with sets: $A cap (B cup C) = (A cap B) cup (A cap C)$ and $X setminus (A cap B) = (X setminus A) cup (X setminus B)$Proofs involving $ (Asetminus B) cup (A cap B)$Prove $Asetminus(B cup C) = (A setminus B) cap (A setminus C)$ using element chasing?How to Prove that $(A ∪ B) setminus (A ∩ B) = (A setminus B) ∪ (B setminus A)$How to show equality $A = (Asetminus B) cup (Asetminus C) cup (A cap B cap C)$Prove that $overlineAsetminus B cap left(overlineAcupoverlineBright)=overlineA$
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$ Asetminus(Bsetminus C) = (Asetminus B)cup (Acap C) $ how to prove? [duplicate]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove the following $A setminus (Bsetminus C) = (Asetminus B)cap (A cup C)$Proving statement - $(A setminus B) cup (A setminus C) = BLeftrightarrow A=B , Ccap B=varnothing$Prove the following $A setminus (Bsetminus C) = (Asetminus B)cap (A cup C)$How to prove $A=(Asetminus B)cup (Acap B)$Show that $Asetminus (Bsetminus C)=(Asetminus B)cup(Acap Bcap C)$Proving equalities with sets: $A cap (B cup C) = (A cap B) cup (A cap C)$ and $X setminus (A cap B) = (X setminus A) cup (X setminus B)$Proofs involving $ (Asetminus B) cup (A cap B)$Prove $Asetminus(B cup C) = (A setminus B) cap (A setminus C)$ using element chasing?How to Prove that $(A ∪ B) setminus (A ∩ B) = (A setminus B) ∪ (B setminus A)$How to show equality $A = (Asetminus B) cup (Asetminus C) cup (A cap B cap C)$Prove that $overlineAsetminus B cap left(overlineAcupoverlineBright)=overlineA$
$begingroup$
This question already has an answer here:
Prove the following $A setminus (Bsetminus C) = (Asetminus B)cap (A cup C)$ [duplicate]
1 answer
how to prove that:
$$
Asetminus(Bsetminus C) = (Asetminus B)cup (Acap C)
$$
My attempts: $$A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C)$$ im new in this, and im pretty stuck here
discrete-mathematics elementary-set-theory
edited Mar 26 at 8:58
Michael Rozenberg
111k1896201
asked Mar 26 at 8:36
SushiSushi
6
$endgroup$
marked as duplicate by Saad, Martin R, Arthur, Asaf Karagila♦ elementary-set-theory
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Mar 26 at 8:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Prove the following $A setminus (Bsetminus C) = (Asetminus B)cap (A cup C)$ [duplicate]
1 answer
how to prove that:
$$
Asetminus(Bsetminus C) = (Asetminus B)cup (Acap C)
$$
My attempts: $$A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C)$$ im new in this, and im pretty stuck here
discrete-mathematics elementary-set-theory
edited Mar 26 at 8:58
Michael Rozenberg
111k1896201
asked Mar 26 at 8:36
SushiSushi
6
$endgroup$
marked as duplicate by Saad, Martin R, Arthur, Asaf Karagila♦ elementary-set-theory
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
What have you tried to do?
$endgroup$
– Jon
Mar 26 at 8:39
$begingroup$
i want to prove that these are equal
$endgroup$
– Sushi
Mar 26 at 8:41
$begingroup$
@Sushi We know what you want to do. That's obvious from reading your question post. We are asking what you have tried, because that's not obvious from your question post. You went through the trouble of signing up here just to ask this question, that can't have been the first thing you tried. What did you try before you came here?
$endgroup$
– Arthur
Mar 26 at 8:42
$begingroup$
@Sushi I solved your problem. If you want to see my solution, show us your trying.
$endgroup$
– Michael Rozenberg
Mar 26 at 8:44
$begingroup$
A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C) im new in this, and im pretty stuck here
$endgroup$
– Sushi
Mar 26 at 8:55
|
show 3 more comments
$begingroup$
This question already has an answer here:
Prove the following $A setminus (Bsetminus C) = (Asetminus B)cap (A cup C)$ [duplicate]
1 answer
how to prove that:
$$
Asetminus(Bsetminus C) = (Asetminus B)cup (Acap C)
$$
My attempts: $$A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C)$$ im new in this, and im pretty stuck here
discrete-mathematics elementary-set-theory
edited Mar 26 at 8:58
Michael Rozenberg
111k1896201
asked Mar 26 at 8:36
SushiSushi
6
$endgroup$
This question already has an answer here:
Prove the following $A setminus (Bsetminus C) = (Asetminus B)cap (A cup C)$ [duplicate]
1 answer
how to prove that:
$$
Asetminus(Bsetminus C) = (Asetminus B)cup (Acap C)
$$
My attempts: $$A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C)$$ im new in this, and im pretty stuck here
This question already has an answer here:
Prove the following $A setminus (Bsetminus C) = (Asetminus B)cap (A cup C)$ [duplicate]
1 answer
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
edited Mar 26 at 8:58
Michael Rozenberg
111k1896201
asked Mar 26 at 8:36
SushiSushi
6
edited Mar 26 at 8:58
Michael Rozenberg
111k1896201
asked Mar 26 at 8:36
SushiSushi
6
edited Mar 26 at 8:58
Michael Rozenberg
111k1896201
edited Mar 26 at 8:58
Michael Rozenberg
111k1896201
edited Mar 26 at 8:58
Michael Rozenberg
111k1896201
111k1896201
asked Mar 26 at 8:36
SushiSushi
6
asked Mar 26 at 8:36
SushiSushi
6
asked Mar 26 at 8:36
SushiSushi
6
6
marked as duplicate by Saad, Martin R, Arthur, Asaf Karagila♦ elementary-set-theory
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marked as duplicate by Saad, Martin R, Arthur, Asaf Karagila♦ elementary-set-theory
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Mar 26 at 8:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
What have you tried to do?
$endgroup$
– Jon
Mar 26 at 8:39
$begingroup$
i want to prove that these are equal
$endgroup$
– Sushi
Mar 26 at 8:41
$begingroup$
@Sushi We know what you want to do. That's obvious from reading your question post. We are asking what you have tried, because that's not obvious from your question post. You went through the trouble of signing up here just to ask this question, that can't have been the first thing you tried. What did you try before you came here?
$endgroup$
– Arthur
Mar 26 at 8:42
$begingroup$
@Sushi I solved your problem. If you want to see my solution, show us your trying.
$endgroup$
– Michael Rozenberg
Mar 26 at 8:44
$begingroup$
A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C) im new in this, and im pretty stuck here
$endgroup$
– Sushi
Mar 26 at 8:55
|
show 3 more comments
2
$begingroup$
What have you tried to do?
$endgroup$
– Jon
Mar 26 at 8:39
$begingroup$
i want to prove that these are equal
$endgroup$
– Sushi
Mar 26 at 8:41
$begingroup$
@Sushi We know what you want to do. That's obvious from reading your question post. We are asking what you have tried, because that's not obvious from your question post. You went through the trouble of signing up here just to ask this question, that can't have been the first thing you tried. What did you try before you came here?
$endgroup$
– Arthur
Mar 26 at 8:42
$begingroup$
@Sushi I solved your problem. If you want to see my solution, show us your trying.
$endgroup$
– Michael Rozenberg
Mar 26 at 8:44
$begingroup$
A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C) im new in this, and im pretty stuck here
$endgroup$
– Sushi
Mar 26 at 8:55
2
2
$begingroup$
What have you tried to do?
$endgroup$
– Jon
Mar 26 at 8:39
$begingroup$
What have you tried to do?
$endgroup$
– Jon
Mar 26 at 8:39
$begingroup$
i want to prove that these are equal
$endgroup$
– Sushi
Mar 26 at 8:41
$begingroup$
i want to prove that these are equal
$endgroup$
– Sushi
Mar 26 at 8:41
$begingroup$
@Sushi We know what you want to do. That's obvious from reading your question post. We are asking what you have tried, because that's not obvious from your question post. You went through the trouble of signing up here just to ask this question, that can't have been the first thing you tried. What did you try before you came here?
$endgroup$
– Arthur
Mar 26 at 8:42
$begingroup$
@Sushi We know what you want to do. That's obvious from reading your question post. We are asking what you have tried, because that's not obvious from your question post. You went through the trouble of signing up here just to ask this question, that can't have been the first thing you tried. What did you try before you came here?
$endgroup$
– Arthur
Mar 26 at 8:42
$begingroup$
@Sushi I solved your problem. If you want to see my solution, show us your trying.
$endgroup$
– Michael Rozenberg
Mar 26 at 8:44
$begingroup$
@Sushi I solved your problem. If you want to see my solution, show us your trying.
$endgroup$
– Michael Rozenberg
Mar 26 at 8:44
$begingroup$
A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C) im new in this, and im pretty stuck here
$endgroup$
– Sushi
Mar 26 at 8:55
$begingroup$
A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C) im new in this, and im pretty stuck here
$endgroup$
– Sushi
Mar 26 at 8:55
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$$xin Asetminus(Bsetminus C)Leftrightarrow xin Awedge xnotin(Bsetminus C)Leftrightarrow xin Awedgeoverlinexin Bwedge xnotin CLeftrightarrow$$
$$Leftrightarrow xin Awedge(xnotin Bvee xin C)Leftrightarrow xin(Asetminus B)vee xin(Acap B)Leftrightarrow xin(Asetminus B)cup(Acap B).$$
answered Mar 26 at 8:49
Michael RozenbergMichael Rozenberg
111k1896201
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$xin Asetminus(Bsetminus C)Leftrightarrow xin Awedge xnotin(Bsetminus C)Leftrightarrow xin Awedgeoverlinexin Bwedge xnotin CLeftrightarrow$$
$$Leftrightarrow xin Awedge(xnotin Bvee xin C)Leftrightarrow xin(Asetminus B)vee xin(Acap B)Leftrightarrow xin(Asetminus B)cup(Acap B).$$
answered Mar 26 at 8:49
Michael RozenbergMichael Rozenberg
111k1896201
$endgroup$
add a comment |
$begingroup$
$$xin Asetminus(Bsetminus C)Leftrightarrow xin Awedge xnotin(Bsetminus C)Leftrightarrow xin Awedgeoverlinexin Bwedge xnotin CLeftrightarrow$$
$$Leftrightarrow xin Awedge(xnotin Bvee xin C)Leftrightarrow xin(Asetminus B)vee xin(Acap B)Leftrightarrow xin(Asetminus B)cup(Acap B).$$
answered Mar 26 at 8:49
Michael RozenbergMichael Rozenberg
111k1896201
$endgroup$
add a comment |
$begingroup$
$$xin Asetminus(Bsetminus C)Leftrightarrow xin Awedge xnotin(Bsetminus C)Leftrightarrow xin Awedgeoverlinexin Bwedge xnotin CLeftrightarrow$$
$$Leftrightarrow xin Awedge(xnotin Bvee xin C)Leftrightarrow xin(Asetminus B)vee xin(Acap B)Leftrightarrow xin(Asetminus B)cup(Acap B).$$
answered Mar 26 at 8:49
Michael RozenbergMichael Rozenberg
111k1896201
$endgroup$
$$xin Asetminus(Bsetminus C)Leftrightarrow xin Awedge xnotin(Bsetminus C)Leftrightarrow xin Awedgeoverlinexin Bwedge xnotin CLeftrightarrow$$
$$Leftrightarrow xin Awedge(xnotin Bvee xin C)Leftrightarrow xin(Asetminus B)vee xin(Acap B)Leftrightarrow xin(Asetminus B)cup(Acap B).$$
answered Mar 26 at 8:49
Michael RozenbergMichael Rozenberg
111k1896201
edited Mar 26 at 9:05
answered Mar 26 at 8:49
Michael RozenbergMichael Rozenberg
111k1896201
answered Mar 26 at 8:49
Michael RozenbergMichael Rozenberg
111k1896201
answered Mar 26 at 8:49
Michael RozenbergMichael Rozenberg
111k1896201
111k1896201
add a comment |
add a comment |
2
$begingroup$
What have you tried to do?
$endgroup$
– Jon
Mar 26 at 8:39
$begingroup$
i want to prove that these are equal
$endgroup$
– Sushi
Mar 26 at 8:41
$begingroup$
@Sushi We know what you want to do. That's obvious from reading your question post. We are asking what you have tried, because that's not obvious from your question post. You went through the trouble of signing up here just to ask this question, that can't have been the first thing you tried. What did you try before you came here?
$endgroup$
– Arthur
Mar 26 at 8:42
$begingroup$
@Sushi I solved your problem. If you want to see my solution, show us your trying.
$endgroup$
– Michael Rozenberg
Mar 26 at 8:44
$begingroup$
A∖(B∖C) = x∈A ∧ x∉B (A∖B)∪(A∩C) = (x∈A ∧ x∉B) ∨ (x∈A ∧ x∈C) im new in this, and im pretty stuck here
$endgroup$
– Sushi
Mar 26 at 8:55