How to show $(A cap B^c cap C^c) cup (A cap B cap C^c) cup (A cap B^c cap C) = Abackslash(B cap C)$?How do I write a rigorous proof of the following problem: $(A triangle B) cup C neq (A cup C) triangle (B cup C)$Prove or disprove $X backslash A cup B = (Xbackslash A) cap (X backslash B)$Show that $A cup B = (A$ $B ) cup (A cap B) cup (B$ $A)$$Abackslash (Bcap C) = (Abackslash B)cup (Abackslash C)$; only one inclusion seems to workProve that $A=emptyset$ iff the equality $bigl(((U backslash A) cap B) cup (A cap (U backslash B))bigr)=B$ holds.How to show equality $A = (Asetminus B) cup (Asetminus C) cup (A cap B cap C)$Show that $(AcapbarB) cup (barAcap barC) = (AcapbarB) cup (barAcap barC) cup (barB cap barC)$ by rewriting$big[(X cap A^c)cup Abig] backslash big[(C cap A^c)cup Bbig] = big[(X cap A^c)backslash (C cap A^c)big] cup (Abackslash B)$?Proving $(A cup B) backslash (A cap B) = (A cup C) backslash (A cap C) implies B = C$Probability set theory: $Omega backslash (A cap B) = (Omega backslash A) cup (Omega backslash B)$?
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How to show $(A cap B^c cap C^c) cup (A cap B cap C^c) cup (A cap B^c cap C) = Abackslash(B cap C)$?
How do I write a rigorous proof of the following problem: $(A triangle B) cup C neq (A cup C) triangle (B cup C)$Prove or disprove $X backslash A cup B = (Xbackslash A) cap (X backslash B)$Show that $A cup B = (A$ $B ) cup (A cap B) cup (B$ $A)$$Abackslash (Bcap C) = (Abackslash B)cup (Abackslash C)$; only one inclusion seems to workProve that $A=emptyset$ iff the equality $bigl(((U backslash A) cap B) cup (A cap (U backslash B))bigr)=B$ holds.How to show equality $A = (Asetminus B) cup (Asetminus C) cup (A cap B cap C)$Show that $(AcapbarB) cup (barAcap barC) = (AcapbarB) cup (barAcap barC) cup (barB cap barC)$ by rewriting$big[(X cap A^c)cup Abig] backslash big[(C cap A^c)cup Bbig] = big[(X cap A^c)backslash (C cap A^c)big] cup (Abackslash B)$?Proving $(A cup B) backslash (A cap B) = (A cup C) backslash (A cap C) implies B = C$Probability set theory: $Omega backslash (A cap B) = (Omega backslash A) cup (Omega backslash B)$?
$begingroup$
I should show the following equation with the rules for sets:
$(A cap B^c cap C^c) cup (A cap B cap C^c) cup (A cap B^c cap C) = Abackslash(B cap C)$
I drew the diagramms for the sets and so it is logical that the equality must hold. But i don't know how to who that with the rules for the sets.
Anyone here who could help me with that?
elementary-set-theory
asked Mar 12 at 9:42
roman11roman11
82
New contributor
roman11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I should show the following equation with the rules for sets:
$(A cap B^c cap C^c) cup (A cap B cap C^c) cup (A cap B^c cap C) = Abackslash(B cap C)$
I drew the diagramms for the sets and so it is logical that the equality must hold. But i don't know how to who that with the rules for the sets.
Anyone here who could help me with that?
elementary-set-theory
asked Mar 12 at 9:42
roman11roman11
82
New contributor
roman11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Mar 12 at 9:46
$begingroup$
Drawing the diagrams is already an excellent start. Do you know the rules how to 'multiply out' unions, ie $(X cap Y) cup Z=(X cup Z)cap (Y cup Z)$?
$endgroup$
– quarague
Mar 12 at 10:17
$begingroup$
I know them with just one set out of the brackets but i have never done that with such a long expression like the left side of my equation.
$endgroup$
– roman11
Mar 12 at 10:19
add a comment |
$begingroup$
I should show the following equation with the rules for sets:
$(A cap B^c cap C^c) cup (A cap B cap C^c) cup (A cap B^c cap C) = Abackslash(B cap C)$
I drew the diagramms for the sets and so it is logical that the equality must hold. But i don't know how to who that with the rules for the sets.
Anyone here who could help me with that?
elementary-set-theory
asked Mar 12 at 9:42
roman11roman11
82
New contributor
roman11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I should show the following equation with the rules for sets:
$(A cap B^c cap C^c) cup (A cap B cap C^c) cup (A cap B^c cap C) = Abackslash(B cap C)$
I drew the diagramms for the sets and so it is logical that the equality must hold. But i don't know how to who that with the rules for the sets.
Anyone here who could help me with that?
elementary-set-theory
elementary-set-theory
asked Mar 12 at 9:42
roman11roman11
82
New contributor
roman11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 12 at 9:42
roman11roman11
82
New contributor
roman11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 12 at 10:03
roman11
asked Mar 12 at 9:42
roman11roman11
82
New contributor
roman11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 12 at 9:42
roman11roman11
82
asked Mar 12 at 9:42
roman11roman11
82
82
New contributor
roman11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
roman11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
roman11 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Mar 12 at 9:46
$begingroup$
Drawing the diagrams is already an excellent start. Do you know the rules how to 'multiply out' unions, ie $(X cap Y) cup Z=(X cup Z)cap (Y cup Z)$?
$endgroup$
– quarague
Mar 12 at 10:17
$begingroup$
I know them with just one set out of the brackets but i have never done that with such a long expression like the left side of my equation.
$endgroup$
– roman11
Mar 12 at 10:19
add a comment |
1
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Mar 12 at 9:46
$begingroup$
Drawing the diagrams is already an excellent start. Do you know the rules how to 'multiply out' unions, ie $(X cap Y) cup Z=(X cup Z)cap (Y cup Z)$?
$endgroup$
– quarague
Mar 12 at 10:17
$begingroup$
I know them with just one set out of the brackets but i have never done that with such a long expression like the left side of my equation.
$endgroup$
– roman11
Mar 12 at 10:19
1
1
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Mar 12 at 9:46
$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Mar 12 at 9:46
$begingroup$
Drawing the diagrams is already an excellent start. Do you know the rules how to 'multiply out' unions, ie $(X cap Y) cup Z=(X cup Z)cap (Y cup Z)$?
$endgroup$
– quarague
Mar 12 at 10:17
$begingroup$
Drawing the diagrams is already an excellent start. Do you know the rules how to 'multiply out' unions, ie $(X cap Y) cup Z=(X cup Z)cap (Y cup Z)$?
$endgroup$
– quarague
Mar 12 at 10:17
$begingroup$
I know them with just one set out of the brackets but i have never done that with such a long expression like the left side of my equation.
$endgroup$
– roman11
Mar 12 at 10:19
$begingroup$
I know them with just one set out of the brackets but i have never done that with such a long expression like the left side of my equation.
$endgroup$
– roman11
Mar 12 at 10:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that
$$(Acap B^ccap C^c)cup(Acap Bcap C^c)cup(Acap B^ccap C)$$
$$=((Acap B^ccap C^c)cup(Acap Bcap C^c))cup((Acap B^ccap C^c)cup(Acap B^ccap C))$$
$$=(Acap C^c)cup(Acap B^c)=(Asetminus C)cup(Asetminus B)=Asetminus(Bcap C)$$
where in the second equation we use the fact that
$(Ucap V)cup(Ucap V^c)=U$.
answered Mar 12 at 11:42
Floris ClaassensFloris Claassens
85116
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Note that
$$(Acap B^ccap C^c)cup(Acap Bcap C^c)cup(Acap B^ccap C)$$
$$=((Acap B^ccap C^c)cup(Acap Bcap C^c))cup((Acap B^ccap C^c)cup(Acap B^ccap C))$$
$$=(Acap C^c)cup(Acap B^c)=(Asetminus C)cup(Asetminus B)=Asetminus(Bcap C)$$
where in the second equation we use the fact that
$(Ucap V)cup(Ucap V^c)=U$.
answered Mar 12 at 11:42
Floris ClaassensFloris Claassens
85116
$endgroup$
add a comment |
$begingroup$
Note that
$$(Acap B^ccap C^c)cup(Acap Bcap C^c)cup(Acap B^ccap C)$$
$$=((Acap B^ccap C^c)cup(Acap Bcap C^c))cup((Acap B^ccap C^c)cup(Acap B^ccap C))$$
$$=(Acap C^c)cup(Acap B^c)=(Asetminus C)cup(Asetminus B)=Asetminus(Bcap C)$$
where in the second equation we use the fact that
$(Ucap V)cup(Ucap V^c)=U$.
answered Mar 12 at 11:42
Floris ClaassensFloris Claassens
85116
$endgroup$
add a comment |
$begingroup$
Note that
$$(Acap B^ccap C^c)cup(Acap Bcap C^c)cup(Acap B^ccap C)$$
$$=((Acap B^ccap C^c)cup(Acap Bcap C^c))cup((Acap B^ccap C^c)cup(Acap B^ccap C))$$
$$=(Acap C^c)cup(Acap B^c)=(Asetminus C)cup(Asetminus B)=Asetminus(Bcap C)$$
where in the second equation we use the fact that
$(Ucap V)cup(Ucap V^c)=U$.
answered Mar 12 at 11:42
Floris ClaassensFloris Claassens
85116
$endgroup$
Note that
$$(Acap B^ccap C^c)cup(Acap Bcap C^c)cup(Acap B^ccap C)$$
$$=((Acap B^ccap C^c)cup(Acap Bcap C^c))cup((Acap B^ccap C^c)cup(Acap B^ccap C))$$
$$=(Acap C^c)cup(Acap B^c)=(Asetminus C)cup(Asetminus B)=Asetminus(Bcap C)$$
where in the second equation we use the fact that
$(Ucap V)cup(Ucap V^c)=U$.
answered Mar 12 at 11:42
Floris ClaassensFloris Claassens
85116
answered Mar 12 at 11:42
Floris ClaassensFloris Claassens
85116
answered Mar 12 at 11:42
Floris ClaassensFloris Claassens
85116
answered Mar 12 at 11:42
Floris ClaassensFloris Claassens
85116
85116
add a comment |
add a comment |
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$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Mar 12 at 9:46
$begingroup$
Drawing the diagrams is already an excellent start. Do you know the rules how to 'multiply out' unions, ie $(X cap Y) cup Z=(X cup Z)cap (Y cup Z)$?
$endgroup$
– quarague
Mar 12 at 10:17
$begingroup$
I know them with just one set out of the brackets but i have never done that with such a long expression like the left side of my equation.
$endgroup$
– roman11
Mar 12 at 10:19