Prove $Acap(A'cap B')'=A$ using set equivalence laws
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I am having trouble using set equivalence laws to prove the following.
$$Acap(A'cap B')'=A$$
Any help would be greatly appreciated.
elementary-set-theory
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up vote
2
down vote
favorite
I am having trouble using set equivalence laws to prove the following.
$$Acap(A'cap B')'=A$$
Any help would be greatly appreciated.
elementary-set-theory
use cap cup to write intersection and union in latex
â Deepesh Meena
Aug 25 at 2:01
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am having trouble using set equivalence laws to prove the following.
$$Acap(A'cap B')'=A$$
Any help would be greatly appreciated.
elementary-set-theory
I am having trouble using set equivalence laws to prove the following.
$$Acap(A'cap B')'=A$$
Any help would be greatly appreciated.
elementary-set-theory
edited Aug 25 at 2:10
zzuussee
2,364524
2,364524
asked Aug 25 at 1:44
Goldmuskhex
233
233
use cap cup to write intersection and union in latex
â Deepesh Meena
Aug 25 at 2:01
add a comment |Â
use cap cup to write intersection and union in latex
â Deepesh Meena
Aug 25 at 2:01
use cap cup to write intersection and union in latex
â Deepesh Meena
Aug 25 at 2:01
use cap cup to write intersection and union in latex
â Deepesh Meena
Aug 25 at 2:01
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
I'll give the algebraic derivation of the statement in supplement to Davids nice pictorials.
By the De Morgan laws for the complement, we first derive
$$Acap (A'cap B')'=Acap ((A')'cup (B')')$$
By elementary properties of the complement, that is $(X')'=X$, we have
$$Acap ((A')'cup (B')')=Acap (Acup B)$$
Now, $cap$ distributes over $cup$, i.e.
$$Acap (Acup B)=(Acap A)cup (Acap B)=Acup (Acap B)$$
as $Acap A=A$. Now, as $Acap Bsubseteq A$, we have $Acup (Acap B)=A$.
1
I would also like to know the reason for the downvote.
â zzuussee
Aug 25 at 2:07
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:41
2
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
Thank you very much, I appreciate it. Glad I could help.
â zzuussee
Aug 25 at 10:26
add a comment |Â
up vote
2
down vote
Sometimes figures are worth 1000 words:
add a comment |Â
up vote
1
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$$A cap (A'cap B')'$$
as by morgan's law $(P cap Q)'= P'cup Q' $
thus $$(A'cap B')'=A cup B$$
$$A cap (A cup B) $$
$$(A cap A) cup (A cap B)$$
$$A cap A =A $$
$$A cup (A cap B)$$
By Venn diagram, it is clear that it is nothing but A
1
why the downvote though I don't understand??
â Deepesh Meena
Aug 25 at 2:02
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:42
1
but this is just crazy on this site if I write a hint I get downvoted or if I write a full answer I get downvoted don't know what to do just pray
â Deepesh Meena
Aug 25 at 2:44
1
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I'll give the algebraic derivation of the statement in supplement to Davids nice pictorials.
By the De Morgan laws for the complement, we first derive
$$Acap (A'cap B')'=Acap ((A')'cup (B')')$$
By elementary properties of the complement, that is $(X')'=X$, we have
$$Acap ((A')'cup (B')')=Acap (Acup B)$$
Now, $cap$ distributes over $cup$, i.e.
$$Acap (Acup B)=(Acap A)cup (Acap B)=Acup (Acap B)$$
as $Acap A=A$. Now, as $Acap Bsubseteq A$, we have $Acup (Acap B)=A$.
1
I would also like to know the reason for the downvote.
â zzuussee
Aug 25 at 2:07
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:41
2
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
Thank you very much, I appreciate it. Glad I could help.
â zzuussee
Aug 25 at 10:26
add a comment |Â
up vote
0
down vote
accepted
I'll give the algebraic derivation of the statement in supplement to Davids nice pictorials.
By the De Morgan laws for the complement, we first derive
$$Acap (A'cap B')'=Acap ((A')'cup (B')')$$
By elementary properties of the complement, that is $(X')'=X$, we have
$$Acap ((A')'cup (B')')=Acap (Acup B)$$
Now, $cap$ distributes over $cup$, i.e.
$$Acap (Acup B)=(Acap A)cup (Acap B)=Acup (Acap B)$$
as $Acap A=A$. Now, as $Acap Bsubseteq A$, we have $Acup (Acap B)=A$.
1
I would also like to know the reason for the downvote.
â zzuussee
Aug 25 at 2:07
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:41
2
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
Thank you very much, I appreciate it. Glad I could help.
â zzuussee
Aug 25 at 10:26
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I'll give the algebraic derivation of the statement in supplement to Davids nice pictorials.
By the De Morgan laws for the complement, we first derive
$$Acap (A'cap B')'=Acap ((A')'cup (B')')$$
By elementary properties of the complement, that is $(X')'=X$, we have
$$Acap ((A')'cup (B')')=Acap (Acup B)$$
Now, $cap$ distributes over $cup$, i.e.
$$Acap (Acup B)=(Acap A)cup (Acap B)=Acup (Acap B)$$
as $Acap A=A$. Now, as $Acap Bsubseteq A$, we have $Acup (Acap B)=A$.
I'll give the algebraic derivation of the statement in supplement to Davids nice pictorials.
By the De Morgan laws for the complement, we first derive
$$Acap (A'cap B')'=Acap ((A')'cup (B')')$$
By elementary properties of the complement, that is $(X')'=X$, we have
$$Acap ((A')'cup (B')')=Acap (Acup B)$$
Now, $cap$ distributes over $cup$, i.e.
$$Acap (Acup B)=(Acap A)cup (Acap B)=Acup (Acap B)$$
as $Acap A=A$. Now, as $Acap Bsubseteq A$, we have $Acup (Acap B)=A$.
answered Aug 25 at 1:55
zzuussee
2,364524
2,364524
1
I would also like to know the reason for the downvote.
â zzuussee
Aug 25 at 2:07
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:41
2
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
Thank you very much, I appreciate it. Glad I could help.
â zzuussee
Aug 25 at 10:26
add a comment |Â
1
I would also like to know the reason for the downvote.
â zzuussee
Aug 25 at 2:07
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:41
2
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
Thank you very much, I appreciate it. Glad I could help.
â zzuussee
Aug 25 at 10:26
1
1
I would also like to know the reason for the downvote.
â zzuussee
Aug 25 at 2:07
I would also like to know the reason for the downvote.
â zzuussee
Aug 25 at 2:07
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:41
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:41
2
2
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
Thank you very much, I appreciate it. Glad I could help.
â zzuussee
Aug 25 at 10:26
Thank you very much, I appreciate it. Glad I could help.
â zzuussee
Aug 25 at 10:26
add a comment |Â
up vote
2
down vote
Sometimes figures are worth 1000 words:
add a comment |Â
up vote
2
down vote
Sometimes figures are worth 1000 words:
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Sometimes figures are worth 1000 words:
Sometimes figures are worth 1000 words:
answered Aug 25 at 1:51
David G. Stork
8,05621232
8,05621232
add a comment |Â
add a comment |Â
up vote
1
down vote
$$A cap (A'cap B')'$$
as by morgan's law $(P cap Q)'= P'cup Q' $
thus $$(A'cap B')'=A cup B$$
$$A cap (A cup B) $$
$$(A cap A) cup (A cap B)$$
$$A cap A =A $$
$$A cup (A cap B)$$
By Venn diagram, it is clear that it is nothing but A
1
why the downvote though I don't understand??
â Deepesh Meena
Aug 25 at 2:02
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:42
1
but this is just crazy on this site if I write a hint I get downvoted or if I write a full answer I get downvoted don't know what to do just pray
â Deepesh Meena
Aug 25 at 2:44
1
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
add a comment |Â
up vote
1
down vote
$$A cap (A'cap B')'$$
as by morgan's law $(P cap Q)'= P'cup Q' $
thus $$(A'cap B')'=A cup B$$
$$A cap (A cup B) $$
$$(A cap A) cup (A cap B)$$
$$A cap A =A $$
$$A cup (A cap B)$$
By Venn diagram, it is clear that it is nothing but A
1
why the downvote though I don't understand??
â Deepesh Meena
Aug 25 at 2:02
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:42
1
but this is just crazy on this site if I write a hint I get downvoted or if I write a full answer I get downvoted don't know what to do just pray
â Deepesh Meena
Aug 25 at 2:44
1
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$A cap (A'cap B')'$$
as by morgan's law $(P cap Q)'= P'cup Q' $
thus $$(A'cap B')'=A cup B$$
$$A cap (A cup B) $$
$$(A cap A) cup (A cap B)$$
$$A cap A =A $$
$$A cup (A cap B)$$
By Venn diagram, it is clear that it is nothing but A
$$A cap (A'cap B')'$$
as by morgan's law $(P cap Q)'= P'cup Q' $
thus $$(A'cap B')'=A cup B$$
$$A cap (A cup B) $$
$$(A cap A) cup (A cap B)$$
$$A cap A =A $$
$$A cup (A cap B)$$
By Venn diagram, it is clear that it is nothing but A
answered Aug 25 at 1:56
Deepesh Meena
2,766721
2,766721
1
why the downvote though I don't understand??
â Deepesh Meena
Aug 25 at 2:02
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:42
1
but this is just crazy on this site if I write a hint I get downvoted or if I write a full answer I get downvoted don't know what to do just pray
â Deepesh Meena
Aug 25 at 2:44
1
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
add a comment |Â
1
why the downvote though I don't understand??
â Deepesh Meena
Aug 25 at 2:02
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:42
1
but this is just crazy on this site if I write a hint I get downvoted or if I write a full answer I get downvoted don't know what to do just pray
â Deepesh Meena
Aug 25 at 2:44
1
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
1
1
why the downvote though I don't understand??
â Deepesh Meena
Aug 25 at 2:02
why the downvote though I don't understand??
â Deepesh Meena
Aug 25 at 2:02
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:42
I think the possible reason for downvote is that you gave a whole answer for a PSQ where the OP has showed no effort.
â user572932
Aug 25 at 2:42
1
1
but this is just crazy on this site if I write a hint I get downvoted or if I write a full answer I get downvoted don't know what to do just pray
â Deepesh Meena
Aug 25 at 2:44
but this is just crazy on this site if I write a hint I get downvoted or if I write a full answer I get downvoted don't know what to do just pray
â Deepesh Meena
Aug 25 at 2:44
1
1
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
I certainly didn't downvote your answer. It was very clear and I very much appreciate the time and effort that you took to answer it for me.
â Goldmuskhex
Aug 25 at 4:02
add a comment |Â
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use cap cup to write intersection and union in latex
â Deepesh Meena
Aug 25 at 2:01