de morgan law $Asetminus (B cap C) = (Asetminus B) cup (Asetminus C) $
$begingroup$
First part :
I want to prove the following De Morgan's law : ref.(dfeuer)
$Asetminus (B cap C) = (Asetminus B) cup (Asetminus C) $
Second part:
Prove that $(Asetminus B) cup (Asetminus C) = Asetminus (B cap C) $
Proof:
Let $yin (Asetminus B) cup (Asetminus C)$
$(Asetminus B) cup (Asetminus C) = (y in A; land y notin
B;) vee (y in A; land y notin C;)$
$;;;;;;;;;;;;;;;;;;;;;;;;;= y in A land (
y notin B; vee ; y notin C ) $
$;;;;;;;;;;;;;;;;;;;;;;;;;= y in A land
(lnot ( yin B) lor lnot( yin C) )$
According to De-Morgan's theorem :
$lnot( B land C) Longleftrightarrow (lnot B lor lnot C)$ thus
$y in A land (lnot ( yin B) lor lnot( yin C) ) = y in A land y notin (B land C)$
We can conclude that $(Asetminus B) cup (Asetminus C) = Asetminus (B cap C)$
proof-verification elementary-set-theory
$endgroup$
add a comment |
$begingroup$
First part :
I want to prove the following De Morgan's law : ref.(dfeuer)
$Asetminus (B cap C) = (Asetminus B) cup (Asetminus C) $
Second part:
Prove that $(Asetminus B) cup (Asetminus C) = Asetminus (B cap C) $
Proof:
Let $yin (Asetminus B) cup (Asetminus C)$
$(Asetminus B) cup (Asetminus C) = (y in A; land y notin
B;) vee (y in A; land y notin C;)$
$;;;;;;;;;;;;;;;;;;;;;;;;;= y in A land (
y notin B; vee ; y notin C ) $
$;;;;;;;;;;;;;;;;;;;;;;;;;= y in A land
(lnot ( yin B) lor lnot( yin C) )$
According to De-Morgan's theorem :
$lnot( B land C) Longleftrightarrow (lnot B lor lnot C)$ thus
$y in A land (lnot ( yin B) lor lnot( yin C) ) = y in A land y notin (B land C)$
We can conclude that $(Asetminus B) cup (Asetminus C) = Asetminus (B cap C)$
proof-verification elementary-set-theory
$endgroup$
$begingroup$
That is not very clear English. Perhaps you could parenthesize it or something?
$endgroup$
– dfeuer
Dec 7 '13 at 23:49
1
$begingroup$
But really, what you generally want to do is to show that for all $x$, if $xin Asetminus (Bcap C)$ then $xin $(Asetminus B)cup (A setminus C)$ and the other way around.
$endgroup$
– dfeuer
Dec 7 '13 at 23:50
1
$begingroup$
$yin Aland (ynotin Blor ynotin C)$ is not equivalent to $yin Aland ynotin (Bcup C)$.
$endgroup$
– dfeuer
Dec 8 '13 at 2:52
$begingroup$
possible duplicate of proof of $ A - left (B cap C right)= left (A - B right) cup left (A - C right)$
$endgroup$
– user91500
Dec 8 '13 at 5:25
add a comment |
$begingroup$
First part :
I want to prove the following De Morgan's law : ref.(dfeuer)
$Asetminus (B cap C) = (Asetminus B) cup (Asetminus C) $
Second part:
Prove that $(Asetminus B) cup (Asetminus C) = Asetminus (B cap C) $
Proof:
Let $yin (Asetminus B) cup (Asetminus C)$
$(Asetminus B) cup (Asetminus C) = (y in A; land y notin
B;) vee (y in A; land y notin C;)$
$;;;;;;;;;;;;;;;;;;;;;;;;;= y in A land (
y notin B; vee ; y notin C ) $
$;;;;;;;;;;;;;;;;;;;;;;;;;= y in A land
(lnot ( yin B) lor lnot( yin C) )$
According to De-Morgan's theorem :
$lnot( B land C) Longleftrightarrow (lnot B lor lnot C)$ thus
$y in A land (lnot ( yin B) lor lnot( yin C) ) = y in A land y notin (B land C)$
We can conclude that $(Asetminus B) cup (Asetminus C) = Asetminus (B cap C)$
proof-verification elementary-set-theory
$endgroup$
First part :
I want to prove the following De Morgan's law : ref.(dfeuer)
$Asetminus (B cap C) = (Asetminus B) cup (Asetminus C) $
Second part:
Prove that $(Asetminus B) cup (Asetminus C) = Asetminus (B cap C) $
Proof:
Let $yin (Asetminus B) cup (Asetminus C)$
$(Asetminus B) cup (Asetminus C) = (y in A; land y notin
B;) vee (y in A; land y notin C;)$
$;;;;;;;;;;;;;;;;;;;;;;;;;= y in A land (
y notin B; vee ; y notin C ) $
$;;;;;;;;;;;;;;;;;;;;;;;;;= y in A land
(lnot ( yin B) lor lnot( yin C) )$
According to De-Morgan's theorem :
$lnot( B land C) Longleftrightarrow (lnot B lor lnot C)$ thus
$y in A land (lnot ( yin B) lor lnot( yin C) ) = y in A land y notin (B land C)$
We can conclude that $(Asetminus B) cup (Asetminus C) = Asetminus (B cap C)$
proof-verification elementary-set-theory
proof-verification elementary-set-theory
edited Dec 10 '18 at 13:47
jiten
1,2411413
1,2411413
asked Dec 7 '13 at 23:45
Hani GotcHani Gotc
318
318
$begingroup$
That is not very clear English. Perhaps you could parenthesize it or something?
$endgroup$
– dfeuer
Dec 7 '13 at 23:49
1
$begingroup$
But really, what you generally want to do is to show that for all $x$, if $xin Asetminus (Bcap C)$ then $xin $(Asetminus B)cup (A setminus C)$ and the other way around.
$endgroup$
– dfeuer
Dec 7 '13 at 23:50
1
$begingroup$
$yin Aland (ynotin Blor ynotin C)$ is not equivalent to $yin Aland ynotin (Bcup C)$.
$endgroup$
– dfeuer
Dec 8 '13 at 2:52
$begingroup$
possible duplicate of proof of $ A - left (B cap C right)= left (A - B right) cup left (A - C right)$
$endgroup$
– user91500
Dec 8 '13 at 5:25
add a comment |
$begingroup$
That is not very clear English. Perhaps you could parenthesize it or something?
$endgroup$
– dfeuer
Dec 7 '13 at 23:49
1
$begingroup$
But really, what you generally want to do is to show that for all $x$, if $xin Asetminus (Bcap C)$ then $xin $(Asetminus B)cup (A setminus C)$ and the other way around.
$endgroup$
– dfeuer
Dec 7 '13 at 23:50
1
$begingroup$
$yin Aland (ynotin Blor ynotin C)$ is not equivalent to $yin Aland ynotin (Bcup C)$.
$endgroup$
– dfeuer
Dec 8 '13 at 2:52
$begingroup$
possible duplicate of proof of $ A - left (B cap C right)= left (A - B right) cup left (A - C right)$
$endgroup$
– user91500
Dec 8 '13 at 5:25
$begingroup$
That is not very clear English. Perhaps you could parenthesize it or something?
$endgroup$
– dfeuer
Dec 7 '13 at 23:49
$begingroup$
That is not very clear English. Perhaps you could parenthesize it or something?
$endgroup$
– dfeuer
Dec 7 '13 at 23:49
1
1
$begingroup$
But really, what you generally want to do is to show that for all $x$, if $xin Asetminus (Bcap C)$ then $xin $(Asetminus B)cup (A setminus C)$ and the other way around.
$endgroup$
– dfeuer
Dec 7 '13 at 23:50
$begingroup$
But really, what you generally want to do is to show that for all $x$, if $xin Asetminus (Bcap C)$ then $xin $(Asetminus B)cup (A setminus C)$ and the other way around.
$endgroup$
– dfeuer
Dec 7 '13 at 23:50
1
1
$begingroup$
$yin Aland (ynotin Blor ynotin C)$ is not equivalent to $yin Aland ynotin (Bcup C)$.
$endgroup$
– dfeuer
Dec 8 '13 at 2:52
$begingroup$
$yin Aland (ynotin Blor ynotin C)$ is not equivalent to $yin Aland ynotin (Bcup C)$.
$endgroup$
– dfeuer
Dec 8 '13 at 2:52
$begingroup$
possible duplicate of proof of $ A - left (B cap C right)= left (A - B right) cup left (A - C right)$
$endgroup$
– user91500
Dec 8 '13 at 5:25
$begingroup$
possible duplicate of proof of $ A - left (B cap C right)= left (A - B right) cup left (A - C right)$
$endgroup$
– user91500
Dec 8 '13 at 5:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I'll guide you through one direction. You will need to figure the other out yourself.
Let $xin A setminus (Bcap C)$.
Then by the definition of $setminus$, $xin A$ and $xnotin Bcap C$. The latter statement can be translated into $lnot (x in B cap C)$.
By the definition of $Bcap C$, $xin Bcap C iff (x in B) land (xin C)$.
Thus we conclude that $lnot ((x in B) land (xin C))$.
By De Morgan's laws for logic, we can rewrite this as
$lnot(xin B)lorlnot(xin C)$.
Suppose that $lnot (xin B)$. Then since $xin A$, $xin Asetminus B$.
Suppose instead that $lnot (x in C)$. Then since $xin A$, $xin A setminus C$.
As one of these must be true, $(xin Asetminus B)lor (xin A setminus C)$.
By the definition of union, $xin (Asetminus B)cup(A setminus C)$.
$endgroup$
$begingroup$
excuse me if it's a silly a question. when we say that $xnotin Bcap C$ we are sure that it's in neither $B$ and $C$. is that right? or it $x$ might belong to one of the two sets
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:06
$begingroup$
Yes I think i get it. I saw many explanations, but it was so clear. your explanation is great
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:11
1
$begingroup$
@HaniGotc, let $B = {1,2}$ and let $C = {2,3}$. Is $1in B$? Is $1in C$? Is $1in Bcap C$? Is it true that $1notin Bcap C$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:13
1
$begingroup$
Thanks for the compliment. Can you explain, perhaps by editing your question, how you would argue in the other direction, to show that if $xin (asetminus B)cup(A setminus C)$ then we can conclude also that $xin A setminus (Bcap C)$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:15
$begingroup$
$ 1 in B; $ yes and $1 notin C;$ yes. So $1;$ doesn't belong to the intersection yes true. OK let me edit it
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:17
add a comment |
$begingroup$
You write,
Can we say that:
$x∈A$ and $x∉B$ and $x∈C$ OR
$x∈A$ and $x∉C$ and $x∈B$
In short, the answer is no, we cannot say that, because you've not told us what $x$ is. You've got to tell us what this $x$ is before you start talking about it.
To show that sets $S$ and $T$ are equal, a typical strategy is to show $S subseteq T$ and $Tsubseteq S$. In order to do this, you're going to let $x$ be some arbitrary member of $S$ and show that this implies that it is in $T$, and then let $y$ be some arbitrary member of $T$ and show that this implies that it is in $S$. Once that's done, you've won.
In your example, $Asetminus (Bcap C)$ is your $S$. So begin by explicitly writing:
"Let $xin Asetminus (Bcap C)$."
Now what do we know about $x$? Well, if $xin Asetminus (Bcap C)$ then $xin A$ but $xnotin (Bcap C)$. Unpacking the second part of that last sentence, if $xnotin (Bcap C)$ then either $xnotin B$ or $xnotin C$—otherwise $x$ would be in their intersection. So this gives us two cases: either $xnotin B$ or $xnotin B$. Suppose that $xnotin B$. We know that $xin A$ so this means that $xin Asetminus B$, and so $xin (Asetminus B)cup (Asetminus C)$. In the second case, we know that $xin A$ and $xnotin C$, which means $xin Asetminus C$, which further implies that $xin (Asetminus B)cup (Asetminus C)$. Since these are the only two possible cases, we conclude that $xin (Asetminus B)cup (Asetminus C)$.
To go to the other direction, again, begin by explicitly writing
"Let $yin (Asetminus B) cup (Asetminus C)$"
And from there, determine what you can say about $y$. I'll let you finish the other direction.
$endgroup$
$begingroup$
I finished the other direction. Can you please just downgrade or tell me if it's right.
$endgroup$
– Hani Gotc
Dec 8 '13 at 2:48
add a comment |
$begingroup$
If $x in A setminus (B cap C ) $. then this occur
$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$
$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$
$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$
$endgroup$
1
$begingroup$
I think that $x notin Bcup C iff xnotin B ,wedge , xnotin C $, because $x notin Bcup C iff (Bcup C)^Ciff B^C cap C^C $
$endgroup$
– Voyager
Dec 8 '13 at 9:13
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'll guide you through one direction. You will need to figure the other out yourself.
Let $xin A setminus (Bcap C)$.
Then by the definition of $setminus$, $xin A$ and $xnotin Bcap C$. The latter statement can be translated into $lnot (x in B cap C)$.
By the definition of $Bcap C$, $xin Bcap C iff (x in B) land (xin C)$.
Thus we conclude that $lnot ((x in B) land (xin C))$.
By De Morgan's laws for logic, we can rewrite this as
$lnot(xin B)lorlnot(xin C)$.
Suppose that $lnot (xin B)$. Then since $xin A$, $xin Asetminus B$.
Suppose instead that $lnot (x in C)$. Then since $xin A$, $xin A setminus C$.
As one of these must be true, $(xin Asetminus B)lor (xin A setminus C)$.
By the definition of union, $xin (Asetminus B)cup(A setminus C)$.
$endgroup$
$begingroup$
excuse me if it's a silly a question. when we say that $xnotin Bcap C$ we are sure that it's in neither $B$ and $C$. is that right? or it $x$ might belong to one of the two sets
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:06
$begingroup$
Yes I think i get it. I saw many explanations, but it was so clear. your explanation is great
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:11
1
$begingroup$
@HaniGotc, let $B = {1,2}$ and let $C = {2,3}$. Is $1in B$? Is $1in C$? Is $1in Bcap C$? Is it true that $1notin Bcap C$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:13
1
$begingroup$
Thanks for the compliment. Can you explain, perhaps by editing your question, how you would argue in the other direction, to show that if $xin (asetminus B)cup(A setminus C)$ then we can conclude also that $xin A setminus (Bcap C)$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:15
$begingroup$
$ 1 in B; $ yes and $1 notin C;$ yes. So $1;$ doesn't belong to the intersection yes true. OK let me edit it
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:17
add a comment |
$begingroup$
I'll guide you through one direction. You will need to figure the other out yourself.
Let $xin A setminus (Bcap C)$.
Then by the definition of $setminus$, $xin A$ and $xnotin Bcap C$. The latter statement can be translated into $lnot (x in B cap C)$.
By the definition of $Bcap C$, $xin Bcap C iff (x in B) land (xin C)$.
Thus we conclude that $lnot ((x in B) land (xin C))$.
By De Morgan's laws for logic, we can rewrite this as
$lnot(xin B)lorlnot(xin C)$.
Suppose that $lnot (xin B)$. Then since $xin A$, $xin Asetminus B$.
Suppose instead that $lnot (x in C)$. Then since $xin A$, $xin A setminus C$.
As one of these must be true, $(xin Asetminus B)lor (xin A setminus C)$.
By the definition of union, $xin (Asetminus B)cup(A setminus C)$.
$endgroup$
$begingroup$
excuse me if it's a silly a question. when we say that $xnotin Bcap C$ we are sure that it's in neither $B$ and $C$. is that right? or it $x$ might belong to one of the two sets
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:06
$begingroup$
Yes I think i get it. I saw many explanations, but it was so clear. your explanation is great
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:11
1
$begingroup$
@HaniGotc, let $B = {1,2}$ and let $C = {2,3}$. Is $1in B$? Is $1in C$? Is $1in Bcap C$? Is it true that $1notin Bcap C$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:13
1
$begingroup$
Thanks for the compliment. Can you explain, perhaps by editing your question, how you would argue in the other direction, to show that if $xin (asetminus B)cup(A setminus C)$ then we can conclude also that $xin A setminus (Bcap C)$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:15
$begingroup$
$ 1 in B; $ yes and $1 notin C;$ yes. So $1;$ doesn't belong to the intersection yes true. OK let me edit it
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:17
add a comment |
$begingroup$
I'll guide you through one direction. You will need to figure the other out yourself.
Let $xin A setminus (Bcap C)$.
Then by the definition of $setminus$, $xin A$ and $xnotin Bcap C$. The latter statement can be translated into $lnot (x in B cap C)$.
By the definition of $Bcap C$, $xin Bcap C iff (x in B) land (xin C)$.
Thus we conclude that $lnot ((x in B) land (xin C))$.
By De Morgan's laws for logic, we can rewrite this as
$lnot(xin B)lorlnot(xin C)$.
Suppose that $lnot (xin B)$. Then since $xin A$, $xin Asetminus B$.
Suppose instead that $lnot (x in C)$. Then since $xin A$, $xin A setminus C$.
As one of these must be true, $(xin Asetminus B)lor (xin A setminus C)$.
By the definition of union, $xin (Asetminus B)cup(A setminus C)$.
$endgroup$
I'll guide you through one direction. You will need to figure the other out yourself.
Let $xin A setminus (Bcap C)$.
Then by the definition of $setminus$, $xin A$ and $xnotin Bcap C$. The latter statement can be translated into $lnot (x in B cap C)$.
By the definition of $Bcap C$, $xin Bcap C iff (x in B) land (xin C)$.
Thus we conclude that $lnot ((x in B) land (xin C))$.
By De Morgan's laws for logic, we can rewrite this as
$lnot(xin B)lorlnot(xin C)$.
Suppose that $lnot (xin B)$. Then since $xin A$, $xin Asetminus B$.
Suppose instead that $lnot (x in C)$. Then since $xin A$, $xin A setminus C$.
As one of these must be true, $(xin Asetminus B)lor (xin A setminus C)$.
By the definition of union, $xin (Asetminus B)cup(A setminus C)$.
edited Jan 17 '17 at 16:06
Martin Sleziak
44.7k9117272
44.7k9117272
answered Dec 7 '13 at 23:59
dfeuerdfeuer
7,69032558
7,69032558
$begingroup$
excuse me if it's a silly a question. when we say that $xnotin Bcap C$ we are sure that it's in neither $B$ and $C$. is that right? or it $x$ might belong to one of the two sets
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:06
$begingroup$
Yes I think i get it. I saw many explanations, but it was so clear. your explanation is great
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:11
1
$begingroup$
@HaniGotc, let $B = {1,2}$ and let $C = {2,3}$. Is $1in B$? Is $1in C$? Is $1in Bcap C$? Is it true that $1notin Bcap C$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:13
1
$begingroup$
Thanks for the compliment. Can you explain, perhaps by editing your question, how you would argue in the other direction, to show that if $xin (asetminus B)cup(A setminus C)$ then we can conclude also that $xin A setminus (Bcap C)$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:15
$begingroup$
$ 1 in B; $ yes and $1 notin C;$ yes. So $1;$ doesn't belong to the intersection yes true. OK let me edit it
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:17
add a comment |
$begingroup$
excuse me if it's a silly a question. when we say that $xnotin Bcap C$ we are sure that it's in neither $B$ and $C$. is that right? or it $x$ might belong to one of the two sets
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:06
$begingroup$
Yes I think i get it. I saw many explanations, but it was so clear. your explanation is great
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:11
1
$begingroup$
@HaniGotc, let $B = {1,2}$ and let $C = {2,3}$. Is $1in B$? Is $1in C$? Is $1in Bcap C$? Is it true that $1notin Bcap C$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:13
1
$begingroup$
Thanks for the compliment. Can you explain, perhaps by editing your question, how you would argue in the other direction, to show that if $xin (asetminus B)cup(A setminus C)$ then we can conclude also that $xin A setminus (Bcap C)$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:15
$begingroup$
$ 1 in B; $ yes and $1 notin C;$ yes. So $1;$ doesn't belong to the intersection yes true. OK let me edit it
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:17
$begingroup$
excuse me if it's a silly a question. when we say that $xnotin Bcap C$ we are sure that it's in neither $B$ and $C$. is that right? or it $x$ might belong to one of the two sets
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:06
$begingroup$
excuse me if it's a silly a question. when we say that $xnotin Bcap C$ we are sure that it's in neither $B$ and $C$. is that right? or it $x$ might belong to one of the two sets
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:06
$begingroup$
Yes I think i get it. I saw many explanations, but it was so clear. your explanation is great
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:11
$begingroup$
Yes I think i get it. I saw many explanations, but it was so clear. your explanation is great
$endgroup$
– Hani Gotc
Dec 8 '13 at 0:11
1
1
$begingroup$
@HaniGotc, let $B = {1,2}$ and let $C = {2,3}$. Is $1in B$? Is $1in C$? Is $1in Bcap C$? Is it true that $1notin Bcap C$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:13
$begingroup$
@HaniGotc, let $B = {1,2}$ and let $C = {2,3}$. Is $1in B$? Is $1in C$? Is $1in Bcap C$? Is it true that $1notin Bcap C$?
$endgroup$
– dfeuer
Dec 8 '13 at 0:13
1
1
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Thanks for the compliment. Can you explain, perhaps by editing your question, how you would argue in the other direction, to show that if $xin (asetminus B)cup(A setminus C)$ then we can conclude also that $xin A setminus (Bcap C)$?
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– dfeuer
Dec 8 '13 at 0:15
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Thanks for the compliment. Can you explain, perhaps by editing your question, how you would argue in the other direction, to show that if $xin (asetminus B)cup(A setminus C)$ then we can conclude also that $xin A setminus (Bcap C)$?
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– dfeuer
Dec 8 '13 at 0:15
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$ 1 in B; $ yes and $1 notin C;$ yes. So $1;$ doesn't belong to the intersection yes true. OK let me edit it
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– Hani Gotc
Dec 8 '13 at 0:17
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$ 1 in B; $ yes and $1 notin C;$ yes. So $1;$ doesn't belong to the intersection yes true. OK let me edit it
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– Hani Gotc
Dec 8 '13 at 0:17
add a comment |
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You write,
Can we say that:
$x∈A$ and $x∉B$ and $x∈C$ OR
$x∈A$ and $x∉C$ and $x∈B$
In short, the answer is no, we cannot say that, because you've not told us what $x$ is. You've got to tell us what this $x$ is before you start talking about it.
To show that sets $S$ and $T$ are equal, a typical strategy is to show $S subseteq T$ and $Tsubseteq S$. In order to do this, you're going to let $x$ be some arbitrary member of $S$ and show that this implies that it is in $T$, and then let $y$ be some arbitrary member of $T$ and show that this implies that it is in $S$. Once that's done, you've won.
In your example, $Asetminus (Bcap C)$ is your $S$. So begin by explicitly writing:
"Let $xin Asetminus (Bcap C)$."
Now what do we know about $x$? Well, if $xin Asetminus (Bcap C)$ then $xin A$ but $xnotin (Bcap C)$. Unpacking the second part of that last sentence, if $xnotin (Bcap C)$ then either $xnotin B$ or $xnotin C$—otherwise $x$ would be in their intersection. So this gives us two cases: either $xnotin B$ or $xnotin B$. Suppose that $xnotin B$. We know that $xin A$ so this means that $xin Asetminus B$, and so $xin (Asetminus B)cup (Asetminus C)$. In the second case, we know that $xin A$ and $xnotin C$, which means $xin Asetminus C$, which further implies that $xin (Asetminus B)cup (Asetminus C)$. Since these are the only two possible cases, we conclude that $xin (Asetminus B)cup (Asetminus C)$.
To go to the other direction, again, begin by explicitly writing
"Let $yin (Asetminus B) cup (Asetminus C)$"
And from there, determine what you can say about $y$. I'll let you finish the other direction.
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I finished the other direction. Can you please just downgrade or tell me if it's right.
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– Hani Gotc
Dec 8 '13 at 2:48
add a comment |
$begingroup$
You write,
Can we say that:
$x∈A$ and $x∉B$ and $x∈C$ OR
$x∈A$ and $x∉C$ and $x∈B$
In short, the answer is no, we cannot say that, because you've not told us what $x$ is. You've got to tell us what this $x$ is before you start talking about it.
To show that sets $S$ and $T$ are equal, a typical strategy is to show $S subseteq T$ and $Tsubseteq S$. In order to do this, you're going to let $x$ be some arbitrary member of $S$ and show that this implies that it is in $T$, and then let $y$ be some arbitrary member of $T$ and show that this implies that it is in $S$. Once that's done, you've won.
In your example, $Asetminus (Bcap C)$ is your $S$. So begin by explicitly writing:
"Let $xin Asetminus (Bcap C)$."
Now what do we know about $x$? Well, if $xin Asetminus (Bcap C)$ then $xin A$ but $xnotin (Bcap C)$. Unpacking the second part of that last sentence, if $xnotin (Bcap C)$ then either $xnotin B$ or $xnotin C$—otherwise $x$ would be in their intersection. So this gives us two cases: either $xnotin B$ or $xnotin B$. Suppose that $xnotin B$. We know that $xin A$ so this means that $xin Asetminus B$, and so $xin (Asetminus B)cup (Asetminus C)$. In the second case, we know that $xin A$ and $xnotin C$, which means $xin Asetminus C$, which further implies that $xin (Asetminus B)cup (Asetminus C)$. Since these are the only two possible cases, we conclude that $xin (Asetminus B)cup (Asetminus C)$.
To go to the other direction, again, begin by explicitly writing
"Let $yin (Asetminus B) cup (Asetminus C)$"
And from there, determine what you can say about $y$. I'll let you finish the other direction.
$endgroup$
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I finished the other direction. Can you please just downgrade or tell me if it's right.
$endgroup$
– Hani Gotc
Dec 8 '13 at 2:48
add a comment |
$begingroup$
You write,
Can we say that:
$x∈A$ and $x∉B$ and $x∈C$ OR
$x∈A$ and $x∉C$ and $x∈B$
In short, the answer is no, we cannot say that, because you've not told us what $x$ is. You've got to tell us what this $x$ is before you start talking about it.
To show that sets $S$ and $T$ are equal, a typical strategy is to show $S subseteq T$ and $Tsubseteq S$. In order to do this, you're going to let $x$ be some arbitrary member of $S$ and show that this implies that it is in $T$, and then let $y$ be some arbitrary member of $T$ and show that this implies that it is in $S$. Once that's done, you've won.
In your example, $Asetminus (Bcap C)$ is your $S$. So begin by explicitly writing:
"Let $xin Asetminus (Bcap C)$."
Now what do we know about $x$? Well, if $xin Asetminus (Bcap C)$ then $xin A$ but $xnotin (Bcap C)$. Unpacking the second part of that last sentence, if $xnotin (Bcap C)$ then either $xnotin B$ or $xnotin C$—otherwise $x$ would be in their intersection. So this gives us two cases: either $xnotin B$ or $xnotin B$. Suppose that $xnotin B$. We know that $xin A$ so this means that $xin Asetminus B$, and so $xin (Asetminus B)cup (Asetminus C)$. In the second case, we know that $xin A$ and $xnotin C$, which means $xin Asetminus C$, which further implies that $xin (Asetminus B)cup (Asetminus C)$. Since these are the only two possible cases, we conclude that $xin (Asetminus B)cup (Asetminus C)$.
To go to the other direction, again, begin by explicitly writing
"Let $yin (Asetminus B) cup (Asetminus C)$"
And from there, determine what you can say about $y$. I'll let you finish the other direction.
$endgroup$
You write,
Can we say that:
$x∈A$ and $x∉B$ and $x∈C$ OR
$x∈A$ and $x∉C$ and $x∈B$
In short, the answer is no, we cannot say that, because you've not told us what $x$ is. You've got to tell us what this $x$ is before you start talking about it.
To show that sets $S$ and $T$ are equal, a typical strategy is to show $S subseteq T$ and $Tsubseteq S$. In order to do this, you're going to let $x$ be some arbitrary member of $S$ and show that this implies that it is in $T$, and then let $y$ be some arbitrary member of $T$ and show that this implies that it is in $S$. Once that's done, you've won.
In your example, $Asetminus (Bcap C)$ is your $S$. So begin by explicitly writing:
"Let $xin Asetminus (Bcap C)$."
Now what do we know about $x$? Well, if $xin Asetminus (Bcap C)$ then $xin A$ but $xnotin (Bcap C)$. Unpacking the second part of that last sentence, if $xnotin (Bcap C)$ then either $xnotin B$ or $xnotin C$—otherwise $x$ would be in their intersection. So this gives us two cases: either $xnotin B$ or $xnotin B$. Suppose that $xnotin B$. We know that $xin A$ so this means that $xin Asetminus B$, and so $xin (Asetminus B)cup (Asetminus C)$. In the second case, we know that $xin A$ and $xnotin C$, which means $xin Asetminus C$, which further implies that $xin (Asetminus B)cup (Asetminus C)$. Since these are the only two possible cases, we conclude that $xin (Asetminus B)cup (Asetminus C)$.
To go to the other direction, again, begin by explicitly writing
"Let $yin (Asetminus B) cup (Asetminus C)$"
And from there, determine what you can say about $y$. I'll let you finish the other direction.
answered Dec 8 '13 at 0:28
crfcrf
3,45612044
3,45612044
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I finished the other direction. Can you please just downgrade or tell me if it's right.
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– Hani Gotc
Dec 8 '13 at 2:48
add a comment |
$begingroup$
I finished the other direction. Can you please just downgrade or tell me if it's right.
$endgroup$
– Hani Gotc
Dec 8 '13 at 2:48
$begingroup$
I finished the other direction. Can you please just downgrade or tell me if it's right.
$endgroup$
– Hani Gotc
Dec 8 '13 at 2:48
$begingroup$
I finished the other direction. Can you please just downgrade or tell me if it's right.
$endgroup$
– Hani Gotc
Dec 8 '13 at 2:48
add a comment |
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If $x in A setminus (B cap C ) $. then this occur
$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$
$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$
$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$
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1
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I think that $x notin Bcup C iff xnotin B ,wedge , xnotin C $, because $x notin Bcup C iff (Bcup C)^Ciff B^C cap C^C $
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– Voyager
Dec 8 '13 at 9:13
add a comment |
$begingroup$
If $x in A setminus (B cap C ) $. then this occur
$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$
$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$
$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$
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1
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I think that $x notin Bcup C iff xnotin B ,wedge , xnotin C $, because $x notin Bcup C iff (Bcup C)^Ciff B^C cap C^C $
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– Voyager
Dec 8 '13 at 9:13
add a comment |
$begingroup$
If $x in A setminus (B cap C ) $. then this occur
$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$
$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$
$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$
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If $x in A setminus (B cap C ) $. then this occur
$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$
$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$
$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$
answered Dec 7 '13 at 23:49
ILoveMathILoveMath
1
1
1
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I think that $x notin Bcup C iff xnotin B ,wedge , xnotin C $, because $x notin Bcup C iff (Bcup C)^Ciff B^C cap C^C $
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– Voyager
Dec 8 '13 at 9:13
add a comment |
1
$begingroup$
I think that $x notin Bcup C iff xnotin B ,wedge , xnotin C $, because $x notin Bcup C iff (Bcup C)^Ciff B^C cap C^C $
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– Voyager
Dec 8 '13 at 9:13
1
1
$begingroup$
I think that $x notin Bcup C iff xnotin B ,wedge , xnotin C $, because $x notin Bcup C iff (Bcup C)^Ciff B^C cap C^C $
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– Voyager
Dec 8 '13 at 9:13
$begingroup$
I think that $x notin Bcup C iff xnotin B ,wedge , xnotin C $, because $x notin Bcup C iff (Bcup C)^Ciff B^C cap C^C $
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– Voyager
Dec 8 '13 at 9:13
add a comment |
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That is not very clear English. Perhaps you could parenthesize it or something?
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– dfeuer
Dec 7 '13 at 23:49
1
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But really, what you generally want to do is to show that for all $x$, if $xin Asetminus (Bcap C)$ then $xin $(Asetminus B)cup (A setminus C)$ and the other way around.
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– dfeuer
Dec 7 '13 at 23:50
1
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$yin Aland (ynotin Blor ynotin C)$ is not equivalent to $yin Aland ynotin (Bcup C)$.
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– dfeuer
Dec 8 '13 at 2:52
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possible duplicate of proof of $ A - left (B cap C right)= left (A - B right) cup left (A - C right)$
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– user91500
Dec 8 '13 at 5:25