de morgan law $Asetminus (B cap C) = (Asetminus B) cup (Asetminus C) $












4












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















4












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














4












4








4


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • $begingroup$
    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










3 Answers
3






active

oldest

votes


















1












$begingroup$

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, $xin (Asetminus B)cup(A setminus C)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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





















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

If $x in A setminus (B cap C ) $. then this occur



$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$



$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$



$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$






share|cite|improve this answer









$endgroup$









  • 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











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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, $xin (Asetminus B)cup(A setminus C)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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, $xin (Asetminus B)cup(A setminus C)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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, $xin (Asetminus B)cup(A setminus C)$.






share|cite|improve this answer











$endgroup$



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, $xin (Asetminus B)cup(A setminus C)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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




















  • $begingroup$
    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




$begingroup$
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




$begingroup$
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






$begingroup$
$ 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













1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '13 at 0:28









crfcrf

3,45612044




3,45612044












  • $begingroup$
    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




$begingroup$
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











0












$begingroup$

If $x in A setminus (B cap C ) $. then this occur



$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$



$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$



$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$






share|cite|improve this answer









$endgroup$









  • 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
















0












$begingroup$

If $x in A setminus (B cap C ) $. then this occur



$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$



$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$



$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$






share|cite|improve this answer









$endgroup$









  • 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














0












0








0





$begingroup$

If $x in A setminus (B cap C ) $. then this occur



$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$



$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$



$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$






share|cite|improve this answer









$endgroup$



If $x in A setminus (B cap C ) $. then this occur



$$ iff x notin Bcup C iff x notin B ; ; text{or} ; ; x notin C iff$$



$$ iff x in A setminus B ; ; text{or} ; ; x in A setminus C iff x in(A setminus B ) cup (A setminus C )$$



$$ therefore A setminus (B cap C ) = (A setminus B ) cup (A setminus C ) $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '13 at 23:49









ILoveMathILoveMath

1




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 $
    $endgroup$
    – Voyager
    Dec 8 '13 at 9:13














  • 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








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 $
$endgroup$
– 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 $
$endgroup$
– Voyager
Dec 8 '13 at 9:13


















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