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Prove-that-If-A-B-and-C-are-subset-of-the-same-universal-set-then-A-B-A-C-A-B-C-




Question Number 7440 by Tawakalitu. last updated on 29/Aug/16
Prove that  If  A, B and C are subset of the same universal set  then (A − B) ∩ (A − C) = A − (B − C)
$${Prove}\:{that} \\ $$$${If}\:\:{A},\:{B}\:{and}\:{C}\:{are}\:{subset}\:{of}\:{the}\:{same}\:{universal}\:{set} \\ $$$${then}\:\left({A}\:−\:{B}\right)\:\cap\:\left({A}\:−\:{C}\right)\:=\:{A}\:−\:\left({B}\:−\:{C}\right) \\ $$
Commented by Yozzia last updated on 29/Aug/16
Let A={1,2,3,4}, B={2,3,4,5}, C={3,4,5,6}.  ⇒A−B={1}, A−C={1,2}  ⇒(A−B)∩(A−C)={1}.  B−C={2}  ⇒A−(B−C)={1,3,4}≠{1}=(A−B)∩(A−C).  The correct identity is (A−B)∩(A−C)=A−(B∪C).
$${Let}\:{A}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right\},\:{B}=\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\},\:{C}=\left\{\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right\}. \\ $$$$\Rightarrow{A}−{B}=\left\{\mathrm{1}\right\},\:{A}−{C}=\left\{\mathrm{1},\mathrm{2}\right\} \\ $$$$\Rightarrow\left({A}−{B}\right)\cap\left({A}−{C}\right)=\left\{\mathrm{1}\right\}. \\ $$$${B}−{C}=\left\{\mathrm{2}\right\} \\ $$$$\Rightarrow{A}−\left({B}−{C}\right)=\left\{\mathrm{1},\mathrm{3},\mathrm{4}\right\}\neq\left\{\mathrm{1}\right\}=\left({A}−{B}\right)\cap\left({A}−{C}\right). \\ $$$${The}\:{correct}\:{identity}\:{is}\:\left({A}−{B}\right)\cap\left({A}−{C}\right)={A}−\left({B}\cup{C}\right). \\ $$
Commented by Rasheed Soomro last updated on 29/Aug/16
Can we say  (A−B)∩(A−C)=A−(B∪C) as  one of the two de margan′s laws? Other law may be                    (A−B)∪(A−C)=A−(B∩C)  ?
$${Can}\:{we}\:{say}\:\:\left({A}−{B}\right)\cap\left({A}−{C}\right)={A}−\left({B}\cup{C}\right)\:{as} \\ $$$${one}\:{of}\:{the}\:{two}\:{de}\:{margan}'{s}\:{laws}?\:{Other}\:{law}\:{may}\:{be} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({A}−{B}\right)\cup\left({A}−{C}\right)={A}−\left({B}\cap{C}\right)\:\:? \\ $$
Commented by Yozzia last updated on 29/Aug/16
I only know (∼a)∧(∼b)=∼(a∨b) and (∼a)∨(∼b)=∼(a∧b)  as De Morgan′s laws in logic. The results you  speak of may have no one′s name attached  to them, as they stem from definitions  and may not be very frequently used.  My text on abstract algebra does not  associate anyone′s name to these results  involving the set difference.
$${I}\:{only}\:{know}\:\left(\sim{a}\right)\wedge\left(\sim{b}\right)=\sim\left({a}\vee{b}\right)\:{and}\:\left(\sim{a}\right)\vee\left(\sim{b}\right)=\sim\left({a}\wedge{b}\right) \\ $$$${as}\:{De}\:{Morgan}'{s}\:{laws}\:{in}\:{logic}.\:{The}\:{results}\:{you} \\ $$$${speak}\:{of}\:{may}\:{have}\:{no}\:{one}'{s}\:{name}\:{attached} \\ $$$${to}\:{them},\:{as}\:{they}\:{stem}\:{from}\:{definitions} \\ $$$${and}\:{may}\:{not}\:{be}\:{very}\:{frequently}\:{used}. \\ $$$${My}\:{text}\:{on}\:{abstract}\:{algebra}\:{does}\:{not} \\ $$$${associate}\:{anyone}'{s}\:{name}\:{to}\:{these}\:{results} \\ $$$${involving}\:{the}\:{set}\:{difference}. \\ $$
Commented by Tawakalitu. last updated on 29/Aug/16
Thank you sir, i understand.
$${Thank}\:{you}\:{sir},\:{i}\:{understand}. \\ $$
Commented by Rasheed Soomro last updated on 30/Aug/16
(A−B)∩(A−C)=A−(B∪C)  (A−B)∪(A−C)=A−(B∩C)  Anyway the above laws laws have close  connection with de-margan laws.  If we let A be universal set U then  (A−B)∩(A−C)=A−(B∪C)⇒(U−B)∩(U−C)=U−(B∪C)  ⇒B ′∩C ′=(B∪C)′⇒(B∪C)′=B ′∩C ′  Similarly (A−B)∪(A−C)=A−(B∩C)⇒(B∩C)′=B ′∪C ′  It seems that the above laws are general and  de-margan laws are special cases! When B,C⊆A and  and A is recognized as U(Universal Set).    I think I read anywhere in Set theory that the  above laws  are “De margan laws ”.Although I cannot confirm it   now.
$$\left({A}−{B}\right)\cap\left({A}−{C}\right)={A}−\left({B}\cup{C}\right) \\ $$$$\left({A}−{B}\right)\cup\left({A}−{C}\right)={A}−\left({B}\cap{C}\right) \\ $$$${Anyway}\:{the}\:{above}\:{laws}\:{laws}\:{have}\:{close} \\ $$$${connection}\:{with}\:{de}-{margan}\:{laws}. \\ $$$${If}\:{we}\:{let}\:{A}\:{be}\:{universal}\:{set}\:\mathbb{U}\:{then} \\ $$$$\left({A}−{B}\right)\cap\left({A}−{C}\right)={A}−\left({B}\cup{C}\right)\Rightarrow\left(\mathbb{U}−{B}\right)\cap\left(\mathbb{U}−{C}\right)=\mathbb{U}−\left({B}\cup{C}\right) \\ $$$$\Rightarrow{B}\:'\cap{C}\:'=\left({B}\cup{C}\right)'\Rightarrow\left({B}\cup{C}\right)'={B}\:'\cap{C}\:' \\ $$$${Similarly}\:\left({A}−{B}\right)\cup\left({A}−{C}\right)={A}−\left({B}\cap{C}\right)\Rightarrow\left({B}\cap{C}\right)'={B}\:'\cup{C}\:' \\ $$$${It}\:{seems}\:{that}\:{the}\:{above}\:{laws}\:{are}\:{general}\:{and} \\ $$$${de}-{margan}\:{laws}\:{are}\:{special}\:{cases}!\:{When}\:{B},{C}\subseteq{A}\:{and} \\ $$$${and}\:{A}\:{is}\:{recognized}\:{as}\:\mathbb{U}\left({Universal}\:{Set}\right). \\ $$$$ \\ $$$${I}\:{think}\:{I}\:{read}\:{anywhere}\:{in}\:{Set}\:{theory}\:{that}\:{the}\:\:{above}\:{laws} \\ $$$${are}\:“{De}\:{margan}\:{laws}\:''.{Although}\:{I}\:{cannot}\:{confirm}\:{it}\: \\ $$$${now}. \\ $$
Commented by Yozzia last updated on 30/Aug/16
Hmm...interesting. Thanks for sharing!
$${Hmm}…{interesting}.\:{Thanks}\:{for}\:{sharing}! \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 31/Aug/16
I really appreciate your effort sirs
$${I}\:{really}\:{appreciate}\:{your}\:{effort}\:{sirs} \\ $$

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