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