Answer to Question #253044 in Discrete Mathematics for Alina

Question #253044

Question#1

Use algebra of sets to prove the following:

i. (𝐵 − 𝐴) ∪ (𝐶 − 𝐴) = (𝐵 ∪ 𝐶) − 𝐴

ii. [(𝐵 − 𝐴) c∩ 𝐴] − 𝐴c= 𝐴

iii. (𝐴𝑐 ∪ 𝐵)c ∩ Ac= ∅


Question#2

Use Mathematical induction to prove the following generalization of one

of De Morgan’s law: ⋃nj=1 𝐴j= ⋃nj=1 Aj

Question#3

Prove that (𝐴 ∪ 𝐵 ∪ 𝐶) ′ = 𝐴′ ∩ 𝐵 ′ ∩ 𝐶 ′



1
Expert's answer
2021-10-19T10:35:41-0400


i. (BC)A=(BC)A=[distribution low]=(B\cup C)-A=(B\cap C)\cap \overline {A}=[distribution\space low]=

(BA)(CA)=(BA)(CA)(B\cap \overline {A})\cup(C\cap \overline{A})=(B-A)\cup(C-A)

ii 

[(BA)cA]Ac=[(BAc)cA](Ac)c=[Morgan low,double complement low]=(Bc(Ac)c)A)A=(BcA)(AA)=[idempotent law]=(BcA)A=A. becauseA(BcA)[(B-A)^c\cap A]-A^c=[(B\cap A^c)^c\cap A]\cap (A^c)^c=[Morgan \space low, double \space complement \space low]=(B^c\cup(A^c)^c)\cap A)\cap A=(B^c\cup A)\cap(A\cap A)=[idempotent \space law]=(B^c\cup A)\cap A=A.\space because A\subset (B^c\cup A)

iii.

(AcB)cAc=[Morgan low]=((Ac)cBcAc=[double complement]=ABcAc=[commutativity and associativity of ]=(AAc)B=[properties of complement]=B=(A^c\cup B)^c\cap A^c=[Morgan\space low]=((A^c)^c\cap B^c\cap A^c=[double\space complement]=A\cap B^c\cap A^c=[commutativity\space and \space associativity \space of \space \cap]=(A\cap A^c)\cap B=[properties \space of\space complement]=\empty\cap B=\empty

Question 2/

1) Basis of induction, n=2

(j=12Aj)c=(A1A2)c=[Morgan law]=A1cA2c=j=12Ajc\left(\bigcup_{j=1}^{2}A_j\right)^c=(A_1\cup A_2)^c=[Morgan\space law]=A_1^c\cap A_2^c=\bigcap_{j=1}^{2}A_j^c

2) Induction step, let the statement is true for n=k, i.e.

(j=1kAj)c=\left(\bigcup_{j=1}^{k}A_j\right)^c= j=1kAjc\bigcap_{j=1}^{k}A_j^c

Considerthe case n=k+1:

(j=1k+1Aj)c=(j=1kAjAk+1)c=[Morgan law]=\left(\bigcup_{j=1}^{k+1}A_j\right)^c=\left(\bigcup_{j=1}^{k}A_j\cup A_{k+1}\right)^c=[Morgan\space law]=

(j=1kAj)cAk+1c=[induction hypothesis]\left(\bigcup_{j=1}^{k}A_j\right)^c\cap A_{k+1}^c=[induction\space hypothesis] =j=1kAjcAk+1c=j=1k+1Ajc\bigcap_{j=1}^{k}A_j^c\cap A_{k+1}^c=\bigcap_{j=1}^{k+1}A_j^c , so statement is true for n=k+1, induction step is verified, therefore the statement is proved by math inductionmethod.

Question 3.

Let A1=A,A2=B,A3=CLet \space A_1=A,A_2=B,A_3=C

Then (ABC)c=(A\cup B\cup C)^c= (j=13Aj)c=[Question 2]=\left(\bigcup_{j=1}^{3}A_j\right)^c=[Question\space 2]= j=13Ajc=(A1 A3)=ABC\bigcap_{j=1}^{3}A_j^c=(A_1\ \cap A_3)=A\cap B\cap C




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