i. (B∪C)−A=(B∩C)∩A=[distribution low]=
(B∩A)∪(C∩A)=(B−A)∪(C−A)
ii
[(B−A)c∩A]−Ac=[(B∩Ac)c∩A]∩(Ac)c=[Morgan low,double complement low]=(Bc∪(Ac)c)∩A)∩A=(Bc∪A)∩(A∩A)=[idempotent law]=(Bc∪A)∩A=A. becauseA⊂(Bc∪A)
iii.
(Ac∪B)c∩Ac=[Morgan low]=((Ac)c∩Bc∩Ac=[double complement]=A∩Bc∩Ac=[commutativity and associativity of ∩]=(A∩Ac)∩B=[properties of complement]=∅∩B=∅
Question 2/
1) Basis of induction, n=2
(⋃j=12Aj)c=(A1∪A2)c=[Morgan law]=A1c∩A2c=⋂j=12Ajc
2) Induction step, let the statement is true for n=k, i.e.
(⋃j=1kAj)c= ⋂j=1kAjc
Considerthe case n=k+1:
(⋃j=1k+1Aj)c=(⋃j=1kAj∪Ak+1)c=[Morgan law]=
(⋃j=1kAj)c∩Ak+1c=[induction hypothesis] =⋂j=1kAjc∩Ak+1c=⋂j=1k+1Ajc , 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=C
Then (A∪B∪C)c= (⋃j=13Aj)c=[Question 2]= ⋂j=13Ajc=(A1 ∩A3)=A∩B∩C
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