Question #252170

Use algebra of sets to prove that,

 [(šµ āˆ’ š“)' ∩ š“] āˆ’ š“' = š“


1
Expert's answer
2021-10-19T08:02:58-0400

Solution:

LHS=[(šµāˆ’š“)ā€²āˆ©š“]āˆ’š“ā€²=[(šµā€²āˆ’š“ā€²)āˆ©š“]āˆ’š“ā€²LHS=[(šµ āˆ’ š“)' ∩ š“] āˆ’ š“' \\=[(šµ' āˆ’ š“')∩ š“] āˆ’ š“'

=[(šµā€²āˆ’š“ā€²)āˆ©š“]∩[š“ā€²]′  [∵Pāˆ’Q=P∩Q′]=[(šµ' āˆ’ š“') ∩ š“] ∩ [š“']' \ \ [\because P-Q=P∩Q']

=[(šµā€²āˆ’š“ā€²)āˆ©š“]∩A=(šµā€²āˆ’š“ā€²)āˆ©š“āˆ©š“=(šµā€²āˆ’š“ā€²)āˆ©š“=[(šµ' āˆ’ š“') ∩ š“] ∩ A \\=(šµ' āˆ’ š“') ∩ š“āˆ© š“ \\=(šµ' āˆ’ š“') ∩ š“

=(šµāˆ’š“)ā€²āˆ©š“=Aāˆ’(Bāˆ’A)=A=RHS\\=(šµ āˆ’ š“)' ∩ š“ \\=A-(B-A) \\=A \\=RHS

Hence, proved.


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