Use algebra of sets to prove that,
[(šµ ā š“)' ā© š“] ā š“' = š“
Solution:
LHS=[(šµāš“)ā²ā©š“]āš“ā²=[(šµā²āš“ā²)ā©š“]āš“ā²LHS=[(šµ ā š“)' ā© š“] ā š“' \\=[(šµ' ā š“')ā© š“] ā š“'LHS=[(BāA)ā²ā©A]āAā²=[(Bā²āAā²)ā©A]āAā²
=[(šµā²āš“ā²)ā©š“]ā©[š“ā²]ā² [āµPāQ=Pā©Qā²]=[(šµ' ā š“') ā© š“] ā© [š“']' \ \ [\because P-Q=Pā©Q']=[(Bā²āAā²)ā©A]ā©[Aā²]ā² [āµPāQ=Pā©Qā²]
=[(šµā²āš“ā²)ā©š“]ā©A=(šµā²āš“ā²)ā©š“ā©š“=(šµā²āš“ā²)ā©š“=[(šµ' ā š“') ā© š“] ā© A \\=(šµ' ā š“') ā© š“ā© š“ \\=(šµ' ā š“') ā© š“=[(Bā²āAā²)ā©A]ā©A=(Bā²āAā²)ā©Aā©A=(Bā²āAā²)ā©A
=(šµāš“)ā²ā©š“=Aā(BāA)=A=RHS\\=(šµ ā š“)' ā© š“ \\=A-(B-A) \\=A \\=RHS=(BāA)ā²ā©A=Aā(BāA)=A=RHS
Hence, proved.
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