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Prove the following Boolean relations: ...

Prove the following Boolean relations:
`AB+bar(B)C+CA=AB+bar(B)C`

Text Solution

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LHS = `AB + bar(B)C + CA = AB + bar(B)C + 1.CA `
`=AB = bar(B)C +(B + bar(B)) CA [because B + bar(B) = 1]`
=`AB + bar(B)C + ABC + Abar(B)C`
`=(AB + ABC) + (bar(B)C+Abar(B)C)`
`= AB(1 + C) + (1 + A ) bar(B)C`
`= AB. 1 + 1. bar(B)C = AB + bar(B)C = RHS`
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