BC + !A!B + !A!C
= BC + !A(!B + !C) distributive law
= BC + !A!(BC) De Morgan's law
= (A + !A)BC + !A!(BC) identity and x OR !x = true
= A(BC) + !A(BC) + !A!(BC) distributive law
= ABC + !A((BC) + !(BC)) distributive law
= ABC + !A x OR !x = true and identity
Simplifying Boolean Algebra [closed]
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10-03-2022 - |
Pergunta
I am trying to prove that BC + !A!B + !A!C = ABC +!A
I have attempted using De Morgan laws, and substituting X for !A!B
and Y for !A!C
, however I made no headway in this.
I"ve alos tried gruopoing the A's like so, !A(!B+!C)
, however again I couldn't get anywhere. If anyone could point me in the right direction, help me solve, show me a tool that can do it, etc. I'd be grateful.
Solução
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