Simplifying Boolean Algebra

Question

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.

Belongs on http://math.stackexchange.com/ ? – Shmiddty Feb 21 '13 at 18:45 — Shmiddty, Feb 21 '13 at 18:45

score 3 · Accepted Answer · answered Feb 21 '13 at 18:53

  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

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