# Reduce using Boolean Algebra [closed]

How to reduce $(\bar B.C + B)$ to $(B + C)$?

Idempotent Law

A * A = A

A + A = A

Associative Law

(A * B) * C = A * (B * C)

(A + B) + C = A + (B + C)

Commutative Law

A * B = B * A

A + B = B + A

Distributive Law

A * (B + C) = A * B + A * C

A + (B * C) = (A + B) * (A + C)

Identity Law

A * 0 = 0 A * 1 = A

A + 1 = 1 A + 0 = A

Complement Law

A * ~A = 0

A + ~A = 1

Involution Law

~(~A) = A

DeMorgan's Law

~(A * B) = ~A + ~B

~(A + B) = ~A * ~B

## closed as off-topic by Bimpelrekkie, Leon Heller, Daniel Grillo, Dave Tweed♦Oct 5 '16 at 13:45

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• I'm voting to close this question because you show no effort in solving this assignment other than copy-pasting all the Laws concerning Boolean logic. – Bimpelrekkie Oct 4 '16 at 8:32
• I simplified Y = A'B'C' + A'B'C + A'BC'+AB'C+ABC'+ABC to A'(C' +B'C) + A(B'C+B), I'm literally at the last step but I just can't figure out why (B'C + B) can be reduced to (B + C) – JavaBeginner Oct 4 '16 at 8:35
• You only apply the laws but you're not using your brain, just switch it on and see what happens. – Bimpelrekkie Oct 4 '16 at 8:37
• I know the end result, I just need the proof – JavaBeginner Oct 4 '16 at 8:39
• Apply the distributive law, then the complement law and see what happens. The answer is literally staring you in the face. – crowie Oct 4 '16 at 8:58

$(\bar BC + B)$ gives this truth table.
Now, I am quite confident that you can see from this table why the $\bar B$ can be removed. Just think about it.
Alternatively, if you expand the original equation (otherwise know as applying Distributive Law), you get $(\bar B+B).(C+B)$, geddit yet?