0
\$\begingroup\$

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

\$\endgroup\$
  • 4
    \$\begingroup\$ 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. \$\endgroup\$ – Bimpelrekkie Oct 4 '16 at 8:32
  • \$\begingroup\$ 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) \$\endgroup\$ – JavaBeginner Oct 4 '16 at 8:35
  • \$\begingroup\$ You only apply the laws but you're not using your brain, just switch it on and see what happens. \$\endgroup\$ – Bimpelrekkie Oct 4 '16 at 8:37
  • \$\begingroup\$ I know the end result, I just need the proof \$\endgroup\$ – JavaBeginner Oct 4 '16 at 8:39
  • 1
    \$\begingroup\$ Apply the distributive law, then the complement law and see what happens. The answer is literally staring you in the face. \$\endgroup\$ – crowie Oct 4 '16 at 8:58
2
\$\begingroup\$

Sometimes the best thing to do is to look at the truth table for the expression you are given.

\$(\bar BC + B)\$ gives this truth table.

enter image description here

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?

| improve this answer | |
\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.