# Boolean Logic Converting DNF to CNF

I'm confused on how to convert DNF to CNF. On the answer sheet my teacher gave me, she just convert it right away with no explanation.

So my teacher convert $F: (A \wedge \neg B) \vee (B \wedge D)$ to CNF form of $(A \vee B) \wedge (\neg B \vee D)$.

How does that go?

Is there a way of converting it without drawing the truth tables?

Any help would be great

De Morgan's Law states $\neg(a + b) \equiv \neg a\neg b$ and $\neg(ab) \equiv \neg a + \neg b$. \begin{aligned}A\neg B + BD \equiv & \neg(\neg(A\neg B)\neg(BD)) \text{ De Morgan's outside} \\ \equiv & \neg((\neg A + B)(\neg B + \neg D)) \text{ De Morgan's inside} \\ \equiv & \neg(\neg A \neg B + \neg A \neg D + B \neg D) \text{ Distributivity} \\ \equiv & \neg(\neg A \neg B + \neg A \neg D (\neg B + B) + B \neg D) \text{ Complementation} \\ \equiv & \neg(\neg A \neg B + \neg A \neg D \neg B + \neg A \neg D B + B \neg D) \text{ Distributivity} \\ \equiv & \neg(\neg A \neg B(1 + \neg D) + B \neg D (1 + \neg A)) \text{ Distributivity} \\ \equiv & \neg(\neg A \neg B + B \neg D) \text{ Annihilator} \\ \equiv & (A + B)(\neg B + D) \text{ De Morgan's outside}\end{aligned}
• I noticed that the two other terms contained $B$ and $\neg B$ which told me that the middle term could be eliminated due to the way the terms were matched. If I'm going to be honest, it was mainly intuition and reasoning logically. It's a trick that you can use sometimes when you're not sure how to eliminate terms. – Chantry Cargill Oct 23 '14 at 8:32