Simplification of the boolean expression using boolean algebra

Simplify the following expressions to the simplest expression using De Morgan's theorem and Boolean algebra.

ABC+A'CD+B'CD

=(AB+A'D+B'D)C

=(AB+(A'+B')D)C

=(AB+(AB)'D)C

can anyone simplify it further and explain how you got there.Thanks in advance.

• We have $AB\lor ABD=AB$ (absorption), and $ABD\lor (AB)'D=D$. So $AB\lor (AB)'D=AB\lor D$. – user10354138 Jun 3 at 8:49

It always helps to expand the terms so they include all variables .. and then reorganize, and recombine in a more efficient way, adding or removing duplicates as needed.

The key principle is:

$$P = PQ + PQ'$$

Starting from your second expression:

$$(AB+A'D+B'D)C=$$

$$(ABD+ABD'+A'BD+A'B'D+AB'D+A'B'D)C=$$

$$(ABD+ABD'+(AB+A'B+AB'+A'B')D)C=$$

$$(AB+(B+B')D)C=$$

$$(AB+D)C$$

• Thanks a lot!This was very helpful! – Jarvis Ferns Jun 3 at 12:04
• @JarvisFerns You're welcome! :) – Bram28 Jun 3 at 12:28