Simplifying $(A'+ D) \times [(B \times C) + D']$ in Boolean algebra

Question

$(A'+ D) \times [(B \times C) + D']$ I can rewrite as $(A' + D) \times (D' + B) \times (D' + C)$. Is there a way of simplifying to get to the step of $D' + D = 1$ or $D' \times D = 0$ or any other simplifying that I can do? Thank you.

Answer 1

The original expression can be expanded to:

A' B C + A' D' + B C D + D D'

D D' cancels to false. Extracting B C we get:

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

If A' D' would be false, either A or D would be true. We can therefore assume for the term on the right that A' is false.

This results in:

A' D'  + B C D

If in doubt, write down the truth table to convince yourself.