How do I prove $A(B+C) = AB+AC$ in Boolean expression?

Question

I need help to prove that $A(B+C) = AB + AC$ in Boolean. It’s been years since I did this and a friend asked for help. I tried but I can’t seem to get it right. Can someone explain how to prove it using Boolean? I was told that my work does not show clearly that it's true.

Answer 1

You could try to use a diagram with $\{A,B,C\}$ and assign values 0 or 1 to each element in the diagram and check when the the statement $A(B + C)$ is true.

For example if $A = 1$, $B=1$ and $C = 0$, which is equivalent to $AB$, $A(B+C)$ is true.

Edit : It is important to know that $0 + 1 = 1 + 0 = 1 + 1 = 1$ ; $0 + 0 = 0$ ;$1*1 = 1$ ;$0 = 0*1=1*0=0*0$

Answer 2

This is just an straight-forward instance of Distribution:

Distribution

$P(Q+R)=PQ+PR$

$P+QR=(P+Q)(P+R)$

But, if you need to prove Distribution itself, you can use a truth-table:

\begin{array}{ccc|c|c} A&B&C&A(B+C)&AB+AC\\ \hline T&T&T&T&T\\ T&T&F&T&T\\ T&F&T&T&T\\ T&F&F&F&F\\ F&T&T&F&F\\ F&T&F&F&F\\ F&F&T&F&F\\ F&F&F&F&F\\ \end{array}