Simplify $(A+C)(AD+AD) + AC + C$ using Boolean algebra

Question

I have solved the equation like this:

 (A + C)(AD + AD) + AC + C
=(A + C)(AD) + (A + C)(AD) + AC + C
=AAD + ACD + AAD + ACD + AC + C
=AAD + AAD + ACD + ACD + AC + C
=AAD + ACD + AC + C 
=AAD + ACD + C(A+1)
=AAD + ACD + C
=AD + ACD + C
=AD(1+C) + C
=AD + C

Am I right?

Answer 1

It looks just fine! Well done!

A bit more simply, $$\begin{align}(A+C)(AD+AD)+AC+C &= (A+C)AD+AC+C\\ &= AAD+CAD+AC+C\\ &= AD+(AD+A+1)C\\ &= AD+C\end{align}$$

Answer 2

I would have taken the following steps: $$ \begin{align} (A + C)(AD + AD) + AC + C &= (A + C)(AD) + (AC + C)\\ &= AD + ADC + C\\ &= AD + C\\ \end{align} $$