DeMorgan's Law - just verifying my answer Can someone please verify if i did this correctly:
if Q = ABC + AC' + C(D + C') + A
show that Q = A + CD
Answer) ---------------------------------
    ABC + AC' + (CD + CC') + A

    ABC + AC(C' + C') + CD + A

    since c' + c' = 1

    ABC + (AC + 1) + CD + A

    AC (B + 1 ) + CD + A

    AC + CD + A

    = AC + A + CD

    A (1 + C) = A

    SO A + CD

Any help would be much appreciated!!!
 A: This

ABC + AC(C' + C') + CD + A
since c' + c' = 1

Is incorrect. It's c + c' = 1. In point of fact, c' + c' = c'. And you can't simply AND in $C'$.

ABC + (AC + 1) + CD + A

This is incorrect because anything +1 simply boils down to 1. So, $ABC + (AC + 1) + CD + A = 1$. What you'd actually get from $AC(C' + C') = ACC' + ACC' = ACC' = A(CC') = 0$, although, to clarify, that's incorrect (because you can't just add the $C'$, there).
What you can do would be $AC = AC(C + C') = ACC + ACC' = AC + 0 = AC$. That's completely fine.
This

ABC + (AC + 1) + CD + A
AC (B + 1 ) + CD + A

Is also incorrect. You are, however, correct in thinking that you can always AND in a 1, so you could have done...
$ABC + AC' + CD + A$
$ABC + AC' + A + CD$
$ABC + AC' + A1 + CD$
$A(BC + C' + 1) + CD$
$A + CD$, because $BC + C' + 1 = 1$.
You don't need DeMorgan's law, here. Just the distributive property of AND. You'd only need to use DeMorgan's law if you wanted to transform something between $(A + B)' = A'B'$ or $(AB)' = A' + B'$.
