I have the question on:
$$f(a,b,c) = (\neg a \wedge \neg b \wedge c) \vee (\neg a \wedge \neg c)$$
I got to the point : $$\neg a \wedge (\neg c \vee (\neg b \wedge c))$$
But on our book, it states it is: $$\neg a \wedge (\neg b \vee \neg c)$$ or $$\neg a \wedge \neg(b \wedge c)$$
What did I do wrong? Thanks!
On
You're on a good path. Here is the identity obtained algebraically:
$$\begin{align} \lnot c \lor \lnot b c & = \lnot c \lor \lnot c \lor \lnot b c \\ & = \lnot c \lor \lnot c 1 \lor \lnot b c \\ & = \lnot c \lor \lnot c (b \lor \lnot b) \lor \lnot b c \\ & = \lnot c \lor \lnot c b \lor \lnot c \lnot b \lor \lnot b c \\ & = \lnot c (1 \lor b) \lor (\lnot c \lor c)\lnot b \\ & = \lnot c 1 \lor 1 \lnot b \\ & = \lnot c \lor \lnot b \end{align}$$
You can see the same result for a standard two-valued Boolean algebra on a Venn diagram:
or in a Karnaugh map:
$$\begin{array}{c|c|c|l} & B=0 & B=1 \\ \hline C=0 & \bbox[2pt, gold]1 & \bbox[2pt, gold]1 & \ \lnot C\\ \hline C=1 & \bbox[2pt, cyan]1 & 0 & \ \lnot BC \end{array}$$
For any $P$ and $Q$, you have $P \vee (\neg P \wedge Q) = P \vee Q$. This is called Reduction. If you don't have Reduction, you can get it as follows:
$P \vee (\neg P \wedge Q) = (P \vee \neg P) \wedge (P \vee Q) = True \wedge (P \vee Q) = P \vee Q$.
So, from where you got to, you can proceed as follows:
$\neg a \wedge (\neg c \vee (\neg b \wedge c)) =$ (Reduction)
$\neg a \wedge (\neg c \vee \neg b)) =$ (DeMorgan)
$\neg a \wedge \neg (c \wedge b)$
So, you weren't doing anything wrong .. you just needed a few more steps.