Truth table boolean algebra

Question

I'm trying to make truth table for this and im terrible in boolean algebra. Can someone help me?

(A'+B')C+ABC'+A'C

Answer 1

Set up:

There are $3$ variables involved, $A$, $B$, and $C$. With $3$ variables, you get $2^3=8$ combination. You can get all those combinations systematically like this: \begin{array}{ccc|c} A&B&C&statement(s) \ to \ be \ evaluated\\ \hline 0&0&0&\\ 0&0&1&\\ 0&1&0&\\ 0&1&1&\\ 1&0&0&\\ 1&0&1&\\ 1&1&0&\\ 1&1&1&\\ \end{array}

See how I did that? I had the $C$ alternate between $0$ and $1$ for every single row, the $B$ alternates every two rows, and the $A$ every four rows.

OK, rather than doing thew whole statement at once, break it down to its smaller terms. So, let's create a column for (A'+B')C, a column for ABC', and one for A'C:

\begin{array}{ccc|c|c|c} A&B&C&(A'+B')C&ABC'&A'C\\ \hline 0&0&0&\\ 0&0&1&\\ 0&1&0&\\ 0&1&1&\\ 1&0&0&\\ 1&0&1&\\ 1&1&0&\\ 1&1&1&\\ \end{array}

OK, the $ABC'$ is easy:

\begin{array}{ccc|c|c|c} A&B&C&(A'+B')C&ABC'&A'C\\ \hline 0&0&0&&0\\ 0&0&1&&0\\ 0&1&0&&0\\ 0&1&1&&0\\ 1&0&0&&0\\ 1&0&1&&0\\ 1&1&0&&1\\ 1&1&1&&0\\ \end{array}

OK, the $'C$ is pretty easy as well:

\begin{array}{ccc|c|c|c} A&B&C&(A'+B')C&ABC'&A'C\\ \hline 0&0&0&&0&0\\ 0&0&1&&0&1\\ 0&1&0&&0&0\\ 0&1&1&&0&1\\ 1&0&0&&0&0\\ 1&0&1&&0&0\\ 1&1&0&&1&0\\ 1&1&1&&0&0\\ \end{array}

Now, if the $(A'+B')C$ is too hard to do in $1$ step, we can just break that one down as well, i.e. let's first figure out ust the $A'+B'$ term:

\begin{array}{ccc|c|c|c|c} A&B&C&A'+B'&(A'+B')C&ABC'&A'C\\ \hline 0&0&0&&&0&0\\ 0&0&1&&&0&1\\ 0&1&0&&&0&0\\ 0&1&1&&&0&1\\ 1&0&0&&&0&0\\ 1&0&1&&&0&0\\ 1&1&0&&&1&0\\ 1&1&1&&&0&0\\ \end{array}

OK, $A'+B'$ is a $1$ as soon as either $A$ or $B$ is a $0$, so:

\begin{array}{ccc|c|c|c|c} A&B&C&A'+B'&(A'+B')C&ABC'&A'C\\ \hline 0&0&0&1&&0&0\\ 0&0&1&1&&0&1\\ 0&1&0&1&&0&0\\ 0&1&1&1&&0&1\\ 1&0&0&1&&0&0\\ 1&0&1&1&&0&0\\ 1&1&0&0&&1&0\\ 1&1&1&0&&0&0\\ \end{array}

And now let's 'multiply' that by $C$:

\begin{array}{ccc|c|c|c|c} A&B&C&A'+B'&(A'+B')C&ABC'&A'C\\ \hline 0&0&0&1&0&0&0\\ 0&0&1&1&1&0&1\\ 0&1&0&1&0&0&0\\ 0&1&1&1&1&0&1\\ 1&0&0&1&0&0&0\\ 1&0&1&1&1&0&0\\ 1&1&0&0&0&1&0\\ 1&1&1&0&0&0&0\\ \end{array}

Finally, let's 'add' the three terms to get the truth-conditions of the statement as a whole:

\begin{array}{ccc|c|c|c|c|c} A&B&C&A'+B'&(A'+B')C&ABC'&A'C&(A'+B')C+ABC'+A'C\\ \hline 0&0&0&1&0&0&0&0\\ 0&0&1&1&1&0&1&1\\ 0&1&0&1&0&0&0&0\\ 0&1&1&1&1&0&1&1\\ 1&0&0&1&0&0&0&0\\ 1&0&1&1&1&0&0&1\\ 1&1&0&0&0&1&0&1\\ 1&1&1&0&0&0&0&0\\ \end{array}