How can I solve for $x$ in terms of A, B and C given the truth table below?
$$\begin{array}{ccc|c}
A & B & C & x ⇔ A\\
\hline
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 1 & 1 & 1\\
1 & 0 & 0 & 0\\
1 & 0 & 1 & 1\\
1 & 1 & 0 & 0\\
1 & 1 & 1 & 0
\end{array}$$
The main way I tried to solve this was by simplifying the truth table into its ANF and then seeing if I could move things around.
So from $(A \land \lnot B \land C) \lor (\lnot A \land B \land C)$ to $(A \land C) \oplus (B \land C)$ but then I got stuck because I didn't know how to get A onto its own in the formula.
The way that I eventually managed to solve it was intuitively but it took forever and it was a lot of guesswork:
$$
(((\lnot A \lor \lnot B) \land (A \lor B) \land \lnot C) ⇔ B) ⇔ A
$$
$$
\therefore x = (((\lnot A \lor \lnot B) \land (A \lor B) \land \lnot C) ⇔ B)
$$
If this question doesn't obey some stylistic convention, I'm happy to edit it. I'm sure it's not professional but I am a hobbyist not a mathematician.
Best Answer
By comparing corresponding truth values for $A$ and $(x{\iff}A)$, you can infer the corresponding truth values for $x$:
If $(x{\iff}A)=1$, then $x=A$, else $x=A'$.
Hence we can extend the truth table to include a column for $x$: $$\begin{array}{ccc|c|c} A & B & C & x{\iff}A&x\\ \hline 0 & 0 & 0 & 0&1\\ 0 & 0 & 1 & 0&1\\ 0 & 1 & 0 & 0&1\\ 0 & 1 & 1 & 1&0\\ 1 & 0 & 0 & 0&0\\ 1 & 0 & 1 & 1&1\\ 1 & 1 & 0 & 0&0\\ 1 & 1 & 1 & 0&0\\ \hline \end{array}$$ which allows us to write $$ x=A'B'C'+A'B'C+A'BC'+AB'C $$ with one term for each of the $4$ rows for which $x=1$.