TITLE: How to solve for $x$ given $x⇔A$ in a truth table?
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: As a footnote to @quasi's answer, which gives a natural and principled way of getting an answer, it is perhaps worth also noting that $x$ won't be unique.
Find one solution $x$, and any wff tautologically equivalent to $x$ will do as well. For example, $(x \land T)   \leftrightarrow A$ has the same truth-table as $x \leftrightarrow A$ for any tautology $T$ at all.