How to solve for $x$ given $x⇔A$ in a truth table? 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.
 A: 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 $x=1$.
A: 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.
A: Give the last column a name, say $Y$.
So, you are given
$$Y = (x \iff A)$$
which is same as
$$Y = (x = A)$$
so
$$Y = \lnot(x \ne A)$$
Unequality of boolean values is their exclusive or:
$$(p\oplus q) \equiv ((p \land\lnot q)\lor(\lnot p\land q))$$
so:
$$Y = \lnot(x \oplus A)$$
hence
$$\lnot Y = x \oplus A$$
and
$$x = \lnot Y \oplus A$$
From the table
$$Y=(A\oplus B)\land C$$
so
$$x = \lnot ((A\oplus B)\land C) \oplus A$$
This expands and then contracts as
$$\begin{align}x & = \lnot ((\lnot AB\lor A\lnot B)\land C) \oplus A \\
& = \lnot (\lnot ABC\lor A\lnot BC) \oplus A \\
& = ((A\lor\lnot B\lor\lnot C)\land (\lnot A\lor B\lor\lnot C)) \oplus A \\
& = \lnot((A\lor\lnot B\lor\lnot C)\land (\lnot A\lor B\lor\lnot C)) \land A \lor
    ((A\lor\lnot B\lor\lnot C)\land (\lnot A\lor B\lor\lnot C)) \land \lnot A\\
& = (\lnot ABC\lor A\lnot BC) \land A \lor
    ((A\lor\lnot B\lor\lnot C)\land (\lnot A\lor B\lor\lnot C)) \land \lnot A\\
& = (\lnot ABCA\lor A\lnot BCA) \lor ((A\lor\lnot B\lor\lnot C)\land \lnot A\\
& = (A\lnot BC) \lor (\lnot A\lnot B\lor\lnot A\lnot C)\\
& = A\lnot BC \lor \lnot A(\lnot B\lor\lnot C)\\
\end{align}
$$
where missing operators are $\land$.
