How can I prove this relation between the elementary set theory and the elementary logic? If you need to prove an equality like $A\Delta B=(B\setminus C)\cup[C\cap (B\Delta A)]$ we can first prove $p\underline{\lor} q\Longleftrightarrow (q\land\overline{r})\lor(q\underline{\lor}p)$ (with a truth table for example) and then use it whith $p=x\in A$, $q=x \in B$ and $r=x \in C$.
So, if we now that $p\underline{\lor} q\Longleftrightarrow (q\land\overline{r})\lor(q\underline{\lor}p)$ then we now that $A\Delta B=(B\setminus C)\cup[C\cap (B\Delta A)]$. But my question is. is true in general (with a general set equation) the reverse implication? that is  $A\Delta B=(B\setminus C)\cup[C\cap (B\Delta A)]$ implies $p\underline{\lor} q\Longleftrightarrow (q\land\overline{r})\lor(q\underline{\lor}p)$.
 A: If we have an equality of sets that is abstract, in the sense that it holds for any sets we choose to plug in for $A, B, C$ etcetera, then the answer is yes.

In order to establish this, we need to investigate truth tables in a more abstract manner. 
Let $P = \{p, q, r, \ldots\}$ be the set of proposition symbols, and let $2 = \{0,1\}$ be the set of "truth values": $0$ for false, $1$ for true. Consider now the set $2^P$ of functions $\nu: P \to 2$. These functions, often called boolean interpretations, correspond to lines of a truth table. E.g. if we define $\nu(x) = 0$ for all $x$, we get the line from the truth table where $p, q, r, \ldots$ are all false.
Using the standard rules of logic, we may extend these $\nu$s to also evaluate terms like $p \land \neg q$; see ProofWiki for a detailed exposition. This is effectively the same as filling the line of the truth table based on the values of the proposition symbols.
Now we can apply our set equalities to sets of boolean interpretations by defining, for every formula $\phi$:
$$[\phi] = \{\nu \in 2^P: \nu(\phi) = 1\}$$
We will prove the following:

Theorem For all formulas $\phi, \psi$, $\phi \iff \psi$ if and only if $[\phi] = [\psi]$.

which intuitively amounts to saying that "two formulas are equivalent iff their truth tables agree".
Basically, the definition of "$\phi$ and $\psi$ are equivalent" is that for every $\nu \in 2^P$ we have $\nu(\phi) = \nu(\psi)$. But then, $[\phi] = [\psi]$ means:
$$\forall \nu: \nu (\phi) = 1 \iff \nu (\psi) = 1$$
Since the only other value any $\nu$ can take is $0$, it follows that $\nu(\phi) = \nu(\psi)$ for all $\nu$. The conclusion of the theorem follows.
Now if we apply a known set-theoretic equality to appropriately chosen sets of the form $[\phi]$, we will obtain the corresponding logical equivalences.

As an example, let us apply the above to $A \cap A^c = \varnothing$, which should correspond to $p \land \neg p = \bot$ (since $\{\nu: \nu(\bot) = 1\} = \varnothing$):
\begin{align*}
[p] \cap [p]^c &= [\bot]\\
[p] \cap \{\nu: \nu(p) =1 \}^c &= [\bot]\\
[p] \cap \{\nu: \nu(p) =0 \} &= [\bot]\\
[p] \cap \{\nu: \nu(\neg p) = 1\} &= [\bot]\\
[p] \cap [\neg p] &= [\bot]\\
\{\nu: \nu(p) = 1 \text{ and } \nu(\neg p) = 1\} &= [\bot]\\
[p \land \neg p] &= [\bot]
\end{align*}
Hence, by the theorem, $p \land \neg p \iff \bot$.
