TODO

Let $p\in \left(X⧵B\right)\cap A$, therefore $p\in X$, $p\notin B$ and $p\in A$, this is impossible since if $p\in A$ then $p\in B$ as $A\subseteq B$, therefore no such $p$ is an element of this set, so that $\left(X⧵B\right)\cap A=\varnothing$

Taking an element $p\in B$ we'd like to show that $p\in X⧵A$, we can see $p\in X$ because $B\subseteq X$, it's also clear that $p\notin A$ since if it were then $A\cap B\ne \varnothing$ which would be a contradiction. Combining this, we have that $p\in X⧵A$ as needed.

$${\displaystyle X\backslash X=\{x\in X:x\notin X\}=\varnothing }$$

TODO

Suppose that $p\in (A\cap X)\backslash B$ therefore $p\in A$, $p\in X$ but $p\notin B$, we can see that $p\in (A\cap X)$ and since $p\notin B$, then $p\notin (B\cap X)$, therefore $p\in (A\cap X)\backslash (B\cap X)$

Now suppose that $p\in (A\cap X)\backslash (B\cap X)$, so that $p\in A\cap X$ and $p\notin B\cap X$, from this we deduct that $p\in X$, so to avoid contradiction, we must have $p\notin B$, since we also know $p\in A$, we can say that $p\in (A\cap X)⧵B$

Suppose that $p\in \left(A\backslash B\right)\cap X$ so $p\in A$, $p\notin B$ and $p\in X$, this means that $p\in A\cap X$ and $p\notin B$, this shows that $p\in (A\cap X)\backslash B$ as needed

Now suppose that $p\in (A\cap X)\backslash B$ so $p\in A$, $p\in X$ but $p\notin B$, recombining this information we can see that $p\in \left(A\backslash B\right)\cap X$, as needed.

We know that $x\in A\backslash (B\cap C)$ is true iff $x\in A$ and $x\notin (B\cap C)$, equivalent to $x\notin B$ or $x\notin C$, since $x\in A$, we can biconditionally re-write this as $x\in A\backslash B$ or $x\in A\backslash C$ iff $x\in \left(A\backslash B\right)\cup \left(A\backslash C\right)$

Since all connectives in the above proof are bi-conditionals, then we know that $A\backslash (B\cap C)=\left(A\backslash B\right)\cup \left(A\backslash C\right)$

Suppose that $x\in A\backslash \left(B\backslash C\right)$ which is true iff $x\in A$ and $x\notin \left(B\backslash C\right)$ which means that it's not true that $(x\in B\text{and}x\notin C)$ so that $x\notin B$ or $x\in C$, since $x\in A$ and we know that $A\subseteq B$ then $x\in B$, which forcs $x\in C$ to avoid a contradiction, therefore

Suppose that $x\in A\backslash B\cap C$, therefore $x\in A$, $x\notin B$ and $x\in C$, thus we can see that $x\notin B\backslash C$ thus $x\in A\backslash \left(B\backslash C\right)$

$x\in A\backslash \left(B\backslash C\right)$ iff $x\in A$ and $x\notin \left(B\backslash C\right)$ equivalent to the following statement not being true: $(x\in B\wedge x\notin C)$ which is $x\notin B$ or $x\in C$, therefore $x\in A\cap C$ or $x\in A⧵B$ so that $x\in (A\cap C)\cup \left(A⧵B\right)$

The above chain of steps can be traversed forwards or backwards since each connective is an iff, therefore the proof is complete

Suppose that $x\in A\backslash \left(A\backslash B\right)$ which is true iff $x\in A$ and $x\notin \left(A\backslash B\right)$ which means that it's not true that $x\in A\wedge x\notin B$ we already know that $x\in A$ holds, so we must have that $x\notin B$ to be false, which means that $x\in B$ so that $x\in A\cap B$

Suppose that $x\in A\cap B$, since $x\in B$, then it's not true that $x\notin B$, therefore $x\notin \left(A\backslash B\right)$ since $x\in A$, this shows that $x\in A\backslash \left(A\backslash B\right)$