Is the following proof of $$(A\setminus B)\cup(A\cap B)=B\implies B\subseteq A$$ correct?
Proof: for all $x$, if $x\in B$ then $x\in A$. By hypothesis,
\begin{align*} x\in(A\setminus B)\cup(A\cap B)&\to x\in A\setminus B\lor x\in A\cap B\tag{1}\\ &\to(x\in A\land x\notin B)\lor(x\in A\land x\in B)\tag{2}\\ &\to x\in A\land(x\notin B\lor x\in B)\tag{3}\\ &\to x\in A\land\mathrm{T}\tag{4}\\ &\to x\in A.\tag{5} \end{align*}
I think the proof is a little confusing when says "for all $x$, if $x\in B$ then $x\in A$. By hypothesis (...)". I think it should say: "for all $x$, $x\in B$. Then by hypothesis, (...)".
Do you agree?
The stuff after "Proof:" is not needed at all, as you do not need to assume that. Instead, you should use the given statement $(A\setminus B)\cup(A\cap B)=B$.
Other than that, every step is clear and is easy to follow, and it looks great :)
Perhaps add a line "since $x \in (\cdots) \implies x \in A$, $\cdots \subseteq A$".
Also can someone teach me how to align equations lol
$$\forall x \in B,$$ $$x \in B \implies x \in (A\setminus B)\cup(A\cap B)$$ $$\implies x\in A\setminus B\lor x\in A\cap B$$ $$\implies (x\in A\land x\notin B)\lor(x\in A\land x\in B)$$ $$\implies x\in A\land(x\notin B\lor x\in B)$$ $$\implies x\in A\land\mathrm{T}$$ $$\implies x\in A$$
$$\because x \in B \implies x \in A$$ $$\therefore B \subseteq A$$