# Prove that $(A \subseteq B) \Leftrightarrow (A \cap B = A)$

I think I need to prove the implication both ways. But how do I do this?

I think ($$\Rightarrow$$) is correct because if $$x$$ belongs to $$A$$ then it belongs to $$B$$ which means that certainly overlap of $$A$$ and $$B$$ will be $$A$$. But is there a formal way to prove this? Both of these implications?

($$\Rightarrow$$) Let $$A\subset B$$. Take $$x\in A\cap B$$. Then $$x \in A$$ and $$x \in B$$. The important is that $$x \in A$$. On the other hand, if $$x\in A$$, since $$A\subset B$$, we have $$x\in B$$, therefore $$x\in A\cap B$$. We conclude that $$A\cap B = A$$.

($$\Leftarrow$$) Assume that $$A\cap B = A$$. Let $$x \in A$$. Since $$A=A\cap B$$, then $$x \in B$$ as well. But $$x\in A \Rightarrow x\in B$$ means that $$A\subset B$$, which finishes the proof.

You can replace terms with their formal definitions and use the formal rules of logic. I'll use a few potential formal definitions to show how it's possible to start proving it, but make sure to use the formal definitions your course or book is using if that matters to you.

### Formal definitions

The formal definitions used in this answer are:

\begin{align} A \subseteq B &: \forall x. x \in A \rightarrow x \in B \\ A \cap B &: \{ x : x \in A \wedge x \in B \} \\ A = B &: \forall x. x \in A \leftrightarrow x \in B \\ \text{Set builder membership} &: x \in \{ y : P(y) \} \leftrightarrow P(x) \end{align}

### Statement rewritten formally

So, we rewrite the statement to prove using its formal definitions and simplify a bit:

\begin{align} (A \subseteq B) &\Leftrightarrow (A \cap B = A) \\ (\forall x. x \in A \rightarrow x \in B) &\Leftrightarrow (\forall x. x \in \{ y : y \in A \wedge y \in B \} \leftrightarrow x \in A) \\ (\forall x. x \in A \rightarrow x \in B) &\Leftrightarrow (\forall x. (x \in A \wedge x \in B) \leftrightarrow x \in A) \end{align}

Now, it can be proven formally fairly easily.