How to deduce two sets are equal when one being subset of another is proven?

Question

Below is a proof that $(A \cap B ) \cup (A \setminus B) = A$:

Let's consider $x \in (A \cap B ) \cup (A \setminus B)$.

1. $ x ∈ (A∩B)∨x ∈ (A \setminus B) $
2. $ x ∈ A∧x ∈ B)∨(x ∈ A∧ x \notin B) $
3. $ x ∈ A∧(x ∈ B∨x \notin B) $

Which means $ x ∈ A $ and thus $ (A∩B)∪(A \setminus B) = A. $

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The proof evidently shows that $ (A∩B)∪(A \setminus B) ⊂ A $ but why does it show the two are equal?

Answer 1

You only need to rewrite the $\setminus$ as complement and use a distributive law:

$ (A \cap B ) \cup (A \setminus B) = (A \cap B ) \cup (A \cap B^C) = A \cap (B \cup B^C) = A $

Answer 2

The proof should look more like this:

\begin{align*} & x \in (A\cap B)\cup (A\setminus B) \\ \Longleftrightarrow\quad& \left( x\in (A\cap B)\right) \vee \left(x\in (A\setminus B)\right) \\ \Longleftrightarrow\quad& \left( x\in A \wedge x\in B\right) \vee \left(x\in A \wedge X \notin B\right) \\ \Longleftrightarrow\quad& x\in A \wedge \left(x\in B \vee X \notin B\right) \\ \Longleftrightarrow\quad& x\in A \wedge \mathrm{true} \\ \Longleftrightarrow\quad& x\in A. \end{align*}

Since each step is an equivalence, you can read it top to bottom to get $\subset$ and bottom to top to get $\supset$.