# Prove that the following three statements are equivalent

1. $A\subset B$
2. $A \cap B^{c} = \emptyset$
3. $A^{c} \cup B = U$

where set $^c$ is the complement of the set

What I have so far is this. The statements are equivalent if they imply each other.

$1\Rightarrow2$: assuming 1 is true, $x \in A$ implies that $x\in B$ which in turn implies that $x\notin B^{c}$. So, by definition the set $A \cap B^{c}$ is always empty and $A\cap B^{c} = \emptyset$

$2\Rightarrow 3$: for this one, I was wondering If I could just say that applying de morgan's laws shows that $(A \cap B^{c})^{c} = \emptyset^{c}$ is logically equivalent to $A^{c} \cup B = U$ showing that 2 implies 3.

If this is incorrect, how would I go about showing that 2 implies 3?

$3\Rightarrow 1$: Assuming that 3 is true, 3 implies that either $x \in B$ or $x \in A^{c}$ and by definition, $x \in A^{c}$ implies that $x \in A$ and so, this implies that $A \subset B$.

$1\Rightarrow 2$ and $2\Rightarrow 3$ seems fine...for $3\Rightarrow 1$ let $x\in A$ then $x\notin A^{c}$ but $A^{c}\cup B=U$ hence $x\in B$ and so $A\subset B$.
Translating the given statements to the logic level leads (by set extensionality and the definitions of $\;\subseteq\;$, $\;\cap\;$, $\;{}^c\;$, $\emptyset\;$, $\cup\;$, and $U\;$) to the following equivalents of your 1, 2, and 3: \begin{align} \tag{1'} & \langle \forall x :: x \in A \;\Rightarrow\; x \in B \rangle \\ \tag{2'} & \langle \forall x :: x \in A \land x \not\in B \;\equiv\; \text{false} \rangle \\ \tag{3'} & \langle \forall x :: x \not\in A \lor x \in B \;\equiv\; \text{true} \rangle \\ \end{align} where $\;x\;$ ranges over your universe $\;U\;$. Now, $\text{(2')}$ and $\text{(3')}$ are equivalent by DeMorgan, and $\text{(1')}$ and $\text{(3')}$ are equivalent by rewriting $\;P \Rightarrow Q\;$ as $\;\lnot P \lor Q\;$. That proves that all of the statements are equivalent.