Proof of $B \subseteq A^C \iff A \cap B = \emptyset$ In my textbook I found the following excercise. There was no solution given. Can anyone tell me, if my proof is correct? Especially my deductions/definition from the statement $A \cap B = \emptyset$. Is there a more rigorous formulation?
Proposition:$B \subseteq A^C \iff A \cap B = \emptyset$
Proof: We first proof that $B \subseteq A^C \implies A \cap B = \emptyset$. To show that $A \cap B = \emptyset$ we need to establish that for every $x$ we have
$$x \in A \implies x \notin B$$
$$x \in b \implies x \notin A$$
From $B \subseteq A^C$ we can deduce that,
$$x \in B \implies x \in A^C \iff x \notin A$$
as well as its contrapositive
$$x \in A \implies x \notin B$$.
This shows that $B \subseteq A^C \implies A \cap B = \emptyset$.
To show that $A \cap B = \emptyset \implies B \subseteq A^C$ we can assume from $A \cap B = \emptyset$ that
$$x \in B \implies x \notin A \iff x \in A^C $$
which completes the proof. $\blacksquare$
 A: The ideas of your proof are basically correct. There is one flaw and a typo.

To show that $A \cap B = \emptyset$ we need to establish that for
every $x$ we have
$$x \in A \implies x \notin B$$ $$x \in b \implies x \notin A$$

$x \in B$ surely.

From $B \subseteq A^C$ we can deduce that,
$$x \in B \implies x \in A^C \iff x \notin A$$
as well as its contrapositive
$$x \in A \implies x \notin B$$.

This part doesn't make perfect sense, you need to have $x\in A \implies x\notin A^c \implies x \notin B$ there, otherwise you're assuming the result you're trying to prove.

This shows that $B \subseteq A^C \implies A \cap B = \emptyset$.

A: Let $A = \{x\in U : p(x) \ \text{is true}\}$ and $B = \{x\in U : q(x) \ \text{is true}\}$.
According to the logical operator properties, we obtain the desired claim:
\begin{align*}
B\subseteq A^{c} & \Longleftrightarrow (\forall x\in U)(q(x)\to \neg p(x) \ \text{is true})\\\\
& \Longleftrightarrow(\forall x\in U)(\neg q(x)\vee \neg p(x) \ \text{is true})\\\\
& \Longleftrightarrow (\forall x\in U)(\neg (q(x)\wedge p(x)) \ \text{is true})\\\\
& \Longleftrightarrow (\forall x\in U)(q(x)\wedge p(x) \ \text{is false})\\\\
& \Longleftrightarrow A\cap B = \varnothing
\end{align*}
Hopefully this helps!
A: Thanks for the answer @Suzu Hirose. Here is my corrected proof:
Proposition:$B \subseteq A^C \iff A \cap B = \emptyset$
Proof: We first proof that $B \subseteq A^C \implies A \cap B = \emptyset$. To show that $A \cap B = \emptyset$ we need to establish that for every $x$ we have
$$x \in A \implies x \notin B$$
$$x \in B \implies x \notin A$$
From $B \subseteq A^C$ we can deduce that,
$$x \in B \implies x \in A^C \iff x \notin A$$
as well as its contrapositive
$$x \in A \iff x \notin A^C \implies x \notin B$$
This shows that $B \subseteq A^C \implies A \cap B = \emptyset$.
To show that $A \cap B = \emptyset \implies B \subseteq A^C$ we can assume from $A \cap B = \emptyset$ that
$$x \in B \implies x \notin A \iff x \in A^C $$
which completes the proof. $\blacksquare$
A: Your proof is already rigorous, but I see some redundancy:

*

*$x \in A \implies x \notin B$ and $x \in B \implies x \notin A$ do sufficiently cover $A \cap B = \emptyset$. As you suggested, they are contrapositives, and thus are equivalent. We can just reduce it to either one statement to mean $A \cap B = \emptyset$.


*It is quite common to prove "$(1) \iff (2)$" by showing the implication then the converse. But we can also do shorter using equivalence chain:
$B \subseteq A^c$ is - not just implies but also - equivalent to $x \in B \implies x \in A^c$. We have seen previously than $A \cap B = \emptyset$ is if and only if $x \in B \implies x \notin A$. By definition of complement set, $x \in A$ is equivalent to $x \notin A$ so we can just swap either statement. We then establish that chain:  $B \subseteq A^c \iff (x \in B \implies x \in A^c) \iff (x \in B \implies x \notin A) \iff A \cap B = \emptyset$, and we are done.
