Prove that $A\subseteq B$ if and only if $A \cap B^c = \emptyset$

Question

Prove that $A\subseteq B$ if and only if $A \cap B^c = \emptyset$

Is this a sufficient proof for the left to right implication, i.e. proving $A\subseteq B \implies A \cap B^c = \emptyset$? I am focused mainly on the second paragraph where I use a contradiction.

Assume that $A\subseteq B$, so $\forall x (x\in A \rightarrow x \in B)$ is true. We show that $A \cap B^c$ and $\emptyset$ are both subsets of each other to prove equality. Note that $\emptyset \subseteq A \cap B^c$ is (vacously) true so now we have to show that $ A \cap B^c\subseteq \emptyset$.

Suppose that $x \in A \cap B^c$. Then $x \in A$ and $x \in B^c \Leftrightarrow x \notin B$. But this contradicts with the assumption $A\subseteq B$ since we have $x \in A \rightarrow x \in B$ false. Thus $x \notin A \cap B^c$ for all $x$ and it follows that $A \cap B^c \subseteq \emptyset $ as $x\in A \cap B^c \rightarrow x \in \emptyset$ is (vacously) true. Therefore, $A \cap B^c = \emptyset$.

I looked here Prove that $A \subseteq B$ if and only if $A \cap \overline{B}=\emptyset.$ and the given answers there use different methods, e.g. using an expression for implication $p \rightarrow q \iff \neg p \lor q$ for the assumption, supposing first that $A \cap B^c \ne \emptyset$, etc. I am more interested in the validity of my method that I used here.

Answer 1

Alternatively to a point by point proof is
a set algebra proof.

If A subset B, then A $\cap$ B$^c$ subset
B $\cap$ B$^c$ = empty set.

Conversely. A = (A $\cap$ B) $\cup$ (A $\cap$ B$^c$)
= A $\cap$ B subset B.

Answer 2

Let's look at how the subset is defined.

$A \subseteq B$ means that $\forall x: x \in A$ $\Rightarrow x \in B$.

Let's prove this statements by a chain of If and only if statements.

1. $A \subseteq B$ $\Leftrightarrow$ $\forall x: x \in A \Rightarrow x \in B$

Now, we can write $A \Rightarrow B$ using only AND-OR-NOT formula like below.

$A \Rightarrow B$ IFF $\neg A \vee B$. (You can simply check this using truth table).

2. $\forall x: x \in A \Rightarrow x \in B \:$ IFF $\: \forall x: \neg(x \in A) \vee (x \in B) \:$ IFF $\:\forall x: \neg((x \in A) \wedge \neg(x \in B))\:$ IFF $\:\forall x: \neg((x \in A) \wedge (x \notin B))\:$ IFF $\:\forall x: \neg((x \in A) \wedge (x \in B^c))\:$ IFF $\:\forall x: \neg(x \in A \cap B^c))\:$ IFF $\:\forall x: (x \notin A \cap B^c))$.
   

This is the end. We've showed that if $A \subseteq B$, then $\forall x: (x \notin A \cap B^c))$, which means that there is not any element in the intersection of $A$ and $B^c$, which is in effect in empty. $\blacksquare$