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.

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    $\begingroup$ looks good to me :) $\endgroup$ – Nosrati Oct 19 '18 at 5:08
  • $\begingroup$ While your logic is correct, perhaps update the presentation. $x\in A \cap \bar B \iff x \not\in B$ is false in general, and a reader may not be paying close enough attention to the assumption that $A\subset B$. $\endgroup$ – David Peterson Oct 19 '18 at 5:10
  • $\begingroup$ I think it is proved $x\in A \cap \bar B \to x\in B$ and $x \not\in B\to x\in\phi$. $\endgroup$ – Nosrati Oct 19 '18 at 5:14
  • $\begingroup$ @holo Please do not use a bar on top to denote complement. This is highly non-standard. $\overset {-} B$ usually denotes the closure of $B$. $\endgroup$ – Kavi Rama Murthy Oct 19 '18 at 5:32

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.


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