Why does $A = X-B$ is equivalent to $A \cup B = X$ and $A \cap B = \emptyset$

Question

I was looking at a solution to the problem and it says that $A = X-B$ is equivalent to $A \cup B = X$ and $A \cap B = \emptyset$. I am wondering why this is true?

Any help would be highly appreciated.

Answer 1

$\begin{align}A \cup B & = (X-B)\cup B \\[1ex] & = \{x: (x\in X \wedge x\notin B)\vee x\in B\} \\[1ex] & =\{x: x\in X \vee x\in B\} \\[1ex] & = X \cup B \\[3ex] A \cap B & = (X-B)\cap B \\[1ex] & = \{x: (x\in X \wedge x\notin B)\wedge x\in B\} \\[1ex] & =\{x: x\in B\wedge x\notin B\} \\[1ex] & = \varnothing\end{align}$

Answer 2

It's not true unless we also know that $B \subseteq X$. Otherwise, for a counterexample, consider $A = \{1, 2\}$ and $B = \{3, 4\}$ and $X = \{1, 2, 3\}$. Then: $$ A = \{1, 2\} = X - B $$ yet on the other hand: $$ A \cup B = \{1,2,3,4\} \neq \{1,2,3\} = X $$