I've been asked to Prove that $A\subseteq B$, $A\cap \overline{B}=\oslash$ and $\overline{A}\cup B=U$ are equivalent. I believe I have done so, but I expect that I have missed something critical.
I have convinced myself that they are equivalent - I just need to prove it.
I appreciate any feedback, if I'm on the right path, if there's a missing portion to the proof, etc..
There are 3 things that need to be proven:
i. $A\subseteq B \equiv A\cap \overline{B} = \oslash$
ii. $A\subseteq B \equiv \overline{A}\cup B=U$
iii. $A\cap \overline{B}=\oslash \equiv \overline{A}\cup B=U$
i: $A\subseteq B$ = $A\cap \overline{B} = \oslash$
$X\in A \implies X\in B$
$X \notin A \vee X \in B$
$X \in A \wedge X \notin B$
$A \cap \overline{B} \equiv A\cap \overline{B} = \oslash$
ii: $A\subseteq B$ = $\overline{A}\cup B = U$
$X\in A \implies X\in B$
$X \notin A \vee X \in B$
Since A contains a subset of B, $\overline{A}$ contains all elements not in A, and thus not in B. So
$X \notin A \vee X \in B = U$
iii: $A\cap \overline{B}=\oslash \equiv \overline{A}\cup B=U$
$X \in A \wedge X \notin B$
$X \notin A \vee X \in B$
Same as above - A contains a subset of B, \overline{A} contains all elements not in A and thus not in B, and so $\overline{A}\cup B=U$
Please pardon the poor TeX syntax. Still trying to learn it a bit.