Prove that $A\subseteq B$, $A\cap \overline{B}=\oslash$ and $\overline{A}\cup B=\mu$ are equivalent I've been asked to Prove that $A\subseteq B$, $A\cap \overline{B}=\oslash$ and $\overline{A}\cup B=\mu$ 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.  $A\subseteq B \equiv \overline{A}\cup B=\mu$
iii. $A\cap \overline{B}=\oslash \equiv \overline{A}\cup B=\mu$
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: $A\subseteq B$ = $\overline{A}\cup B = \mu$
$X\in A \implies X\in B$
$X \notin A \vee X \in B$
Since A contains a subset of B, $\overline{A}$ contains all elements not in A, and thus not in B.  So
$X \notin A \vee X \in B = \mu$
iii: $A\cap \overline{B}=\oslash \equiv \overline{A}\cup B=\mu$
$X \in A \wedge X \notin B$
$X \notin A \vee X \in B$
Same as above - A contains a subset of B, \overline{A} contains all elements not in A and thus not in B, and so  $\overline{A}\cup B=\mu$
Please pardon the poor TeX syntax.  Still trying to learn it a bit.
 A: You only need to transform each statement using $\subseteq$ , $=$, $\cap$ and $\cup$ into their exact logical formulas of first-order-language that characterize their definitions, and start using logical equivalences.
Here i'm using $A^c$ as the U \ A.
i: $ A \subseteq B \equiv A \cap B^c = \emptyset $ 
$    A \subseteq B   \\
\equiv \forall x ( x \in A  \to x \in B )  \,\,\,\,\,\,\,\,\, \text{ [Definition of} \subseteq\text{]}\\\equiv 
y \in A \to y \in B  \, \, \,\,\,\,\,\text{  [  Universal Specification]}\\ 
\equiv \neg( y \in A \wedge   y \notin B ) \, \, \,\,\,\,\,\,\,\,\,\,\text{  [ Equivalence of implication]} \\\equiv (y \in A \wedge y\notin B) \leftrightarrow F \,\,\,\,\,\,\,\,\,\,\,\text{ [ F is used to represent  any contradiction in Set Theory]} \\\equiv  (y \in A \wedge  y \notin B) \leftrightarrow  (y \in \emptyset ) \,\,\,\,\,\, \text{[}  y \in \emptyset \text{ is a contradiction in Set Theory]}\\ \equiv   \forall x(x \in A \wedge  y \notin B \leftrightarrow x \in \emptyset ) \,\,\, \text{[Universal generalization, definition of equality between sets]} \\\equiv  A \cap B^c  = \emptyset    $
ii: $A \subseteq B \equiv  A^c \cup B = U $
$A \subseteq B  \\
\equiv \forall x(x \in A \to x \in B) \\
    \equiv y \in A \to y \in B \\
\equiv y \notin A  \vee  y \in B \\
\equiv y \notin A \vee y \in B  \leftrightarrow  T  \,\,\,\,\,\, \text{ [T is used to represent a valid formula in set theory]}  \\
\equiv y \notin A \vee y \in B \leftrightarrow  y \in U  \,\,\,\,\,\,\,\, \text{   [} y \in U\text{ is a valid formula in Set Theory ]} \\
\equiv  \forall x(x \notin A \vee x \in B \leftrightarrow x \in U ) \\
\equiv  A^c \cup B = U $
iii:  $A \cap B^c    = \emptyset  \equiv A^c \cup B = U $
$A \cap B^c = \emptyset  \\
\equiv \forall x( x \in A \wedge  x \notin B \leftrightarrow x \in \emptyset ) \\
\equiv y \in A \wedge  y \notin B \leftrightarrow y \in \emptyset  \\
\equiv ( y \in A \wedge y \notin B \leftrightarrow F ) \\
\equiv \neg( y \in A \wedge y \notin B) \leftrightarrow T \\
\equiv  y \notin A  \vee  y \in B \leftrightarrow y \in U \\
\equiv \forall x ( x \notin A \vee x \in B \leftrightarrow x \in U ) \\
\equiv A^c \vee B = U  $
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Tag}[1]{\text{(#1)}}
\newcommand{\c}[1]{\overline{#1}}
\newcommand{\univ}{\mu}
\newcommand{\false}{\text{false}}
\newcommand{\true}{\text{true}}
$My approach would be to translate your three statements from the set level to the logic level, by expanding the definitions, and then simplifying.
(In this answer $\;x\;$ ranges over our 'universe' $\;\univ\;$.)
For the first statement, we simply have the definition of $\;\subseteq\;$:
$$\calc
A \subseteq B
\calcop{\equiv}{expand definition of $\;\subseteq\;$}
\langle \forall x :: x \in A \Rightarrow x \in B \rangle
\calcop{\equiv}{logic: $\;\lnot P \lor Q\;$ is another way to write $\;P \Rightarrow Q\;$}
\tag{*} \langle \forall x :: x \not\in A \lor x \in B \rangle
\endcalc$$
For the second statement, we get
$$\calc
A \cap \c B = \varnothing
\calcop{\equiv}{expand definitions of $\;=, \cap, \c{\phantom\square}, \varnothing\;$}
\langle \forall x :: x \in A \land x \not\in B \;\equiv\; \false \rangle
\calcop{\equiv}{logic: simplify $\;P \equiv \false\;$ to $\;\lnot P\;$; DeMorgan}
\tag{*} \langle \forall x :: x \not\in A \lor x \in B \rangle
\endcalc$$
Finally, the third statement gives us
$$\calc
\c A \cup B = \univ
\calcop{\equiv}{expand definitions of $\;=, \cup, \c{\phantom\square}, \univ\;$}
\langle \forall x :: x \not\in A \lor x \in B \;\equiv\; \true \rangle
\calcop{\equiv}{logic: simplify $\;P \equiv \true\;$ to $\;P\;$}
\tag{*} \langle \forall x :: x \not\in A \lor x \in B \rangle
\endcalc$$
Since all $\Tag{*}$ are equivalent, the original statements also are equivalent.
