Answer
See step by step work for solution
Work Step by Step
To proof: $A \subseteq B$ if and only if $\bar B \subseteq \bar A$
Proof:
By the definition of a subset.
$A \subseteq B$ is logically equivalent with all elements for which:
if $x$ is an element of $A$, then $x$ is an element of $B$.
$A \subseteq B \equiv \forall x(x \in A \rightarrow x \in B)$
Use logical equivalence $(p\rightarrow q) \equiv (\neg q\rightarrow \neg p)$
$ \equiv \forall x(\neg (x \in B) \rightarrow \neg (x \in A))$
By the definition of the complement:
$ \equiv \forall x((x \in \bar B) \rightarrow (x \in \bar A))$
By the definition of the set
$\equiv \bar B \subseteq \bar A$
Since $A \subseteq B$ is logically equivalent with $\bar B \subseteq \bar A$,
$A \subseteq B \leftrightarrow \bar B \subseteq \bar A$ is a tautology,
$$A \subseteq B \leftrightarrow \bar B \subseteq \bar A$$
Hence, $A \subseteq B$ if and only if $\bar B \subseteq \bar A$
