Double Complement of a set proof

Let $A$ be a subset of some universe $U$.

We'll prove this by proving that the two sets are subsets of each other (and must therefore be equal). We'll use nothing but the definition that if $x \in X$, then $x \notin X^C$ (and what clearly follows: if $x \notin X$, then $x \in X^C$).

Take $a \in A$. Then $a \notin A^C$. Then $a \in (A^C)^C$, by definition of the complement. So $A \subseteq (A^C)^C$.

Now take $a \in (A^C)^C$. That means $a \notin A^C$. Therefore $a \in A$. So $(A^C)^C \subseteq A$.

As $A \subseteq (A^C)^C$ and $(A^C)^C \subseteq A$, we conclude that $A = (A^C)^C$.

For other techniques, see here. This is, in some sense, a duplicate question.

I know it's old, but here's how I solved it using logical equivalences. For people who coming from the future, I was required to solve it like this.

We know that: $A=\{x|x\in A\}$, which reads: there is an x in the set of A. And we know that the complement of A is $A^c=\{x| x\notin A\}$, which reads: there is an x which is not in A.

We start with $(A^c)^c = \{x|x\notin A^c\}$ is read as: there is an x that is not in the complement set of A. That means that if an x is not in the complement set of A (and the complement set of A is the set that doesn't have A), then it must be in A. All we do is use negation rules.

$(A^c)^c = \{x|x\notin A^c\}$

$(A^c)^c = \{x|\neg (x \in A^c)\}$

$(A^c)^c = \{x|\neg (x \notin A)\}$

$(A^c)^c = \{x|\neg (\neg (x \in A))\}$

$(A^c)^c = \{x| x \in A\}$

Therefore

$(A^c)^c=A$