In a the poset \\( P(X) \\), the complement of the top element \\( X \\) defines negation,
$$
\neg(X): P(X) \rightarrow P(X) \\
x \mapsto X \setminus x.
$$

Proof.

Let \\( x=X \\), then \\(X \setminus X = \varnothing \\).
Let \\( x=\varnothing \\), then \\(X \setminus \varnothing = X\\).
Finally, let \\( x=y,\\) for some \\(y, \\, \varnothing \subsetneq y \subsetneq X \\), then there will exist some subset \\(z, \varnothing \subsetneq z \subsetneq X \\) where \\(z \neq y \\) and \\( z = X \setminus y \\).