TITLE: How to show the double negation law in Boolean algebra
QUESTION [5 upvotes]: I want to show the double negation law
$\lnot \lnot s = s \tag{0}$
where $s$ is an element of Boolean algebra.
And $\lnot$ is defined as $\lnot s := s \rightarrow 0$.
Boolean algebra is a Heyting algebra with following two laws.
$s \lor \lnot s = 1 \tag{1}$
$s \land \lnot s = 0 \tag{2}$
And Heyting algebra is a lattice with $0$, $1$, and $\rightarrow$.
I want to show (0) using only (1), (2), and the nature of Heyting algebra.
I could show the following law using the nature of Heyting algebra.
$s \le \lnot \lnot s \tag{3}$
To prove (0), the following law need to be proved. But I feel it is difficult.
$s \ge \lnot \lnot s \tag{4}$

REPLY [4 votes]: From (1) and (2) applied to $\neg s$, that is,
$$\neg s \vee \neg\neg s = 1 \qquad\text{and}\qquad \neg s \wedge \neg\neg s = 0$$
you get that $\neg\neg s$ is a complement of $\neg s$ in the underlying distributive lattice.
From (1) and (2), $s$ is also a complement of $\neg s$.  
In a distributive lattice, an element has, at most, one complement.
Therefore $s = \neg\neg s$.