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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. $s \ge \lnot \lnot s \tag{4}$

konyonyo
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1 Answers1

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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 $s = \neg\neg s$.

amrsa
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