= Solution
Suppose every <regular open element> is complemented. In a <Heyting algebra>, a complemented element is regular: if $a\vee b=1$ and $a\wedge b=0$, then $b=\neg a$ and $a=\neg b$. Finite joins of complemented elements are complemented. Hence $\neg\neg a\vee\neg\neg b$ is regular, and it is the least regular upper bound of $a\vee b$. Therefore
$$
\neg\neg(a\vee b)=\neg\neg a\vee\neg\neg b.
$$
This proves \b[(ii) implies (iii)]. The identification of the <regular open elements> with interiors of closures was established in the preceding part.
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