Join with Complement is Top

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Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 2.


$\exists \top \in S: \forall a \in S: a \vee \neg a = \top$

where $\wedge$ denotes the meet operation in $S$.

This element $\top$ is unique for any given $S$, and is named top.


Let $\exists r, s \in S: r \vee \neg r = a, \ s \vee \neg s = b$


\(\ds a\) \(=\) \(\ds r \vee \neg r\) by hypothesis
\(\ds \) \(=\) \(\ds \paren {s \vee \neg s} \wedge \paren {r \vee \neg r}\) Boolean Algebra: Axiom $(\text {BA}_2 \ 5)$
\(\ds \) \(=\) \(\ds \paren {r \vee \neg r} \wedge \paren {s \vee \neg s}\) Boolean Algebra: Axiom $(\text {BA}_2 \ 1)$
\(\ds \) \(=\) \(\ds s \vee \neg s\) Boolean Algebra: Axiom $(\text {BA}_2 \ 5)$
\(\ds \) \(=\) \(\ds b\) by hypothesis

Thus, whatever $r$ and $s$ may be:

$r \vee \neg r = s \vee \neg s$

This unique element can be assigned the symbol $\top$ and named top as required.