Join with Complement is Top

Theorem

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

Then:

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

where $\vee$ denotes the join operation in $S$.

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

Proof

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

Then:

\(\ds a\)

\(=\)

\(\ds r \vee \neg r\)

by hypothesis

\(\ds \)

\(=\)

\(\ds \paren {s \vee \neg s} \wedge \paren {r \vee \neg r}\)

Boolean Algebra Axiom $(\text {BA}_2 5)$: Identity Elements

\(\ds \)

\(=\)

\(\ds \paren {r \vee \neg r} \wedge \paren {s \vee \neg s}\)

Boolean Algebra Axiom $(\text {BA}_2 1)$: Commutativity

\(\ds \)

\(=\)

\(\ds s \vee \neg s\)

Boolean Algebra Axiom $(\text {BA}_2 5)$: Identity Elements

\(\ds \)

\(=\)

\(\ds b\)

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.

$\blacksquare$

Sources

• 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.8$: Problems: $1 \ \text{B}$