# 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 $\wedge$ denotes the meet 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:

 $\displaystyle a$ $=$ $\displaystyle r \vee \neg r$ by hypothesis $\displaystyle$ $=$ $\displaystyle \paren {s \vee \neg s} \wedge \paren {r \vee \neg r}$ Boolean Algebra: Axiom $(BA_2 \ 5)$ $\displaystyle$ $=$ $\displaystyle \paren {r \vee \neg r} \wedge \paren {s \vee \neg s}$ Boolean Algebra: Axiom $(BA_2 \ 1)$ $\displaystyle$ $=$ $\displaystyle s \vee \neg s$ Boolean Algebra: Axiom $(BA_2 \ 5)$ $\displaystyle$ $=$ $\displaystyle b$ 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.

$\blacksquare$