# Element of Toset has at most One Immediate Predecessor

## Theorem

Let $\struct {S, \preceq}$ be a toset.

Let $a \in S$.

Then $a$ has at most one immediate predecessor.

## Proof

Let $b, b' \in S$ be immediate predecessors of $a$.

We have that $\preceq$ is a total ordering.

$b \preceq b'$

By virtue of $b$ being a immediate predecessor of $a$:

$\neg \exists c \in S: b \prec c \prec a$

However, since $b'$ is also an immediate predecessor:

$b' \prec a$

Hence, it cannot be the case that $b \prec b'$.

Since $b \preceq b'$, it follows that $b = b'$.

Hence the result.

$\blacksquare$