Nonlimit Ordinal Cofinal to One

Theorem
Let $x$ be a nonlimit nonempty ordinal.

Let $\operatorname{cof}$ denote the cofinal relation. Let $\le$ denote the subset relation.

Let $1$ denote the ordinal one.

Then:


 * $\operatorname{cof} \left({ x,1 }\right)$

Proof
Since $1 = 0^+$, $1$ is not a limit ordinal.

Therefore:


 * $0 < 1 \le x$ follows by Ordinal Less than Successor and No Ordinal Between Set and Successor.

$\operatorname{cof} \left({ x,1 }\right)$ follows by Condition for Cofinal Nonlimit Ordinals