Theorem elom

Description: Membership in omega. The left conjunct can be eliminated if we assume the Axiom of Infinity; see elom3 . (Contributed by NM, 15-May-1994)

	Ref	Expression
Assertion	elom	$$⊢A\in \mathrm{\omega }\leftrightarrow \left(A\in \mathrm{On}\wedge \forall x\left(\mathrm{Lim}x\to A\in x\right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		eleq1	$$⊢y=A\to \left(y\in x\leftrightarrow A\in x\right)$$
2	1	imbi2d	$$⊢y=A\to \left(\left(\mathrm{Lim}x\to y\in x\right)\leftrightarrow \left(\mathrm{Lim}x\to A\in x\right)\right)$$
3	2	albidv	$$⊢y=A\to \left(\forall x\left(\mathrm{Lim}x\to y\in x\right)\leftrightarrow \forall x\left(\mathrm{Lim}x\to A\in x\right)\right)$$
4		df-om	$$⊢\mathrm{\omega }=\left\{y\in \mathrm{On}|\forall x\left(\mathrm{Lim}x\to y\in x\right)\right\}$$
5	3 4	elrab2	$$⊢A\in \mathrm{\omega }\leftrightarrow \left(A\in \mathrm{On}\wedge \forall x\left(\mathrm{Lim}x\to A\in x\right)\right)$$