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Metamath Proof Explorer

Theorem omsson

Description: Omega is a subset of On . (Contributed by NM, 13-Jun-1994) (Proof shortened by Andrew Salmon, 27-Aug-2011)

		Ref	Expression
	Assertion	omsson	$⊢ ω \subseteq On$

Step	Hyp	Ref	Expression
1		df-om	$⊢ ω = \{x \in On \| \forall y (Lim y \to x \in y)\}$
2	1	ssrab3	$⊢ ω \subseteq On$