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Metamath Proof Explorer

Theorem r1rankidb

Description: Any set is a subset of the hierarchy of its rank. (Contributed by Mario Carneiro, 3-Jun-2013) (Revised by Mario Carneiro, 17-Nov-2014)

		Ref	Expression
	Assertion	r1rankidb	$⊢ A \in ⋃ R_{1} [On] \to A \subseteq R_{1} (rank (A))$

Proof

Step	Hyp	Ref	Expression
1		ssid	$⊢ rank (A) \subseteq rank (A)$
2		rankdmr1	$⊢ rank (A) \in dom R_{1}$
3		rankr1bg	$⊢ (A \in ⋃ R_{1} [On] \land rank (A) \in dom R_{1}) \to (A \subseteq R_{1} (rank (A)) \leftrightarrow rank (A) \subseteq rank (A))$
4	2 3	mpan2	$⊢ A \in ⋃ R_{1} [On] \to (A \subseteq R_{1} (rank (A)) \leftrightarrow rank (A) \subseteq rank (A))$
5	1 4	mpbiri	$⊢ A \in ⋃ R_{1} [On] \to A \subseteq R_{1} (rank (A))$