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Metamath Proof Explorer

Theorem recsfnon

Description: Strong transfinite recursion defines a function on ordinals. (Contributed by Stefan O'Rear, 18-Jan-2015)

		Ref	Expression
	Assertion	recsfnon	$⊢ recs (F) Fn On$

Step	Hyp	Ref	Expression
1		eqid	$⊢ recs (F) = recs (F)$
2	1	tfr1	$⊢ recs (F) Fn On$