Theorem recsfnon

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

	Ref	Expression
Assertion	recsfnon	$$⊢\mathrm{recs}\left(F\right)Fn\mathrm{On}$$

Proof

Step	Hyp	Ref	Expression
1		eqid	$$⊢\mathrm{recs}\left(F\right)=\mathrm{recs}\left(F\right)$$
2	1	tfr1	$$⊢\mathrm{recs}\left(F\right)Fn\mathrm{On}$$