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

Proof

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