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