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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

Proof

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