Metamath Proof Explorer


Theorem tfrlem6

Description: Lemma for transfinite recursion. The union of all acceptable functions is a relation. (Contributed by NM, 8-Aug-1994) (Revised by Mario Carneiro, 9-May-2015)

Ref Expression
Hypothesis tfrlem.1 𝐴 = { 𝑓 ∣ ∃ 𝑥 ∈ On ( 𝑓 Fn 𝑥 ∧ ∀ 𝑦𝑥 ( 𝑓𝑦 ) = ( 𝐹 ‘ ( 𝑓𝑦 ) ) ) }
Assertion tfrlem6 Rel recs ( 𝐹 )

Proof

Step Hyp Ref Expression
1 tfrlem.1 𝐴 = { 𝑓 ∣ ∃ 𝑥 ∈ On ( 𝑓 Fn 𝑥 ∧ ∀ 𝑦𝑥 ( 𝑓𝑦 ) = ( 𝐹 ‘ ( 𝑓𝑦 ) ) ) }
2 reluni ( Rel 𝐴 ↔ ∀ 𝑔𝐴 Rel 𝑔 )
3 1 tfrlem4 ( 𝑔𝐴 → Fun 𝑔 )
4 funrel ( Fun 𝑔 → Rel 𝑔 )
5 3 4 syl ( 𝑔𝐴 → Rel 𝑔 )
6 2 5 mprgbir Rel 𝐴
7 1 recsfval recs ( 𝐹 ) = 𝐴
8 7 releqi ( Rel recs ( 𝐹 ) ↔ Rel 𝐴 )
9 6 8 mpbir Rel recs ( 𝐹 )