Metamath Proof Explorer


Theorem tfrlem14

Description: Lemma for transfinite recursion. Assuming ax-rep , dom recs e.V <-> recs e. V , so since dom recs is an ordinal, it must be equal to On . (Contributed by NM, 14-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 tfrlem14 dom recs ( 𝐹 ) = On

Proof

Step Hyp Ref Expression
1 tfrlem.1 𝐴 = { 𝑓 ∣ ∃ 𝑥 ∈ On ( 𝑓 Fn 𝑥 ∧ ∀ 𝑦𝑥 ( 𝑓𝑦 ) = ( 𝐹 ‘ ( 𝑓𝑦 ) ) ) }
2 1 tfrlem13 ¬ recs ( 𝐹 ) ∈ V
3 1 tfrlem7 Fun recs ( 𝐹 )
4 funex ( ( Fun recs ( 𝐹 ) ∧ dom recs ( 𝐹 ) ∈ On ) → recs ( 𝐹 ) ∈ V )
5 3 4 mpan ( dom recs ( 𝐹 ) ∈ On → recs ( 𝐹 ) ∈ V )
6 2 5 mto ¬ dom recs ( 𝐹 ) ∈ On
7 1 tfrlem8 Ord dom recs ( 𝐹 )
8 ordeleqon ( Ord dom recs ( 𝐹 ) ↔ ( dom recs ( 𝐹 ) ∈ On ∨ dom recs ( 𝐹 ) = On ) )
9 7 8 mpbi ( dom recs ( 𝐹 ) ∈ On ∨ dom recs ( 𝐹 ) = On )
10 6 9 mtpor dom recs ( 𝐹 ) = On