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 ∧ ∀ 𝑦 ∈ 𝑥 ( 𝑓 ‘ 𝑦 ) = ( 𝐹 ‘ ( 𝑓 ↾ 𝑦 ) ) ) }
Assertion tfrlem14 dom recs ( 𝐹 ) = On

Proof

Step Hyp Ref Expression
1 tfrlem.1 ⊢ 𝐴 = { 𝑓 ∣ ∃ 𝑥 ∈ On ( 𝑓 Fn 𝑥 ∧ ∀ 𝑦 ∈ 𝑥 ( 𝑓 ‘ 𝑦 ) = ( 𝐹 ‘ ( 𝑓 ↾ 𝑦 ) ) ) }
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