Metamath Proof Explorer


Theorem tfrlem8

Description: Lemma for transfinite recursion. The domain of recs is an ordinal. (Contributed by NM, 14-Aug-1994) (Proof shortened by Alan Sare, 11-Mar-2008)

Ref Expression
Hypothesis tfrlem.1 𝐴 = { 𝑓 ∣ ∃ 𝑥 ∈ On ( 𝑓 Fn 𝑥 ∧ ∀ 𝑦𝑥 ( 𝑓𝑦 ) = ( 𝐹 ‘ ( 𝑓𝑦 ) ) ) }
Assertion tfrlem8 Ord dom recs ( 𝐹 )

Proof

Step Hyp Ref Expression
1 tfrlem.1 𝐴 = { 𝑓 ∣ ∃ 𝑥 ∈ On ( 𝑓 Fn 𝑥 ∧ ∀ 𝑦𝑥 ( 𝑓𝑦 ) = ( 𝐹 ‘ ( 𝑓𝑦 ) ) ) }
2 1 tfrlem3 𝐴 = { 𝑔 ∣ ∃ 𝑧 ∈ On ( 𝑔 Fn 𝑧 ∧ ∀ 𝑤𝑧 ( 𝑔𝑤 ) = ( 𝐹 ‘ ( 𝑔𝑤 ) ) ) }
3 2 abeq2i ( 𝑔𝐴 ↔ ∃ 𝑧 ∈ On ( 𝑔 Fn 𝑧 ∧ ∀ 𝑤𝑧 ( 𝑔𝑤 ) = ( 𝐹 ‘ ( 𝑔𝑤 ) ) ) )
4 fndm ( 𝑔 Fn 𝑧 → dom 𝑔 = 𝑧 )
5 4 adantr ( ( 𝑔 Fn 𝑧 ∧ ∀ 𝑤𝑧 ( 𝑔𝑤 ) = ( 𝐹 ‘ ( 𝑔𝑤 ) ) ) → dom 𝑔 = 𝑧 )
6 5 eleq1d ( ( 𝑔 Fn 𝑧 ∧ ∀ 𝑤𝑧 ( 𝑔𝑤 ) = ( 𝐹 ‘ ( 𝑔𝑤 ) ) ) → ( dom 𝑔 ∈ On ↔ 𝑧 ∈ On ) )
7 6 biimprcd ( 𝑧 ∈ On → ( ( 𝑔 Fn 𝑧 ∧ ∀ 𝑤𝑧 ( 𝑔𝑤 ) = ( 𝐹 ‘ ( 𝑔𝑤 ) ) ) → dom 𝑔 ∈ On ) )
8 7 rexlimiv ( ∃ 𝑧 ∈ On ( 𝑔 Fn 𝑧 ∧ ∀ 𝑤𝑧 ( 𝑔𝑤 ) = ( 𝐹 ‘ ( 𝑔𝑤 ) ) ) → dom 𝑔 ∈ On )
9 3 8 sylbi ( 𝑔𝐴 → dom 𝑔 ∈ On )
10 eleq1a ( dom 𝑔 ∈ On → ( 𝑧 = dom 𝑔𝑧 ∈ On ) )
11 9 10 syl ( 𝑔𝐴 → ( 𝑧 = dom 𝑔𝑧 ∈ On ) )
12 11 rexlimiv ( ∃ 𝑔𝐴 𝑧 = dom 𝑔𝑧 ∈ On )
13 12 abssi { 𝑧 ∣ ∃ 𝑔𝐴 𝑧 = dom 𝑔 } ⊆ On
14 ssorduni ( { 𝑧 ∣ ∃ 𝑔𝐴 𝑧 = dom 𝑔 } ⊆ On → Ord { 𝑧 ∣ ∃ 𝑔𝐴 𝑧 = dom 𝑔 } )
15 13 14 ax-mp Ord { 𝑧 ∣ ∃ 𝑔𝐴 𝑧 = dom 𝑔 }
16 1 recsfval recs ( 𝐹 ) = 𝐴
17 16 dmeqi dom recs ( 𝐹 ) = dom 𝐴
18 dmuni dom 𝐴 = 𝑔𝐴 dom 𝑔
19 vex 𝑔 ∈ V
20 19 dmex dom 𝑔 ∈ V
21 20 dfiun2 𝑔𝐴 dom 𝑔 = { 𝑧 ∣ ∃ 𝑔𝐴 𝑧 = dom 𝑔 }
22 17 18 21 3eqtri dom recs ( 𝐹 ) = { 𝑧 ∣ ∃ 𝑔𝐴 𝑧 = dom 𝑔 }
23 ordeq ( dom recs ( 𝐹 ) = { 𝑧 ∣ ∃ 𝑔𝐴 𝑧 = dom 𝑔 } → ( Ord dom recs ( 𝐹 ) ↔ Ord { 𝑧 ∣ ∃ 𝑔𝐴 𝑧 = dom 𝑔 } ) )
24 22 23 ax-mp ( Ord dom recs ( 𝐹 ) ↔ Ord { 𝑧 ∣ ∃ 𝑔𝐴 𝑧 = dom 𝑔 } )
25 15 24 mpbir Ord dom recs ( 𝐹 )