Metamath Proof Explorer


Theorem tfrlem13

Description: Lemma for transfinite recursion. If recs is a set function, then C is acceptable, and thus a subset of recs , but dom C is bigger than dom recs . This is a contradiction, so recs must be a proper class function. (Contributed by NM, 14-Aug-1994) (Revised by Mario Carneiro, 14-Nov-2014)

Ref Expression
Hypothesis tfrlem.1 A = f | x On f Fn x y x f y = F f y
Assertion tfrlem13 ¬ recs F V

Proof

Step Hyp Ref Expression
1 tfrlem.1 A = f | x On f Fn x y x f y = F f y
2 1 tfrlem8 Ord dom recs F
3 ordirr Ord dom recs F ¬ dom recs F dom recs F
4 2 3 ax-mp ¬ dom recs F dom recs F
5 eqid recs F dom recs F F recs F = recs F dom recs F F recs F
6 1 5 tfrlem12 recs F V recs F dom recs F F recs F A
7 elssuni recs F dom recs F F recs F A recs F dom recs F F recs F A
8 1 recsfval recs F = A
9 7 8 sseqtrrdi recs F dom recs F F recs F A recs F dom recs F F recs F recs F
10 dmss recs F dom recs F F recs F recs F dom recs F dom recs F F recs F dom recs F
11 6 9 10 3syl recs F V dom recs F dom recs F F recs F dom recs F
12 2 a1i recs F V Ord dom recs F
13 dmexg recs F V dom recs F V
14 elon2 dom recs F On Ord dom recs F dom recs F V
15 12 13 14 sylanbrc recs F V dom recs F On
16 sucidg dom recs F On dom recs F suc dom recs F
17 15 16 syl recs F V dom recs F suc dom recs F
18 1 5 tfrlem10 dom recs F On recs F dom recs F F recs F Fn suc dom recs F
19 fndm recs F dom recs F F recs F Fn suc dom recs F dom recs F dom recs F F recs F = suc dom recs F
20 15 18 19 3syl recs F V dom recs F dom recs F F recs F = suc dom recs F
21 17 20 eleqtrrd recs F V dom recs F dom recs F dom recs F F recs F
22 11 21 sseldd recs F V dom recs F dom recs F
23 4 22 mto ¬ recs F V