Metamath Proof Explorer


Theorem frrlem9

Description: Lemma for founded recursion. Show that the founded recursive generator produces a function. Hypothesis three will be eliminated using different induction rules depending on if we use partial ordering or Infinity. (Contributed by Scott Fenton, 27-Aug-2022)

Ref Expression
Hypotheses frrlem9.1 B = f | x f Fn x x A y x Pred R A y x y x f y = y G f Pred R A y
frrlem9.2 F = frecs R A G
frrlem9.3 φ g B h B x g u x h v u = v
Assertion frrlem9 φ Fun F

Proof

Step Hyp Ref Expression
1 frrlem9.1 B = f | x f Fn x x A y x Pred R A y x y x f y = y G f Pred R A y
2 frrlem9.2 F = frecs R A G
3 frrlem9.3 φ g B h B x g u x h v u = v
4 eluni2 x u B g B x u g
5 df-br x F u x u F
6 1 2 frrlem5 F = B
7 6 eleq2i x u F x u B
8 5 7 bitri x F u x u B
9 df-br x g u x u g
10 9 rexbii g B x g u g B x u g
11 4 8 10 3bitr4i x F u g B x g u
12 eluni2 x v B h B x v h
13 df-br x F v x v F
14 6 eleq2i x v F x v B
15 13 14 bitri x F v x v B
16 df-br x h v x v h
17 16 rexbii h B x h v h B x v h
18 12 15 17 3bitr4i x F v h B x h v
19 11 18 anbi12i x F u x F v g B x g u h B x h v
20 reeanv g B h B x g u x h v g B x g u h B x h v
21 19 20 bitr4i x F u x F v g B h B x g u x h v
22 3 rexlimdvva φ g B h B x g u x h v u = v
23 21 22 syl5bi φ x F u x F v u = v
24 23 alrimiv φ v x F u x F v u = v
25 24 alrimivv φ x u v x F u x F v u = v
26 1 2 frrlem6 Rel F
27 dffun2 Fun F Rel F x u v x F u x F v u = v
28 26 27 mpbiran Fun F x u v x F u x F v u = v
29 25 28 sylibr φ Fun F