Metamath Proof Explorer


Theorem tfrlem9a

Description: Lemma for transfinite recursion. Without using ax-rep , show that all the restrictions of recs are sets. (Contributed by Mario Carneiro, 16-Nov-2014)

Ref Expression
Hypothesis tfrlem.1 A=f|xOnfFnxyxfy=Ffy
Assertion tfrlem9a BdomrecsFrecsFBV

Proof

Step Hyp Ref Expression
1 tfrlem.1 A=f|xOnfFnxyxfy=Ffy
2 1 tfrlem7 FunrecsF
3 funfvop FunrecsFBdomrecsFBrecsFBrecsF
4 2 3 mpan BdomrecsFBrecsFBrecsF
5 1 recsfval recsF=A
6 5 eleq2i BrecsFBrecsFBrecsFBA
7 eluni BrecsFBAgBrecsFBggA
8 6 7 bitri BrecsFBrecsFgBrecsFBggA
9 4 8 sylib BdomrecsFgBrecsFBggA
10 simprr BdomrecsFBrecsFBggAgA
11 vex gV
12 1 11 tfrlem3a gAzOngFnzazga=Fga
13 10 12 sylib BdomrecsFBrecsFBggAzOngFnzazga=Fga
14 2 a1i BdomrecsFBrecsFBggAzOngFnzFunrecsF
15 simplrr BdomrecsFBrecsFBggAzOngFnzgA
16 elssuni gAgA
17 15 16 syl BdomrecsFBrecsFBggAzOngFnzgA
18 17 5 sseqtrrdi BdomrecsFBrecsFBggAzOngFnzgrecsF
19 fndm gFnzdomg=z
20 19 ad2antll BdomrecsFBrecsFBggAzOngFnzdomg=z
21 simprl BdomrecsFBrecsFBggAzOngFnzzOn
22 20 21 eqeltrd BdomrecsFBrecsFBggAzOngFnzdomgOn
23 eloni domgOnOrddomg
24 22 23 syl BdomrecsFBrecsFBggAzOngFnzOrddomg
25 simpll BdomrecsFBrecsFBggAzOngFnzBdomrecsF
26 fvexd BdomrecsFBrecsFBggAzOngFnzrecsFBV
27 simplrl BdomrecsFBrecsFBggAzOngFnzBrecsFBg
28 df-br BgrecsFBBrecsFBg
29 27 28 sylibr BdomrecsFBrecsFBggAzOngFnzBgrecsFB
30 breldmg BdomrecsFrecsFBVBgrecsFBBdomg
31 25 26 29 30 syl3anc BdomrecsFBrecsFBggAzOngFnzBdomg
32 ordelss OrddomgBdomgBdomg
33 24 31 32 syl2anc BdomrecsFBrecsFBggAzOngFnzBdomg
34 fun2ssres FunrecsFgrecsFBdomgrecsFB=gB
35 14 18 33 34 syl3anc BdomrecsFBrecsFBggAzOngFnzrecsFB=gB
36 11 resex gBV
37 36 a1i BdomrecsFBrecsFBggAzOngFnzgBV
38 35 37 eqeltrd BdomrecsFBrecsFBggAzOngFnzrecsFBV
39 38 expr BdomrecsFBrecsFBggAzOngFnzrecsFBV
40 39 adantrd BdomrecsFBrecsFBggAzOngFnzazga=FgarecsFBV
41 40 rexlimdva BdomrecsFBrecsFBggAzOngFnzazga=FgarecsFBV
42 13 41 mpd BdomrecsFBrecsFBggArecsFBV
43 9 42 exlimddv BdomrecsFrecsFBV