Metamath Proof Explorer


Theorem tfrlem16

Description: Lemma for finite recursion. Without assuming ax-rep , we can show that the domain of the constructed function is a limit ordinal, and hence contains all the finite ordinals. (Contributed by Mario Carneiro, 14-Nov-2014)

Ref Expression
Hypothesis tfrlem.1 A=f|xOnfFnxyxfy=Ffy
Assertion tfrlem16 LimdomrecsF

Proof

Step Hyp Ref Expression
1 tfrlem.1 A=f|xOnfFnxyxfy=Ffy
2 1 tfrlem8 OrddomrecsF
3 ordzsl OrddomrecsFdomrecsF=zOndomrecsF=suczLimdomrecsF
4 2 3 mpbi domrecsF=zOndomrecsF=suczLimdomrecsF
5 res0 recsF=
6 0ex V
7 5 6 eqeltri recsFV
8 0elon On
9 1 tfrlem15 OndomrecsFrecsFV
10 8 9 ax-mp domrecsFrecsFV
11 7 10 mpbir domrecsF
12 11 n0ii ¬domrecsF=
13 12 pm2.21i domrecsF=LimdomrecsF
14 1 tfrlem13 ¬recsFV
15 simpr zOndomrecsF=suczdomrecsF=sucz
16 df-suc sucz=zz
17 15 16 eqtrdi zOndomrecsF=suczdomrecsF=zz
18 17 reseq2d zOndomrecsF=suczrecsFdomrecsF=recsFzz
19 1 tfrlem6 RelrecsF
20 resdm RelrecsFrecsFdomrecsF=recsF
21 19 20 ax-mp recsFdomrecsF=recsF
22 resundi recsFzz=recsFzrecsFz
23 18 21 22 3eqtr3g zOndomrecsF=suczrecsF=recsFzrecsFz
24 vex zV
25 24 sucid zsucz
26 25 15 eleqtrrid zOndomrecsF=suczzdomrecsF
27 1 tfrlem9a zdomrecsFrecsFzV
28 26 27 syl zOndomrecsF=suczrecsFzV
29 snex zrecsFzV
30 1 tfrlem7 FunrecsF
31 funressn FunrecsFrecsFzzrecsFz
32 30 31 ax-mp recsFzzrecsFz
33 29 32 ssexi recsFzV
34 unexg recsFzVrecsFzVrecsFzrecsFzV
35 28 33 34 sylancl zOndomrecsF=suczrecsFzrecsFzV
36 23 35 eqeltrd zOndomrecsF=suczrecsFV
37 36 rexlimiva zOndomrecsF=suczrecsFV
38 14 37 mto ¬zOndomrecsF=sucz
39 38 pm2.21i zOndomrecsF=suczLimdomrecsF
40 id LimdomrecsFLimdomrecsF
41 13 39 40 3jaoi domrecsF=zOndomrecsF=suczLimdomrecsFLimdomrecsF
42 4 41 ax-mp LimdomrecsF