Metamath Proof Explorer


Theorem dfac11

Description: The right-hand side of this theorem (compare with ac4 ), sometimes known as the "axiom of multiple choice", is a choice equivalent. Curiously, this statement cannot be proved without ax-reg , despite not mentioning the cumulative hierarchy in any way as most consequences of regularity do.

This is definition (MC) of Schechter p. 141.EDITORIAL: the proof is not original with me of course but I lost my reference sometime after writing it.

A multiple choice function allows any total order to be extended to a choice function, which in turn defines a well-ordering. Since a well-ordering on a set defines a simple ordering of the power set, this allows the trivial well-ordering of the empty set to be transfinitely bootstrapped up the cumulative hierarchy to any desired level. (Contributed by Stefan O'Rear, 20-Jan-2015) (Revised by Stefan O'Rear, 1-Jun-2015)

Ref Expression
Assertion dfac11 CHOICE x f z x z f z 𝒫 z Fin

Proof

Step Hyp Ref Expression
1 dfac3 CHOICE a c d a d c d d
2 raleq a = x d a d c d d d x d c d d
3 2 exbidv a = x c d a d c d d c d x d c d d
4 3 cbvalvw a c d a d c d d x c d x d c d d
5 neeq1 d = z d z
6 fveq2 d = z c d = c z
7 id d = z d = z
8 6 7 eleq12d d = z c d d c z z
9 5 8 imbi12d d = z d c d d z c z z
10 9 cbvralvw d x d c d d z x z c z z
11 fveq2 b = z c b = c z
12 11 sneqd b = z c b = c z
13 eqid b x c b = b x c b
14 snex c z V
15 12 13 14 fvmpt z x b x c b z = c z
16 15 3ad2ant1 z x z c z z b x c b z = c z
17 simp3 z x z c z z c z z
18 17 snssd z x z c z z c z z
19 14 elpw c z 𝒫 z c z z
20 18 19 sylibr z x z c z z c z 𝒫 z
21 snfi c z Fin
22 21 a1i z x z c z z c z Fin
23 20 22 elind z x z c z z c z 𝒫 z Fin
24 fvex c z V
25 24 snnz c z
26 25 a1i z x z c z z c z
27 eldifsn c z 𝒫 z Fin c z 𝒫 z Fin c z
28 23 26 27 sylanbrc z x z c z z c z 𝒫 z Fin
29 16 28 eqeltrd z x z c z z b x c b z 𝒫 z Fin
30 29 3exp z x z c z z b x c b z 𝒫 z Fin
31 30 a2d z x z c z z z b x c b z 𝒫 z Fin
32 31 ralimia z x z c z z z x z b x c b z 𝒫 z Fin
33 10 32 sylbi d x d c d d z x z b x c b z 𝒫 z Fin
34 vex x V
35 34 mptex b x c b V
36 fveq1 f = b x c b f z = b x c b z
37 36 eleq1d f = b x c b f z 𝒫 z Fin b x c b z 𝒫 z Fin
38 37 imbi2d f = b x c b z f z 𝒫 z Fin z b x c b z 𝒫 z Fin
39 38 ralbidv f = b x c b z x z f z 𝒫 z Fin z x z b x c b z 𝒫 z Fin
40 35 39 spcev z x z b x c b z 𝒫 z Fin f z x z f z 𝒫 z Fin
41 33 40 syl d x d c d d f z x z f z 𝒫 z Fin
42 41 exlimiv c d x d c d d f z x z f z 𝒫 z Fin
43 42 alimi x c d x d c d d x f z x z f z 𝒫 z Fin
44 4 43 sylbi a c d a d c d d x f z x z f z 𝒫 z Fin
45 1 44 sylbi CHOICE x f z x z f z 𝒫 z Fin
46 fvex R1 rank a V
47 46 pwex 𝒫 R1 rank a V
48 raleq x = 𝒫 R1 rank a z x z f z 𝒫 z Fin z 𝒫 R1 rank a z f z 𝒫 z Fin
49 48 exbidv x = 𝒫 R1 rank a f z x z f z 𝒫 z Fin f z 𝒫 R1 rank a z f z 𝒫 z Fin
50 47 49 spcv x f z x z f z 𝒫 z Fin f z 𝒫 R1 rank a z f z 𝒫 z Fin
51 rankon rank a On
52 51 a1i z 𝒫 R1 rank a z f z 𝒫 z Fin rank a On
53 id z 𝒫 R1 rank a z f z 𝒫 z Fin z 𝒫 R1 rank a z f z 𝒫 z Fin
54 52 53 aomclem8 z 𝒫 R1 rank a z f z 𝒫 z Fin b b We R1 rank a
55 54 exlimiv f z 𝒫 R1 rank a z f z 𝒫 z Fin b b We R1 rank a
56 vex a V
57 r1rankid a V a R1 rank a
58 wess a R1 rank a b We R1 rank a b We a
59 58 eximdv a R1 rank a b b We R1 rank a b b We a
60 56 57 59 mp2b b b We R1 rank a b b We a
61 50 55 60 3syl x f z x z f z 𝒫 z Fin b b We a
62 61 alrimiv x f z x z f z 𝒫 z Fin a b b We a
63 dfac8 CHOICE a b b We a
64 62 63 sylibr x f z x z f z 𝒫 z Fin CHOICE
65 45 64 impbii CHOICE x f z x z f z 𝒫 z Fin