Metamath Proof Explorer


Theorem fodomacn

Description: A version of fodom that doesn't require the Axiom of Choice ax-ac . If A has choice sequences of length B , then any surjection from A to B can be inverted to an injection the other way. (Contributed by Mario Carneiro, 31-Aug-2015)

Ref Expression
Assertion fodomacn A AC _ B F : A onto B B A

Proof

Step Hyp Ref Expression
1 foelrn F : A onto B x B y A x = F y
2 1 ralrimiva F : A onto B x B y A x = F y
3 fveq2 y = f x F y = F f x
4 3 eqeq2d y = f x x = F y x = F f x
5 4 acni3 A AC _ B x B y A x = F y f f : B A x B x = F f x
6 2 5 sylan2 A AC _ B F : A onto B f f : B A x B x = F f x
7 simpll A AC _ B F : A onto B f : B A x B x = F f x A AC _ B
8 acnrcl A AC _ B B V
9 7 8 syl A AC _ B F : A onto B f : B A x B x = F f x B V
10 simprl A AC _ B F : A onto B f : B A x B x = F f x f : B A
11 fveq2 f y = f z F f y = F f z
12 simprr A AC _ B F : A onto B f : B A x B x = F f x x B x = F f x
13 id x = y x = y
14 2fveq3 x = y F f x = F f y
15 13 14 eqeq12d x = y x = F f x y = F f y
16 15 rspccva x B x = F f x y B y = F f y
17 id x = z x = z
18 2fveq3 x = z F f x = F f z
19 17 18 eqeq12d x = z x = F f x z = F f z
20 19 rspccva x B x = F f x z B z = F f z
21 16 20 eqeqan12d x B x = F f x y B x B x = F f x z B y = z F f y = F f z
22 21 anandis x B x = F f x y B z B y = z F f y = F f z
23 12 22 sylan A AC _ B F : A onto B f : B A x B x = F f x y B z B y = z F f y = F f z
24 11 23 syl5ibr A AC _ B F : A onto B f : B A x B x = F f x y B z B f y = f z y = z
25 24 ralrimivva A AC _ B F : A onto B f : B A x B x = F f x y B z B f y = f z y = z
26 dff13 f : B 1-1 A f : B A y B z B f y = f z y = z
27 10 25 26 sylanbrc A AC _ B F : A onto B f : B A x B x = F f x f : B 1-1 A
28 f1dom2g B V A AC _ B f : B 1-1 A B A
29 9 7 27 28 syl3anc A AC _ B F : A onto B f : B A x B x = F f x B A
30 6 29 exlimddv A AC _ B F : A onto B B A
31 30 ex A AC _ B F : A onto B B A