Metamath Proof Explorer


Theorem domssex2

Description: A corollary of disjenex . If F is an injection from A to B then there is a right inverse g of F from B to a superset of A . (Contributed by Mario Carneiro, 7-Feb-2015) (Revised by Mario Carneiro, 24-Jun-2015)

Ref Expression
Assertion domssex2 F : A 1-1 B A V B W g g : B 1-1 V g F = I A

Proof

Step Hyp Ref Expression
1 f1f F : A 1-1 B F : A B
2 fex2 F : A B A V B W F V
3 1 2 syl3an1 F : A 1-1 B A V B W F V
4 f1stres 1 st B ran F × 𝒫 ran A : B ran F × 𝒫 ran A B ran F
5 difexg B W B ran F V
6 5 3ad2ant3 F : A 1-1 B A V B W B ran F V
7 snex 𝒫 ran A V
8 xpexg B ran F V 𝒫 ran A V B ran F × 𝒫 ran A V
9 6 7 8 sylancl F : A 1-1 B A V B W B ran F × 𝒫 ran A V
10 fex2 1 st B ran F × 𝒫 ran A : B ran F × 𝒫 ran A B ran F B ran F × 𝒫 ran A V B ran F V 1 st B ran F × 𝒫 ran A V
11 4 9 6 10 mp3an2i F : A 1-1 B A V B W 1 st B ran F × 𝒫 ran A V
12 unexg F V 1 st B ran F × 𝒫 ran A V F 1 st B ran F × 𝒫 ran A V
13 3 11 12 syl2anc F : A 1-1 B A V B W F 1 st B ran F × 𝒫 ran A V
14 cnvexg F 1 st B ran F × 𝒫 ran A V F 1 st B ran F × 𝒫 ran A -1 V
15 13 14 syl F : A 1-1 B A V B W F 1 st B ran F × 𝒫 ran A -1 V
16 eqid F 1 st B ran F × 𝒫 ran A -1 = F 1 st B ran F × 𝒫 ran A -1
17 16 domss2 F : A 1-1 B A V B W F 1 st B ran F × 𝒫 ran A -1 : B 1-1 onto ran F 1 st B ran F × 𝒫 ran A -1 A ran F 1 st B ran F × 𝒫 ran A -1 F 1 st B ran F × 𝒫 ran A -1 F = I A
18 17 simp1d F : A 1-1 B A V B W F 1 st B ran F × 𝒫 ran A -1 : B 1-1 onto ran F 1 st B ran F × 𝒫 ran A -1
19 f1of1 F 1 st B ran F × 𝒫 ran A -1 : B 1-1 onto ran F 1 st B ran F × 𝒫 ran A -1 F 1 st B ran F × 𝒫 ran A -1 : B 1-1 ran F 1 st B ran F × 𝒫 ran A -1
20 18 19 syl F : A 1-1 B A V B W F 1 st B ran F × 𝒫 ran A -1 : B 1-1 ran F 1 st B ran F × 𝒫 ran A -1
21 ssv ran F 1 st B ran F × 𝒫 ran A -1 V
22 f1ss F 1 st B ran F × 𝒫 ran A -1 : B 1-1 ran F 1 st B ran F × 𝒫 ran A -1 ran F 1 st B ran F × 𝒫 ran A -1 V F 1 st B ran F × 𝒫 ran A -1 : B 1-1 V
23 20 21 22 sylancl F : A 1-1 B A V B W F 1 st B ran F × 𝒫 ran A -1 : B 1-1 V
24 17 simp3d F : A 1-1 B A V B W F 1 st B ran F × 𝒫 ran A -1 F = I A
25 23 24 jca F : A 1-1 B A V B W F 1 st B ran F × 𝒫 ran A -1 : B 1-1 V F 1 st B ran F × 𝒫 ran A -1 F = I A
26 f1eq1 g = F 1 st B ran F × 𝒫 ran A -1 g : B 1-1 V F 1 st B ran F × 𝒫 ran A -1 : B 1-1 V
27 coeq1 g = F 1 st B ran F × 𝒫 ran A -1 g F = F 1 st B ran F × 𝒫 ran A -1 F
28 27 eqeq1d g = F 1 st B ran F × 𝒫 ran A -1 g F = I A F 1 st B ran F × 𝒫 ran A -1 F = I A
29 26 28 anbi12d g = F 1 st B ran F × 𝒫 ran A -1 g : B 1-1 V g F = I A F 1 st B ran F × 𝒫 ran A -1 : B 1-1 V F 1 st B ran F × 𝒫 ran A -1 F = I A
30 15 25 29 spcedv F : A 1-1 B A V B W g g : B 1-1 V g F = I A