Metamath Proof Explorer


Theorem aciunf1

Description: Choice in an index union. (Contributed by Thierry Arnoux, 4-May-2020)

Ref Expression
Hypotheses aciunf1.0 φ A V
aciunf1.1 φ j A B W
Assertion aciunf1 φ f f : j A B 1-1 j A j × B k j A B 2 nd f k = k

Proof

Step Hyp Ref Expression
1 aciunf1.0 φ A V
2 aciunf1.1 φ j A B W
3 ssrab2 j A | B A
4 ssexg j A | B A A V j A | B V
5 3 1 4 sylancr φ j A | B V
6 rabid j j A | B j A B
7 6 bilani φ j j A | B j A B
8 7 simprd φ j j A | B B
9 nfrab1 _ j j A | B
10 7 simpld φ j j A | B j A
11 10 2 syldan φ j j A | B B W
12 5 8 9 11 aciunf1lem φ f f : j j A | B B 1-1 j j A | B j × B k j j A | B B 2 nd f k = k
13 eqidd φ f = f
14 nfv j φ
15 nfcv _ j A
16 nfrab1 _ j j A | B =
17 15 16 nfdif _ j A j A | B =
18 difrab j A | j A | B = = j A | ¬ B =
19 15 rabtru j A | = A
20 19 difeq1i j A | j A | B = = A j A | B =
21 truan ¬ B = ¬ B =
22 df-ne B ¬ B =
23 21 22 bitr4i ¬ B = B
24 23 rabbii j A | ¬ B = = j A | B
25 18 20 24 3eqtr3i A j A | B = = j A | B
26 25 a1i φ A j A | B = = j A | B
27 eqidd φ B = B
28 14 17 9 26 27 iuneq12df φ j A j A | B = B = j j A | B B
29 rabid j j A | B = j A B =
30 29 bilani φ j j A | B = j A B =
31 30 simprd φ j j A | B = B =
32 31 ralrimiva φ j j A | B = B =
33 16 iunxdif3 j j A | B = B = j A j A | B = B = j A B
34 32 33 syl φ j A j A | B = B = j A B
35 28 34 eqtr3d φ j j A | B B = j A B
36 eqidd φ j × B = j × B
37 14 17 9 26 36 iuneq12df φ j A j A | B = j × B = j j A | B j × B
38 31 xpeq2d φ j j A | B = j × B = j ×
39 xp0 j × =
40 38 39 eqtrdi φ j j A | B = j × B =
41 40 ralrimiva φ j j A | B = j × B =
42 16 iunxdif3 j j A | B = j × B = j A j A | B = j × B = j A j × B
43 41 42 syl φ j A j A | B = j × B = j A j × B
44 37 43 eqtr3d φ j j A | B j × B = j A j × B
45 13 35 44 f1eq123d φ f : j j A | B B 1-1 j j A | B j × B f : j A B 1-1 j A j × B
46 35 raleqdv φ k j j A | B B 2 nd f k = k k j A B 2 nd f k = k
47 45 46 anbi12d φ f : j j A | B B 1-1 j j A | B j × B k j j A | B B 2 nd f k = k f : j A B 1-1 j A j × B k j A B 2 nd f k = k
48 47 exbidv φ f f : j j A | B B 1-1 j j A | B j × B k j j A | B B 2 nd f k = k f f : j A B 1-1 j A j × B k j A B 2 nd f k = k
49 12 48 mpbid φ f f : j A B 1-1 j A j × B k j A B 2 nd f k = k