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 biimpi j j A | B j A B
8 7 adantl φ j j A | B j A B
9 8 simprd φ j j A | B B
10 nfrab1 _ j j A | B
11 8 simpld φ j j A | B j A
12 11 2 syldan φ j j A | B B W
13 5 9 10 12 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
14 eqidd φ f = f
15 nfv j φ
16 nfcv _ j A
17 nfrab1 _ j j A | B =
18 16 17 nfdif _ j A j A | B =
19 difrab j A | j A | B = = j A | ¬ B =
20 16 rabtru j A | = A
21 20 difeq1i j A | j A | B = = A j A | B =
22 truan ¬ B = ¬ B =
23 df-ne B ¬ B =
24 22 23 bitr4i ¬ B = B
25 24 rabbii j A | ¬ B = = j A | B
26 19 21 25 3eqtr3i A j A | B = = j A | B
27 26 a1i φ A j A | B = = j A | B
28 eqidd φ B = B
29 15 18 10 27 28 iuneq12df φ j A j A | B = B = j j A | B B
30 rabid j j A | B = j A B =
31 30 biimpi j j A | B = j A B =
32 31 adantl φ j j A | B = j A B =
33 32 simprd φ j j A | B = B =
34 33 ralrimiva φ j j A | B = B =
35 17 iunxdif3 j j A | B = B = j A j A | B = B = j A B
36 34 35 syl φ j A j A | B = B = j A B
37 29 36 eqtr3d φ j j A | B B = j A B
38 eqidd φ j × B = j × B
39 15 18 10 27 38 iuneq12df φ j A j A | B = j × B = j j A | B j × B
40 33 xpeq2d φ j j A | B = j × B = j ×
41 xp0 j × =
42 40 41 eqtrdi φ j j A | B = j × B =
43 42 ralrimiva φ j j A | B = j × B =
44 17 iunxdif3 j j A | B = j × B = j A j A | B = j × B = j A j × B
45 43 44 syl φ j A j A | B = j × B = j A j × B
46 39 45 eqtr3d φ j j A | B j × B = j A j × B
47 14 37 46 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
48 37 raleqdv φ k j j A | B B 2 nd f k = k k j A B 2 nd f k = k
49 47 48 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
50 49 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
51 13 50 mpbid φ f f : j A B 1-1 j A j × B k j A B 2 nd f k = k