Metamath Proof Explorer


Theorem metakunt17

Description: The union of three disjoint bijections is a bijection. (Contributed by metakunt, 28-May-2024)

Ref Expression
Hypotheses metakunt17.1 φ G : A 1-1 onto X
metakunt17.2 φ H : B 1-1 onto Y
metakunt17.3 φ I : C 1-1 onto Z
metakunt17.4 φ A B =
metakunt17.5 φ A C =
metakunt17.6 φ B C =
metakunt17.7 φ X Y =
metakunt17.8 φ X Z =
metakunt17.9 φ Y Z =
metakunt17.10 φ F = G H I
metakunt17.11 φ D = A B C
metakunt17.12 φ W = X Y Z
Assertion metakunt17 φ F : D 1-1 onto W

Proof

Step Hyp Ref Expression
1 metakunt17.1 φ G : A 1-1 onto X
2 metakunt17.2 φ H : B 1-1 onto Y
3 metakunt17.3 φ I : C 1-1 onto Z
4 metakunt17.4 φ A B =
5 metakunt17.5 φ A C =
6 metakunt17.6 φ B C =
7 metakunt17.7 φ X Y =
8 metakunt17.8 φ X Z =
9 metakunt17.9 φ Y Z =
10 metakunt17.10 φ F = G H I
11 metakunt17.11 φ D = A B C
12 metakunt17.12 φ W = X Y Z
13 4 7 jca φ A B = X Y =
14 1 2 13 jca31 φ G : A 1-1 onto X H : B 1-1 onto Y A B = X Y =
15 f1oun G : A 1-1 onto X H : B 1-1 onto Y A B = X Y = G H : A B 1-1 onto X Y
16 14 15 syl φ G H : A B 1-1 onto X Y
17 indir A B C = A C B C
18 5 6 uneq12d φ A C B C =
19 0un =
20 19 a1i φ =
21 18 20 eqtrd φ A C B C =
22 17 21 syl5eq φ A B C =
23 indir X Y Z = X Z Y Z
24 8 9 uneq12d φ X Z Y Z =
25 24 20 eqtrd φ X Z Y Z =
26 23 25 syl5eq φ X Y Z =
27 22 26 jca φ A B C = X Y Z =
28 16 3 27 jca31 φ G H : A B 1-1 onto X Y I : C 1-1 onto Z A B C = X Y Z =
29 f1oun G H : A B 1-1 onto X Y I : C 1-1 onto Z A B C = X Y Z = G H I : A B C 1-1 onto X Y Z
30 28 29 syl φ G H I : A B C 1-1 onto X Y Z
31 10 11 12 f1oeq123d φ F : D 1-1 onto W G H I : A B C 1-1 onto X Y Z
32 30 31 mpbird φ F : D 1-1 onto W