Metamath Proof Explorer


Theorem unielxp

Description: The membership relation for a Cartesian product is inherited by union. (Contributed by NM, 16-Sep-2006)

Ref Expression
Assertion unielxp A B × C A B × C

Proof

Step Hyp Ref Expression
1 elxp7 A B × C A V × V 1 st A B 2 nd A C
2 elvvuni A V × V A A
3 2 adantr A V × V 1 st A B 2 nd A C A A
4 simprl A A A V × V 1 st A B 2 nd A C A V × V
5 eleq2 x = A A x A A
6 eleq1 x = A x V × V A V × V
7 fveq2 x = A 1 st x = 1 st A
8 7 eleq1d x = A 1 st x B 1 st A B
9 fveq2 x = A 2 nd x = 2 nd A
10 9 eleq1d x = A 2 nd x C 2 nd A C
11 8 10 anbi12d x = A 1 st x B 2 nd x C 1 st A B 2 nd A C
12 6 11 anbi12d x = A x V × V 1 st x B 2 nd x C A V × V 1 st A B 2 nd A C
13 5 12 anbi12d x = A A x x V × V 1 st x B 2 nd x C A A A V × V 1 st A B 2 nd A C
14 13 spcegv A V × V A A A V × V 1 st A B 2 nd A C x A x x V × V 1 st x B 2 nd x C
15 4 14 mpcom A A A V × V 1 st A B 2 nd A C x A x x V × V 1 st x B 2 nd x C
16 eluniab A x | x V × V 1 st x B 2 nd x C x A x x V × V 1 st x B 2 nd x C
17 15 16 sylibr A A A V × V 1 st A B 2 nd A C A x | x V × V 1 st x B 2 nd x C
18 xp2 B × C = x V × V | 1 st x B 2 nd x C
19 df-rab x V × V | 1 st x B 2 nd x C = x | x V × V 1 st x B 2 nd x C
20 18 19 eqtri B × C = x | x V × V 1 st x B 2 nd x C
21 20 unieqi B × C = x | x V × V 1 st x B 2 nd x C
22 17 21 eleqtrrdi A A A V × V 1 st A B 2 nd A C A B × C
23 3 22 mpancom A V × V 1 st A B 2 nd A C A B × C
24 1 23 sylbi A B × C A B × C