Metamath Proof Explorer


Theorem indi

Description: Distributive law for intersection over union. Exercise 10 of TakeutiZaring p. 17. (Contributed by NM, 30-Sep-2002) (Proof shortened by Andrew Salmon, 26-Jun-2011)

Ref Expression
Assertion indi A B C = A B A C

Proof

Step Hyp Ref Expression
1 andi x A x B x C x A x B x A x C
2 elin x A B x A x B
3 elin x A C x A x C
4 2 3 orbi12i x A B x A C x A x B x A x C
5 1 4 bitr4i x A x B x C x A B x A C
6 elun x B C x B x C
7 6 anbi2i x A x B C x A x B x C
8 elun x A B A C x A B x A C
9 5 7 8 3bitr4i x A x B C x A B A C
10 9 ineqri A B C = A B A C