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