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 ) )