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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 ) )