Metamath Proof Explorer


Theorem iinin2

Description: Indexed intersection of intersection. Generalization of half of theorem "Distributive laws" in Enderton p. 30. Use intiin to recover Enderton's theorem. (Contributed by Mario Carneiro, 19-Mar-2015)

Ref Expression
Assertion iinin2 ( 𝐴 ≠ ∅ → ∩ 𝑥 ∈ 𝐴 ( 𝐵 ∩ 𝐶 ) = ( 𝐵 ∩ ∩ 𝑥 ∈ 𝐴 𝐶 ) )

Proof

Step Hyp Ref Expression
1 r19.28zv ⊢ ( 𝐴 ≠ ∅ → ( ∀ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶 ) ↔ ( 𝑦 ∈ 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ) ) )
2 elin ⊢ ( 𝑦 ∈ ( 𝐵 ∩ 𝐶 ) ↔ ( 𝑦 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶 ) )
3 2 ralbii ⊢ ( ∀ 𝑥 ∈ 𝐴 𝑦 ∈ ( 𝐵 ∩ 𝐶 ) ↔ ∀ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶 ) )
4 eliin ⊢ ( 𝑦 ∈ V → ( 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 𝐶 ↔ ∀ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ) )
5 4 elv ⊢ ( 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 𝐶 ↔ ∀ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 )
6 5 anbi2i ⊢ ( ( 𝑦 ∈ 𝐵 ∧ 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 𝐶 ) ↔ ( 𝑦 ∈ 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 𝑦 ∈ 𝐶 ) )
7 1 3 6 3bitr4g ⊢ ( 𝐴 ≠ ∅ → ( ∀ 𝑥 ∈ 𝐴 𝑦 ∈ ( 𝐵 ∩ 𝐶 ) ↔ ( 𝑦 ∈ 𝐵 ∧ 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 𝐶 ) ) )
8 eliin ⊢ ( 𝑦 ∈ V → ( 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 ( 𝐵 ∩ 𝐶 ) ↔ ∀ 𝑥 ∈ 𝐴 𝑦 ∈ ( 𝐵 ∩ 𝐶 ) ) )
9 8 elv ⊢ ( 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 ( 𝐵 ∩ 𝐶 ) ↔ ∀ 𝑥 ∈ 𝐴 𝑦 ∈ ( 𝐵 ∩ 𝐶 ) )
10 elin ⊢ ( 𝑦 ∈ ( 𝐵 ∩ ∩ 𝑥 ∈ 𝐴 𝐶 ) ↔ ( 𝑦 ∈ 𝐵 ∧ 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 𝐶 ) )
11 7 9 10 3bitr4g ⊢ ( 𝐴 ≠ ∅ → ( 𝑦 ∈ ∩ 𝑥 ∈ 𝐴 ( 𝐵 ∩ 𝐶 ) ↔ 𝑦 ∈ ( 𝐵 ∩ ∩ 𝑥 ∈ 𝐴 𝐶 ) ) )
12 11 eqrdv ⊢ ( 𝐴 ≠ ∅ → ∩ 𝑥 ∈ 𝐴 ( 𝐵 ∩ 𝐶 ) = ( 𝐵 ∩ ∩ 𝑥 ∈ 𝐴 𝐶 ) )