Metamath Proof Explorer


Theorem iineq12dv

Description: Equality deduction for indexed intersection. (Contributed by Glauco Siliprandi, 26-Jun-2021) Remove DV conditions. (Revised by GG, 1-Sep-2025)

Ref Expression
Hypotheses iineq12dv.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
iineq12dv.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → 𝐶 = 𝐷 )
Assertion iineq12dv ( 𝜑 → ∩ 𝑥 ∈ 𝐴 𝐶 = ∩ 𝑥 ∈ 𝐵 𝐷 )

Proof

Step Hyp Ref Expression
1 iineq12dv.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 iineq12dv.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → 𝐶 = 𝐷 )
3 1 eleq2d ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
4 3 imbi1d ⊢ ( 𝜑 → ( ( 𝑥 ∈ 𝐴 → 𝑡 ∈ 𝐶 ) ↔ ( 𝑥 ∈ 𝐵 → 𝑡 ∈ 𝐶 ) ) )
5 4 ralbidv2 ⊢ ( 𝜑 → ( ∀ 𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 ↔ ∀ 𝑥 ∈ 𝐵 𝑡 ∈ 𝐶 ) )
6 5 abbidv ⊢ ( 𝜑 → { 𝑡 ∣ ∀ 𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 } = { 𝑡 ∣ ∀ 𝑥 ∈ 𝐵 𝑡 ∈ 𝐶 } )
7 df-iin ⊢ ∩ 𝑥 ∈ 𝐴 𝐶 = { 𝑡 ∣ ∀ 𝑥 ∈ 𝐴 𝑡 ∈ 𝐶 }
8 df-iin ⊢ ∩ 𝑥 ∈ 𝐵 𝐶 = { 𝑡 ∣ ∀ 𝑥 ∈ 𝐵 𝑡 ∈ 𝐶 }
9 6 7 8 3eqtr4g ⊢ ( 𝜑 → ∩ 𝑥 ∈ 𝐴 𝐶 = ∩ 𝑥 ∈ 𝐵 𝐶 )
10 2 iineq2dv ⊢ ( 𝜑 → ∩ 𝑥 ∈ 𝐵 𝐶 = ∩ 𝑥 ∈ 𝐵 𝐷 )
11 9 10 eqtrd ⊢ ( 𝜑 → ∩ 𝑥 ∈ 𝐴 𝐶 = ∩ 𝑥 ∈ 𝐵 𝐷 )