Metamath Proof Explorer


Theorem rexcom

Description: Commutation of restricted existential quantifiers. (Contributed by NM, 19-Nov-1995) (Revised by Mario Carneiro, 14-Oct-2016) (Proof shortened by BJ, 26-Aug-2023) (Proof shortened by Wolf Lammen, 8-Dec-2024)

Ref Expression
Assertion rexcom ( ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∃ 𝑦 ∈ 𝐵 ∃ 𝑥 ∈ 𝐴 𝜑 )

Proof

Step Hyp Ref Expression
1 ralcom ⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ∀ 𝑦 ∈ 𝐵 ∀ 𝑥 ∈ 𝐴 ¬ 𝜑 )
2 ralnex2 ⊢ ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ¬ ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 )
3 ralnex2 ⊢ ( ∀ 𝑦 ∈ 𝐵 ∀ 𝑥 ∈ 𝐴 ¬ 𝜑 ↔ ¬ ∃ 𝑦 ∈ 𝐵 ∃ 𝑥 ∈ 𝐴 𝜑 )
4 1 2 3 3bitr3i ⊢ ( ¬ ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ¬ ∃ 𝑦 ∈ 𝐵 ∃ 𝑥 ∈ 𝐴 𝜑 )
5 4 con4bii ⊢ ( ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∃ 𝑦 ∈ 𝐵 ∃ 𝑥 ∈ 𝐴 𝜑 )