Theorem rexcomOLD

Description: Obsolete version of rexcom as of 8-Dec-2024. (Contributed by NM, 19-Nov-1995) (Revised by Mario Carneiro, 14-Oct-2016) (Proof shortened by BJ, 26-Aug-2023) (New usage is discouraged.) (Proof modification is discouraged.)

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

Proof

Step	Hyp	Ref	Expression
1		df-rex	⊢ ( ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∃ 𝑦 ( 𝑦 ∈ 𝐵 ∧ 𝜑 ) )
2	1	rexbii	⊢ ( ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ( 𝑦 ∈ 𝐵 ∧ 𝜑 ) )
3		rexcom4	⊢ ( ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ( 𝑦 ∈ 𝐵 ∧ 𝜑 ) ↔ ∃ 𝑦 ∃ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 ∧ 𝜑 ) )
4		r19.42v	⊢ ( ∃ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 ∧ 𝜑 ) ↔ ( 𝑦 ∈ 𝐵 ∧ ∃ 𝑥 ∈ 𝐴 𝜑 ) )
5	4	exbii	⊢ ( ∃ 𝑦 ∃ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 ∧ 𝜑 ) ↔ ∃ 𝑦 ( 𝑦 ∈ 𝐵 ∧ ∃ 𝑥 ∈ 𝐴 𝜑 ) )
6		df-rex	⊢ ( ∃ 𝑦 ∈ 𝐵 ∃ 𝑥 ∈ 𝐴 𝜑 ↔ ∃ 𝑦 ( 𝑦 ∈ 𝐵 ∧ ∃ 𝑥 ∈ 𝐴 𝜑 ) )
7	5 6	bitr4i	⊢ ( ∃ 𝑦 ∃ 𝑥 ∈ 𝐴 ( 𝑦 ∈ 𝐵 ∧ 𝜑 ) ↔ ∃ 𝑦 ∈ 𝐵 ∃ 𝑥 ∈ 𝐴 𝜑 )
8	2 3 7	3bitri	⊢ ( ∃ 𝑥 ∈ 𝐴 ∃ 𝑦 ∈ 𝐵 𝜑 ↔ ∃ 𝑦 ∈ 𝐵 ∃ 𝑥 ∈ 𝐴 𝜑 )