Theorem reeanlem 3365

Description: Lemma factoring out common proof steps of reeanv 3367 and reean 3366. (Contributed by Wolf Lammen, 20-Aug-2023.)

Hypothesis

Ref	Expression
reeanlem.1	⊢ (∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑦 ∈ 𝐵 ∧ 𝜓)) ↔ (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ∧ ∃𝑦(𝑦 ∈ 𝐵 ∧ 𝜓)))

Assertion

Ref	Expression
reeanlem	⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∃𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑦 ∈ 𝐵 𝜓))

Distinct variable groups:   𝑦,𝐴   𝑥,𝐵   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)   𝐴(𝑥)   𝐵(𝑦)

Proof of Theorem reeanlem

Step	Hyp	Ref	Expression
1		an4 654	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑦 ∈ 𝐵 ∧ 𝜓)))
2	1	2exbii 1849	. . 3 ⊢ (∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)) ↔ ∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑦 ∈ 𝐵 ∧ 𝜓)))
3		reeanlem.1	. . 3 ⊢ (∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑦 ∈ 𝐵 ∧ 𝜓)) ↔ (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ∧ ∃𝑦(𝑦 ∈ 𝐵 ∧ 𝜓)))
4	2, 3	bitri 277	. 2 ⊢ (∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)) ↔ (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ∧ ∃𝑦(𝑦 ∈ 𝐵 ∧ 𝜓)))
5		r2ex 3303	. 2 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ ∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)))
6		df-rex 3144	. . 3 ⊢ (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
7		df-rex 3144	. . 3 ⊢ (∃𝑦 ∈ 𝐵 𝜓 ↔ ∃𝑦(𝑦 ∈ 𝐵 ∧ 𝜓))
8	6, 7	anbi12i 628	. 2 ⊢ ((∃𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑦 ∈ 𝐵 𝜓) ↔ (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ∧ ∃𝑦(𝑦 ∈ 𝐵 ∧ 𝜓)))
9	4, 5, 8	3bitr4i 305	1 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∃𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑦 ∈ 𝐵 𝜓))

Syntax hints:   ↔ wb 208   ∧ wa 398  ∃wex 1780   ∈ wcel 2114  ∃wrex 3139

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1781  df-ral 3143  df-rex 3144

This theorem is referenced by:  reean  3366  reeanv  3367