Theorem reeanlem 3234

Description: Lemma factoring out common proof steps of reeanv 3235 and reean 3313. (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,𝑦

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

Proof of Theorem reeanlem

Step	Hyp	Ref	Expression
1		an4 669	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (𝑦 ∈ 𝐵 ∧ 𝜓)))
2	1	2exbii 1882	. . 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 278	. 2 ⊢ (∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)) ↔ (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ∧ ∃𝑦(𝑦 ∈ 𝐵 ∧ 𝜓)))
5		r2ex 3200	. 2 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ ∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ (𝜑 ∧ 𝜓)))
6		df-rex 3088	. . 3 ⊢ (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
7		df-rex 3088	. . 3 ⊢ (∃𝑦 ∈ 𝐵 𝜓 ↔ ∃𝑦(𝑦 ∈ 𝐵 ∧ 𝜓))
8	6, 7	anbi12i 640	. 2 ⊢ ((∃𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑦 ∈ 𝐵 𝜓) ↔ (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ∧ ∃𝑦(𝑦 ∈ 𝐵 ∧ 𝜓)))
9	4, 5, 8	3bitr4i 306	1 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∃𝑥 ∈ 𝐴 𝜑 ∧ ∃𝑦 ∈ 𝐵 𝜓))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401  ∃wex 1812   ∈ wcel 2145  ∃wrex 3087

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-ral 3078  df-rex 3088

This theorem is used by:  reeanv  3235  reean  3313