Theorem raaan2 4481

Description: Rearrange restricted quantifiers with two different restricting classes, analogous to raaan 4477. It is necessary that either both restricting classes are empty or both are not empty. (Contributed by Alexander van der Vekens, 29-Jun-2017.)

Hypotheses

Ref	Expression
raaan2.1	⊢ Ⅎ𝑦𝜑
raaan2.2	⊢ Ⅎ𝑥𝜓

Assertion

Ref	Expression
raaan2	⊢ ((𝐴 = ∅ ↔ 𝐵 = ∅) → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))

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

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

Proof of Theorem raaan2

Step	Hyp	Ref	Expression
1		dfbi3 1065	. 2 ⊢ ((𝐴 = ∅ ↔ 𝐵 = ∅) ↔ ((𝐴 = ∅ ∧ 𝐵 = ∅) ∨ (¬ 𝐴 = ∅ ∧ ¬ 𝐵 = ∅)))
2		rzal 4453	. . . . 5 ⊢ (𝐴 = ∅ → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓))
3	2	adantr 486	. . . 4 ⊢ ((𝐴 = ∅ ∧ 𝐵 = ∅) → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓))
4		rzal 4453	. . . . 5 ⊢ (𝐴 = ∅ → ∀𝑥 ∈ 𝐴 𝜑)
5	4	adantr 486	. . . 4 ⊢ ((𝐴 = ∅ ∧ 𝐵 = ∅) → ∀𝑥 ∈ 𝐴 𝜑)
6		rzal 4453	. . . . 5 ⊢ (𝐵 = ∅ → ∀𝑦 ∈ 𝐵 𝜓)
7	6	adantl 487	. . . 4 ⊢ ((𝐴 = ∅ ∧ 𝐵 = ∅) → ∀𝑦 ∈ 𝐵 𝜓)
8		pm5.1 836	. . . 4 ⊢ ((∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ∧ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)) → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
9	3, 5, 7, 8	syl12anc 850	. . 3 ⊢ ((𝐴 = ∅ ∧ 𝐵 = ∅) → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
10		df-ne 2958	. . . . 5 ⊢ (𝐵 ≠ ∅ ↔ ¬ 𝐵 = ∅)
11		raaan2.1	. . . . . . 7 ⊢ Ⅎ𝑦𝜑
12	11	r19.28z 4461	. . . . . 6 ⊢ (𝐵 ≠ ∅ → (∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
13	12	ralbidv 3187	. . . . 5 ⊢ (𝐵 ≠ ∅ → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ ∀𝑥 ∈ 𝐴 (𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
14	10, 13	sylbir 238	. . . 4 ⊢ (¬ 𝐵 = ∅ → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ ∀𝑥 ∈ 𝐴 (𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
15		df-ne 2958	. . . . 5 ⊢ (𝐴 ≠ ∅ ↔ ¬ 𝐴 = ∅)
16		nfcv 2924	. . . . . . 7 ⊢ Ⅎ𝑥𝐵
17		raaan2.2	. . . . . . 7 ⊢ Ⅎ𝑥𝜓
18	16, 17	nfralw 3311	. . . . . 6 ⊢ Ⅎ𝑥∀𝑦 ∈ 𝐵 𝜓
19	18	r19.27z 4469	. . . . 5 ⊢ (𝐴 ≠ ∅ → (∀𝑥 ∈ 𝐴 (𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
20	15, 19	sylbir 238	. . . 4 ⊢ (¬ 𝐴 = ∅ → (∀𝑥 ∈ 𝐴 (𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
21	14, 20	sylan9bbr 520	. . 3 ⊢ ((¬ 𝐴 = ∅ ∧ ¬ 𝐵 = ∅) → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
22	9, 21	jaoi 871	. 2 ⊢ (((𝐴 = ∅ ∧ 𝐵 = ∅) ∨ (¬ 𝐴 = ∅ ∧ ¬ 𝐵 = ∅)) → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
23	1, 22	sylbi 220	1 ⊢ ((𝐴 = ∅ ↔ 𝐵 = ∅) → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (∀𝑥 ∈ 𝐴 𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570  Ⅎwnf 1816   ≠ wne 2957  ∀wral 3078  ∅c0 4282

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  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ne 2958  df-ral 3079  df-dif 3905  df-nul 4283

This theorem is used by:  2reu4lem  4482