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 2960	. . . . 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 3189	. . . . 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 2960	. . . . 5 ⊢ (𝐴 ≠ ∅ ↔ ¬ 𝐴 = ∅)
16		nfcv 2926	. . . . . . 7 ⊢ Ⅎ𝑥𝐵
17		raaan2.2	. . . . . . 7 ⊢ Ⅎ𝑥𝜓
18	16, 17	nfralw 3313	. . . . . 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 2959  ∀wral 3080  ∅c0 4281

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-dif 4050  df-nul 4282

This theorem is used by:  2reu4lem  4482