Theorem raaan2 4478

Description: Rearrange restricted quantifiers with two different restricting classes, analogous to raaan 4474. 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 4450	. . . . 5 ⊢ (𝐴 = ∅ → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓))
3	2	adantr 486	. . . 4 ⊢ ((𝐴 = ∅ ∧ 𝐵 = ∅) → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓))
4		rzal 4450	. . . . 5 ⊢ (𝐴 = ∅ → ∀𝑥 ∈ 𝐴 𝜑)
5	4	adantr 486	. . . 4 ⊢ ((𝐴 = ∅ ∧ 𝐵 = ∅) → ∀𝑥 ∈ 𝐴 𝜑)
6		rzal 4450	. . . . 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 2956	. . . . 5 ⊢ (𝐵 ≠ ∅ ↔ ¬ 𝐵 = ∅)
11		raaan2.1	. . . . . . 7 ⊢ Ⅎ𝑦𝜑
12	11	r19.28z 4458	. . . . . 6 ⊢ (𝐵 ≠ ∅ → (∀𝑦 ∈ 𝐵 (𝜑 ∧ 𝜓) ↔ (𝜑 ∧ ∀𝑦 ∈ 𝐵 𝜓)))
13	12	ralbidv 3185	. . . . 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 2956	. . . . 5 ⊢ (𝐴 ≠ ∅ ↔ ¬ 𝐴 = ∅)
16		nfcv 2922	. . . . . . 7 ⊢ Ⅎ𝑥𝐵
17		raaan2.2	. . . . . . 7 ⊢ Ⅎ𝑥𝜓
18	16, 17	nfralw 3309	. . . . . 6 ⊢ Ⅎ𝑥∀𝑦 ∈ 𝐵 𝜓
19	18	r19.27z 4466	. . . . 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 2955  ∀wral 3076  ∅c0 4279

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 2213  ax-ext 2732

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 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-dif 3902  df-nul 4280

This theorem is used by:  2reu4lem  4479