Theorem reusv2lem5 5364

Description: Lemma for reusv2 5365. (Contributed by NM, 4-Jan-2013.) (Proof shortened by Mario Carneiro, 19-Nov-2016.)

Assertion

Ref	Expression
reusv2lem5	⊢ ((∀𝑦 ∈ 𝐵 𝐶 ∈ 𝐴 ∧ 𝐵 ≠ ∅) → (∃!𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 = 𝐶 ↔ ∃!𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝑥 = 𝐶))

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

Allowed substitution hint:   𝐶(𝑦)

Proof of Theorem reusv2lem5

Step	Hyp	Ref	Expression
1		tru 1574	. . . . . . . . 9 ⊢ ⊤
2		biimt 363	. . . . . . . . 9 ⊢ ((𝐶 ∈ 𝐴 ∧ ⊤) → (𝑥 = 𝐶 ↔ ((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶)))
3	1, 2	mpan2 704	. . . . . . . 8 ⊢ (𝐶 ∈ 𝐴 → (𝑥 = 𝐶 ↔ ((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶)))
4		ibar 538	. . . . . . . 8 ⊢ (𝐶 ∈ 𝐴 → (𝑥 = 𝐶 ↔ (𝐶 ∈ 𝐴 ∧ 𝑥 = 𝐶)))
5	3, 4	bitr3d 284	. . . . . . 7 ⊢ (𝐶 ∈ 𝐴 → (((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶) ↔ (𝐶 ∈ 𝐴 ∧ 𝑥 = 𝐶)))
6		eleq1 2849	. . . . . . . 8 ⊢ (𝑥 = 𝐶 → (𝑥 ∈ 𝐴 ↔ 𝐶 ∈ 𝐴))
7	6	pm5.32ri 586	. . . . . . 7 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 = 𝐶) ↔ (𝐶 ∈ 𝐴 ∧ 𝑥 = 𝐶))
8	5, 7	bitr4di 292	. . . . . 6 ⊢ (𝐶 ∈ 𝐴 → (((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 = 𝐶)))
9	8	ralimi 3100	. . . . 5 ⊢ (∀𝑦 ∈ 𝐵 𝐶 ∈ 𝐴 → ∀𝑦 ∈ 𝐵 (((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 = 𝐶)))
10		ralbi 3118	. . . . 5 ⊢ (∀𝑦 ∈ 𝐵 (((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 = 𝐶)) → (∀𝑦 ∈ 𝐵 ((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶) ↔ ∀𝑦 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝑥 = 𝐶)))
11	9, 10	syl 18	. . . 4 ⊢ (∀𝑦 ∈ 𝐵 𝐶 ∈ 𝐴 → (∀𝑦 ∈ 𝐵 ((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶) ↔ ∀𝑦 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝑥 = 𝐶)))
12	11	eubidv 2612	. . 3 ⊢ (∀𝑦 ∈ 𝐵 𝐶 ∈ 𝐴 → (∃!𝑥∀𝑦 ∈ 𝐵 ((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶) ↔ ∃!𝑥∀𝑦 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝑥 = 𝐶)))
13		r19.28zv 4462	. . . 4 ⊢ (𝐵 ≠ ∅ → (∀𝑦 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝑥 = 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ ∀𝑦 ∈ 𝐵 𝑥 = 𝐶)))
14	13	eubidv 2612	. . 3 ⊢ (𝐵 ≠ ∅ → (∃!𝑥∀𝑦 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝑥 = 𝐶) ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ∀𝑦 ∈ 𝐵 𝑥 = 𝐶)))
15	12, 14	sylan9bb 519	. 2 ⊢ ((∀𝑦 ∈ 𝐵 𝐶 ∈ 𝐴 ∧ 𝐵 ≠ ∅) → (∃!𝑥∀𝑦 ∈ 𝐵 ((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶) ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ∀𝑦 ∈ 𝐵 𝑥 = 𝐶)))
16	1	biantrur 540	. . . . 5 ⊢ (𝑥 = 𝐶 ↔ (⊤ ∧ 𝑥 = 𝐶))
17	16	rexbii 3110	. . . 4 ⊢ (∃𝑦 ∈ 𝐵 𝑥 = 𝐶 ↔ ∃𝑦 ∈ 𝐵 (⊤ ∧ 𝑥 = 𝐶))
18	17	reubii 3375	. . 3 ⊢ (∃!𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 = 𝐶 ↔ ∃!𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (⊤ ∧ 𝑥 = 𝐶))
19		reusv2lem4 5363	. . 3 ⊢ (∃!𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (⊤ ∧ 𝑥 = 𝐶) ↔ ∃!𝑥∀𝑦 ∈ 𝐵 ((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶))
20	18, 19	bitri 278	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 = 𝐶 ↔ ∃!𝑥∀𝑦 ∈ 𝐵 ((𝐶 ∈ 𝐴 ∧ ⊤) → 𝑥 = 𝐶))
21		df-reu 3367	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝑥 = 𝐶 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ ∀𝑦 ∈ 𝐵 𝑥 = 𝐶))
22	15, 20, 21	3bitr4g 317	1 ⊢ ((∀𝑦 ∈ 𝐵 𝐶 ∈ 𝐴 ∧ 𝐵 ≠ ∅) → (∃!𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 = 𝐶 ↔ ∃!𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝑥 = 𝐶))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  ⊤wtru 1571   ∈ wcel 2145  ∃!weu 2594   ≠ wne 2956  ∀wral 3077  ∃wrex 3087  ∃!wreu 3364  ∅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 2733  ax-nul 5260  ax-pow 5327

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-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ne 2957  df-ral 3078  df-rex 3088  df-reu 3367  df-rab 3414  df-v 3453  df-sbc 3740  df-csb 3848  df-dif 3902  df-nul 4280

This theorem is used by:  reusv2  5365