Theorem isref 23400

Description: The property of being a refinement of a cover. Dr. Nyikos once commented in class that the term "refinement" is actually misleading and that people are inclined to confuse it with the notion defined in isfne 35759. On the other hand, the two concepts do seem to have a dual relationship. (Contributed by Jeff Hankins, 18-Jan-2010.) (Revised by Thierry Arnoux, 3-Feb-2020.)

Hypotheses

Ref	Expression
isref.1	⊢ 𝑋 = ∪ 𝐴
isref.2	⊢ 𝑌 = ∪ 𝐵

Assertion

Ref	Expression
isref	⊢ (𝐴 ∈ 𝐶 → (𝐴Ref𝐵 ↔ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)))

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

Allowed substitution hints:   𝐴(𝑦)   𝐶(𝑥,𝑦)   𝑋(𝑥,𝑦)   𝑌(𝑥,𝑦)

Proof of Theorem isref

Dummy variables 𝑎 𝑏 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		refrel 23399	. . . 4 ⊢ Rel Ref
2	1	brrelex2i 5729	. . 3 ⊢ (𝐴Ref𝐵 → 𝐵 ∈ V)
3	2	anim2i 616	. 2 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝐴Ref𝐵) → (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ V))
4		simpl 482	. . 3 ⊢ ((𝐴 ∈ 𝐶 ∧ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)) → 𝐴 ∈ 𝐶)
5		simpr 484	. . . . . . 7 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → 𝑌 = 𝑋)
6		isref.2	. . . . . . 7 ⊢ 𝑌 = ∪ 𝐵
7		isref.1	. . . . . . 7 ⊢ 𝑋 = ∪ 𝐴
8	5, 6, 7	3eqtr3g 2790	. . . . . 6 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → ∪ 𝐵 = ∪ 𝐴)
9		uniexg 7739	. . . . . . 7 ⊢ (𝐴 ∈ 𝐶 → ∪ 𝐴 ∈ V)
10	9	adantr 480	. . . . . 6 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → ∪ 𝐴 ∈ V)
11	8, 10	eqeltrd 2828	. . . . 5 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → ∪ 𝐵 ∈ V)
12		uniexb 7760	. . . . 5 ⊢ (𝐵 ∈ V ↔ ∪ 𝐵 ∈ V)
13	11, 12	sylibr 233	. . . 4 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → 𝐵 ∈ V)
14	13	adantrr 716	. . 3 ⊢ ((𝐴 ∈ 𝐶 ∧ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)) → 𝐵 ∈ V)
15	4, 14	jca 511	. 2 ⊢ ((𝐴 ∈ 𝐶 ∧ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)) → (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ V))
16		unieq 4914	. . . . . 6 ⊢ (𝑎 = 𝐴 → ∪ 𝑎 = ∪ 𝐴)
17	16, 7	eqtr4di 2785	. . . . 5 ⊢ (𝑎 = 𝐴 → ∪ 𝑎 = 𝑋)
18	17	eqeq2d 2738	. . . 4 ⊢ (𝑎 = 𝐴 → (∪ 𝑏 = ∪ 𝑎 ↔ ∪ 𝑏 = 𝑋))
19		raleq 3317	. . . 4 ⊢ (𝑎 = 𝐴 → (∀𝑥 ∈ 𝑎 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦 ↔ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦))
20	18, 19	anbi12d 630	. . 3 ⊢ (𝑎 = 𝐴 → ((∪ 𝑏 = ∪ 𝑎 ∧ ∀𝑥 ∈ 𝑎 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦) ↔ (∪ 𝑏 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦)))
21		unieq 4914	. . . . . 6 ⊢ (𝑏 = 𝐵 → ∪ 𝑏 = ∪ 𝐵)
22	21, 6	eqtr4di 2785	. . . . 5 ⊢ (𝑏 = 𝐵 → ∪ 𝑏 = 𝑌)
23	22	eqeq1d 2729	. . . 4 ⊢ (𝑏 = 𝐵 → (∪ 𝑏 = 𝑋 ↔ 𝑌 = 𝑋))
24		rexeq 3316	. . . . 5 ⊢ (𝑏 = 𝐵 → (∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦 ↔ ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦))
25	24	ralbidv 3172	. . . 4 ⊢ (𝑏 = 𝐵 → (∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦 ↔ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦))
26	23, 25	anbi12d 630	. . 3 ⊢ (𝑏 = 𝐵 → ((∪ 𝑏 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦) ↔ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)))
27		df-ref 23396	. . 3 ⊢ Ref = {⟨𝑎, 𝑏⟩ ∣ (∪ 𝑏 = ∪ 𝑎 ∧ ∀𝑥 ∈ 𝑎 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦)}
28	20, 26, 27	brabg 5535	. 2 ⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ V) → (𝐴Ref𝐵 ↔ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)))
29	3, 15, 28	pm5.21nd 801	1 ⊢ (𝐴 ∈ 𝐶 → (𝐴Ref𝐵 ↔ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1534   ∈ wcel 2099  ∀wral 3056  ∃wrex 3065  Vcvv 3469   ⊆ wss 3944  ∪ cuni 4903   class class class wbr 5142  Refcref 23393

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2698  ax-sep 5293  ax-nul 5300  ax-pow 5359  ax-pr 5423  ax-un 7734

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-3an 1087  df-tru 1537  df-fal 1547  df-ex 1775  df-sb 2061  df-clab 2705  df-cleq 2719  df-clel 2805  df-ral 3057  df-rex 3066  df-rab 3428  df-v 3471  df-dif 3947  df-un 3949  df-in 3951  df-ss 3961  df-nul 4319  df-if 4525  df-pw 4600  df-sn 4625  df-pr 4627  df-op 4631  df-uni 4904  df-br 5143  df-opab 5205  df-xp 5678  df-rel 5679  df-ref 23396

This theorem is referenced by:  refbas  23401  refssex  23402  ssref  23403  refref  23404  reftr  23405  refun0  23406  dissnref  23419  reff  33376  locfinreflem  33377  cmpcref  33387  fnessref  35777  refssfne  35778