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Theorem isref 23790
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 37049. 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.
StepHypRef Expression
1 refrel 23789 . . . 4 Rel Ref
21brrelex2i 5704 . . 3 (𝐴Ref𝐵 → 𝐵 ∈ V)
32anim2i 629 . 2 ((𝐴 ∈ 𝐶 ∧ 𝐴Ref𝐵) → (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ V))
4 simpl 488 . . 3 ((𝐴 ∈ 𝐶 ∧ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)) → 𝐴 ∈ 𝐶)
5 simpr 490 . . . . . . 7 ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → 𝑌 = 𝑋)
6 isref.2 . . . . . . 7 𝑌 = ∪ 𝐵
7 isref.1 . . . . . . 7 𝑋 = ∪ 𝐴
85, 6, 73eqtr3g 2818 . . . . . 6 ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → ∪ 𝐵 = ∪ 𝐴)
9 uniexg 7740 . . . . . . 7 (𝐴 ∈ 𝐶 → ∪ 𝐴 ∈ V)
109adantr 486 . . . . . 6 ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → ∪ 𝐴 ∈ V)
118, 10eqeltrd 2860 . . . . 5 ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → ∪ 𝐵 ∈ V)
12 uniexb 7761 . . . . 5 (𝐵 ∈ V ↔ ∪ 𝐵 ∈ V)
1311, 12sylibr 237 . . . 4 ((𝐴 ∈ 𝐶 ∧ 𝑌 = 𝑋) → 𝐵 ∈ V)
1413adantrr 730 . . 3 ((𝐴 ∈ 𝐶 ∧ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)) → 𝐵 ∈ V)
154, 14jca 521 . 2 ((𝐴 ∈ 𝐶 ∧ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)) → (𝐴 ∈ 𝐶 ∧ 𝐵 ∈ V))
16 unieq 4877 . . . . . 6 (𝑎 = 𝐴 → ∪ 𝑎 = ∪ 𝐴)
1716, 7eqtr4di 2813 . . . . 5 (𝑎 = 𝐴 → ∪ 𝑎 = 𝑋)
1817eqeq2d 2771 . . . 4 (𝑎 = 𝐴 → (∪ 𝑏 = ∪ 𝑎 ↔ ∪ 𝑏 = 𝑋))
19 raleq 3316 . . . 4 (𝑎 = 𝐴 → (∀𝑥 ∈ 𝑎 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦 ↔ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦))
2018, 19anbi12d 644 . . 3 (𝑎 = 𝐴 → ((∪ 𝑏 = ∪ 𝑎 ∧ ∀𝑥 ∈ 𝑎 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦) ↔ (∪ 𝑏 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦)))
21 unieq 4877 . . . . . 6 (𝑏 = 𝐵 → ∪ 𝑏 = ∪ 𝐵)
2221, 6eqtr4di 2813 . . . . 5 (𝑏 = 𝐵 → ∪ 𝑏 = 𝑌)
2322eqeq1d 2762 . . . 4 (𝑏 = 𝐵 → (∪ 𝑏 = 𝑋 ↔ 𝑌 = 𝑋))
24 rexeq 3315 . . . . 5 (𝑏 = 𝐵 → (∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦 ↔ ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦))
2524ralbidv 3185 . . . 4 (𝑏 = 𝐵 → (∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦 ↔ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦))
2623, 25anbi12d 644 . . 3 (𝑏 = 𝐵 → ((∪ 𝑏 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦) ↔ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)))
27 df-ref 23786 . . 3 Ref = {⟨𝑎, 𝑏⟩ ∣ (∪ 𝑏 = ∪ 𝑎 ∧ ∀𝑥 ∈ 𝑎 ∃𝑦 ∈ 𝑏 𝑥 ⊆ 𝑦)}
2820, 26, 27brabg 5510 . 2 ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ V) → (𝐴Ref𝐵 ↔ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)))
293, 15, 28pm5.21nd 814 1 (𝐴 ∈ 𝐶 → (𝐴Ref𝐵 ↔ (𝑌 = 𝑋 ∧ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝑥 ⊆ 𝑦)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  ∀wral 3076  ∃wrex 3086  Vcvv 3450   ⊆ wss 3898  ∪ cuni 4866   class class class wbr 5102  Refcref 23783
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-ext 2732  ax-sep 5248  ax-pow 5326  ax-pr 5390  ax-un 7734
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4279  df-if 4482  df-pw 4558  df-sn 4584  df-pr 4586  df-op 4590  df-uni 4867  df-br 5103  df-opab 5167  df-xp 5653  df-rel 5654  df-ref 23786
This theorem is used by:  refbas  23791  refssex  23792  ssref  23793  refref  23794  reftr  23795  refun0  23796  dissnref  23809  reff  34405  locfinreflem  34406  cmpcref  34416  fnessref  37067  refssfne  37068
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