Theorem refssex 23398

Description: Every set in a refinement has a superset in the original cover. (Contributed by Jeff Hankins, 18-Jan-2010.) (Revised by Thierry Arnoux, 3-Feb-2020.)

Assertion

Ref	Expression
refssex	⊢ ((𝐴Ref𝐵 ∧ 𝑆 ∈ 𝐴) → ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥)

Distinct variable groups:   𝑥,𝐵   𝑥,𝑆

Allowed substitution hint:   𝐴(𝑥)

Proof of Theorem refssex

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		refrel 23395	. . . . 5 ⊢ Rel Ref
2	1	brrelex1i 5694	. . . 4 ⊢ (𝐴Ref𝐵 → 𝐴 ∈ V)
3		eqid 2729	. . . . . 6 ⊢ ∪ 𝐴 = ∪ 𝐴
4		eqid 2729	. . . . . 6 ⊢ ∪ 𝐵 = ∪ 𝐵
5	3, 4	isref 23396	. . . . 5 ⊢ (𝐴 ∈ V → (𝐴Ref𝐵 ↔ (∪ 𝐵 = ∪ 𝐴 ∧ ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)))
6	5	simplbda 499	. . . 4 ⊢ ((𝐴 ∈ V ∧ 𝐴Ref𝐵) → ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)
7	2, 6	mpancom 688	. . 3 ⊢ (𝐴Ref𝐵 → ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)
8		sseq1 3972	. . . . 5 ⊢ (𝑦 = 𝑆 → (𝑦 ⊆ 𝑥 ↔ 𝑆 ⊆ 𝑥))
9	8	rexbidv 3157	. . . 4 ⊢ (𝑦 = 𝑆 → (∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥 ↔ ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥))
10	9	rspccv 3585	. . 3 ⊢ (∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥 → (𝑆 ∈ 𝐴 → ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥))
11	7, 10	syl 17	. 2 ⊢ (𝐴Ref𝐵 → (𝑆 ∈ 𝐴 → ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥))
12	11	imp 406	1 ⊢ ((𝐴Ref𝐵 ∧ 𝑆 ∈ 𝐴) → ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥)

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1540   ∈ wcel 2109  ∀wral 3044  ∃wrex 3053  Vcvv 3447   ⊆ wss 3914  ∪ cuni 4871   class class class wbr 5107  Refcref 23389

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701  ax-sep 5251  ax-nul 5261  ax-pow 5320  ax-pr 5387  ax-un 7711

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-ral 3045  df-rex 3054  df-rab 3406  df-v 3449  df-dif 3917  df-un 3919  df-in 3921  df-ss 3931  df-nul 4297  df-if 4489  df-pw 4565  df-sn 4590  df-pr 4592  df-op 4596  df-uni 4872  df-br 5108  df-opab 5170  df-xp 5644  df-rel 5645  df-ref 23392

This theorem is referenced by:  reftr  23401  refun0  23402  refssfne  36346