Theorem refssex 23454

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 23451	. . . . 5 ⊢ Rel Ref
2	1	brrelex1i 5715	. . . 4 ⊢ (𝐴Ref𝐵 → 𝐴 ∈ V)
3		eqid 2736	. . . . . 6 ⊢ ∪ 𝐴 = ∪ 𝐴
4		eqid 2736	. . . . . 6 ⊢ ∪ 𝐵 = ∪ 𝐵
5	3, 4	isref 23452	. . . . 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 3989	. . . . 5 ⊢ (𝑦 = 𝑆 → (𝑦 ⊆ 𝑥 ↔ 𝑆 ⊆ 𝑥))
9	8	rexbidv 3165	. . . 4 ⊢ (𝑦 = 𝑆 → (∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥 ↔ ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥))
10	9	rspccv 3603	. . 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 3052  ∃wrex 3061  Vcvv 3464   ⊆ wss 3931  ∪ cuni 4888   class class class wbr 5124  Refcref 23445

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 2708  ax-sep 5271  ax-nul 5281  ax-pow 5340  ax-pr 5407  ax-un 7734

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 2715  df-cleq 2728  df-clel 2810  df-ral 3053  df-rex 3062  df-rab 3421  df-v 3466  df-dif 3934  df-un 3936  df-in 3938  df-ss 3948  df-nul 4314  df-if 4506  df-pw 4582  df-sn 4607  df-pr 4609  df-op 4613  df-uni 4889  df-br 5125  df-opab 5187  df-xp 5665  df-rel 5666  df-ref 23448

This theorem is referenced by:  reftr  23457  refun0  23458  refssfne  36381