Theorem refssex 23235

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 23232	. . . . 5 ⊢ Rel Ref
2	1	brrelex1i 5731	. . . 4 ⊢ (𝐴Ref𝐵 → 𝐴 ∈ V)
3		eqid 2730	. . . . . 6 ⊢ ∪ 𝐴 = ∪ 𝐴
4		eqid 2730	. . . . . 6 ⊢ ∪ 𝐵 = ∪ 𝐵
5	3, 4	isref 23233	. . . . 5 ⊢ (𝐴 ∈ V → (𝐴Ref𝐵 ↔ (∪ 𝐵 = ∪ 𝐴 ∧ ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)))
6	5	simplbda 498	. . . 4 ⊢ ((𝐴 ∈ V ∧ 𝐴Ref𝐵) → ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)
7	2, 6	mpancom 684	. . 3 ⊢ (𝐴Ref𝐵 → ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)
8		sseq1 4006	. . . . 5 ⊢ (𝑦 = 𝑆 → (𝑦 ⊆ 𝑥 ↔ 𝑆 ⊆ 𝑥))
9	8	rexbidv 3176	. . . 4 ⊢ (𝑦 = 𝑆 → (∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥 ↔ ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥))
10	9	rspccv 3608	. . 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 405	1 ⊢ ((𝐴Ref𝐵 ∧ 𝑆 ∈ 𝐴) → ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥)

Syntax hints:   → wi 4   ∧ wa 394   = wceq 1539   ∈ wcel 2104  ∀wral 3059  ∃wrex 3068  Vcvv 3472   ⊆ wss 3947  ∪ cuni 4907   class class class wbr 5147  Refcref 23226

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 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2701  ax-sep 5298  ax-nul 5305  ax-pow 5362  ax-pr 5426  ax-un 7727

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-ral 3060  df-rex 3069  df-rab 3431  df-v 3474  df-dif 3950  df-un 3952  df-in 3954  df-ss 3964  df-nul 4322  df-if 4528  df-pw 4603  df-sn 4628  df-pr 4630  df-op 4634  df-uni 4908  df-br 5148  df-opab 5210  df-xp 5681  df-rel 5682  df-ref 23229

This theorem is referenced by:  reftr  23238  refun0  23239  refssfne  35546