Theorem refssex 23578

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 23575	. . . . 5 ⊢ Rel Ref
2	1	brrelex1i 5704	. . . 4 ⊢ (𝐴Ref𝐵 → 𝐴 ∈ V)
3		eqid 2763	. . . . . 6 ⊢ ∪ 𝐴 = ∪ 𝐴
4		eqid 2763	. . . . . 6 ⊢ ∪ 𝐵 = ∪ 𝐵
5	3, 4	isref 23576	. . . . 5 ⊢ (𝐴 ∈ V → (𝐴Ref𝐵 ↔ (∪ 𝐵 = ∪ 𝐴 ∧ ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)))
6	5	simplbda 503	. . . 4 ⊢ ((𝐴 ∈ V ∧ 𝐴Ref𝐵) → ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)
7	2, 6	mpancom 698	. . 3 ⊢ (𝐴Ref𝐵 → ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)
8		sseq1 3962	. . . . 5 ⊢ (𝑦 = 𝑆 → (𝑦 ⊆ 𝑥 ↔ 𝑆 ⊆ 𝑥))
9	8	rexbidv 3187	. . . 4 ⊢ (𝑦 = 𝑆 → (∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥 ↔ ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥))
10	9	rspccv 3579	. . 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 410	1 ⊢ ((𝐴Ref𝐵 ∧ 𝑆 ∈ 𝐴) → ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥)

Syntax hints:   → wi 4   ∧ wa 399   = wceq 1561   ∈ wcel 2143  ∀wral 3077  ∃wrex 3087  Vcvv 3455   ⊆ wss 3905  ∪ cuni 4866   class class class wbr 5101  Refcref 23569

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1816  ax-4 1830  ax-5 1931  ax-6 1988  ax-7 2029  ax-8 2145  ax-9 2153  ax-ext 2735  ax-sep 5247  ax-pow 5323  ax-pr 5391  ax-un 7718

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1101  df-tru 1564  df-fal 1574  df-ex 1801  df-sb 2092  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3078  df-rex 3088  df-rab 3416  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4482  df-pw 4558  df-sn 4584  df-pr 4586  df-op 4590  df-uni 4867  df-br 5102  df-opab 5164  df-xp 5654  df-rel 5655  df-ref 23572

This theorem is referenced by:  reftr  23581  refun0  23582  refssfne  36723