Theorem refssex 23647

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 23644	. . . . 5 ⊢ Rel Ref
2	1	brrelex1i 5717	. . . 4 ⊢ (𝐴Ref𝐵 → 𝐴 ∈ V)
3		eqid 2761	. . . . . 6 ⊢ ∪ 𝐴 = ∪ 𝐴
4		eqid 2761	. . . . . 6 ⊢ ∪ 𝐵 = ∪ 𝐵
5	3, 4	isref 23645	. . . . 5 ⊢ (𝐴 ∈ V → (𝐴Ref𝐵 ↔ (∪ 𝐵 = ∪ 𝐴 ∧ ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)))
6	5	simplbda 504	. . . 4 ⊢ ((𝐴 ∈ V ∧ 𝐴Ref𝐵) → ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)
7	2, 6	mpancom 700	. . 3 ⊢ (𝐴Ref𝐵 → ∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥)
8		sseq1 3961	. . . . 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 3577	. . 3 ⊢ (∀𝑦 ∈ 𝐴 ∃𝑥 ∈ 𝐵 𝑦 ⊆ 𝑥 → (𝑆 ∈ 𝐴 → ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥))
11	7, 10	syl 18	. 2 ⊢ (𝐴Ref𝐵 → (𝑆 ∈ 𝐴 → ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥))
12	11	imp 411	1 ⊢ ((𝐴Ref𝐵 ∧ 𝑆 ∈ 𝐴) → ∃𝑥 ∈ 𝐵 𝑆 ⊆ 𝑥)

Syntax hints:   → wi 4   ∧ wa 400   = wceq 1568   ∈ wcel 2141  ∀wral 3077  ∃wrex 3087  Vcvv 3453   ⊆ wss 3904  ∪ cuni 4871   class class class wbr 5108  Refcref 23638

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2143  ax-9 2151  ax-ext 2733  ax-sep 5256  ax-pow 5336  ax-pr 5404  ax-un 7732

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1571  df-fal 1581  df-ex 1808  df-sb 2095  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-rex 3088  df-rab 3415  df-v 3455  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4487  df-pw 4563  df-sn 4589  df-pr 4591  df-op 4595  df-uni 4872  df-br 5109  df-opab 5173  df-xp 5667  df-rel 5668  df-ref 23641

This theorem is referenced by:  reftr  23650  refun0  23651  refssfne  36835