Theorem crefss 34270

Description: The "every open cover has an 𝐴 refinement" predicate respects inclusion. (Contributed by Thierry Arnoux, 7-Jan-2020.)

Assertion

Ref	Expression
crefss	⊢ (𝐴 ⊆ 𝐵 → CovHasRef𝐴 ⊆ CovHasRef𝐵)

Proof of Theorem crefss

Dummy variables 𝑗 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		sslin 4198	. . . . . . 7 ⊢ (𝐴 ⊆ 𝐵 → (𝒫 𝑗 ∩ 𝐴) ⊆ (𝒫 𝑗 ∩ 𝐵))
2		ssrexv 4010	. . . . . . 7 ⊢ ((𝒫 𝑗 ∩ 𝐴) ⊆ (𝒫 𝑗 ∩ 𝐵) → (∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦))
3	1, 2	syl 18	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦))
4	3	imim2d 58	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → ((∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦) → (∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)))
5	4	ralimdv 3182	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦) → ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)))
6	5	anim2d 624	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦)) → (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦))))
7		eqid 2766	. . . 4 ⊢ ∪ 𝑗 = ∪ 𝑗
8	7	iscref 34265	. . 3 ⊢ (𝑗 ∈ CovHasRef𝐴 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦)))
9	7	iscref 34265	. . 3 ⊢ (𝑗 ∈ CovHasRef𝐵 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)))
10	6, 8, 9	3imtr4g 299	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝑗 ∈ CovHasRef𝐴 → 𝑗 ∈ CovHasRef𝐵))
11	10	ssrdv 3946	1 ⊢ (𝐴 ⊆ 𝐵 → CovHasRef𝐴 ⊆ CovHasRef𝐵)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2146  ∀wral 3082  ∃wrex 3092   ∩ cin 3907   ⊆ wss 3908  𝒫 cpw 4567  ∪ cuni 4877   class class class wbr 5114  Topctop 23087  Refcref 23696  CovHasRefccref 34263

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-in 3915  df-ss 3925  df-pw 4569  df-uni 4878  df-cref 34264

This theorem is used by: (None)