Theorem crefss 34107

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 4194	. . . . . . 7 ⊢ (𝐴 ⊆ 𝐵 → (𝒫 𝑗 ∩ 𝐴) ⊆ (𝒫 𝑗 ∩ 𝐵))
2		ssrexv 4006	. . . . . . 7 ⊢ ((𝒫 𝑗 ∩ 𝐴) ⊆ (𝒫 𝑗 ∩ 𝐵) → (∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦))
3	1, 2	syl 17	. . . . . 6 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦))
4	3	imim2d 57	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → ((∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦) → (∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)))
5	4	ralimdv 3175	. . . 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 621	. . 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 2761	. . . 4 ⊢ ∪ 𝑗 = ∪ 𝑗
8	7	iscref 34102	. . 3 ⊢ (𝑗 ∈ CovHasRef𝐴 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦)))
9	7	iscref 34102	. . 3 ⊢ (𝑗 ∈ CovHasRef𝐵 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)))
10	6, 8, 9	3imtr4g 298	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝑗 ∈ CovHasRef𝐴 → 𝑗 ∈ CovHasRef𝐵))
11	10	ssrdv 3942	1 ⊢ (𝐴 ⊆ 𝐵 → CovHasRef𝐴 ⊆ CovHasRef𝐵)

Syntax hints:   → wi 4   ∧ wa 399   = wceq 1559   ∈ wcel 2141  ∀wral 3075  ∃wrex 3085   ∩ cin 3903   ⊆ wss 3904  𝒫 cpw 4554  ∪ cuni 4864   class class class wbr 5099  Topctop 22933  Refcref 23542  CovHasRefccref 34100

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1562  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3076  df-rex 3086  df-rab 3414  df-v 3455  df-in 3911  df-ss 3921  df-pw 4556  df-uni 4865  df-cref 34101

This theorem is referenced by: (None)