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Theorem crefss 33795
Description: The "every open cover has an 𝐴 refinement" predicate respects inclusion. (Contributed by Thierry Arnoux, 7-Jan-2020.)
Assertion
Ref Expression
crefss (𝐴𝐵 → CovHasRef𝐴 ⊆ CovHasRef𝐵)

Proof of Theorem crefss
Dummy variables 𝑗 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 sslin 4264 . . . . . . 7 (𝐴𝐵 → (𝒫 𝑗𝐴) ⊆ (𝒫 𝑗𝐵))
2 ssrexv 4078 . . . . . . 7 ((𝒫 𝑗𝐴) ⊆ (𝒫 𝑗𝐵) → (∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦))
31, 2syl 17 . . . . . 6 (𝐴𝐵 → (∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦))
43imim2d 57 . . . . 5 (𝐴𝐵 → (( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦) → ( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)))
54ralimdv 3175 . . . 4 (𝐴𝐵 → (∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦) → ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)))
65anim2d 611 . . 3 (𝐴𝐵 → ((𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦)) → (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦))))
7 eqid 2740 . . . 4 𝑗 = 𝑗
87iscref 33790 . . 3 (𝑗 ∈ CovHasRef𝐴 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦)))
97iscref 33790 . . 3 (𝑗 ∈ CovHasRef𝐵 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)))
106, 8, 93imtr4g 296 . 2 (𝐴𝐵 → (𝑗 ∈ CovHasRef𝐴𝑗 ∈ CovHasRef𝐵))
1110ssrdv 4014 1 (𝐴𝐵 → CovHasRef𝐴 ⊆ CovHasRef𝐵)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1537  wcel 2108  wral 3067  wrex 3076  cin 3975  wss 3976  𝒫 cpw 4622   cuni 4931   class class class wbr 5166  Topctop 22920  Refcref 23531  CovHasRefccref 33788
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ral 3068  df-rex 3077  df-rab 3444  df-v 3490  df-in 3983  df-ss 3993  df-pw 4624  df-uni 4932  df-cref 33789
This theorem is referenced by: (None)
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