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Theorem crefss 32487
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 4195 . . . . . . 7 (𝐴𝐵 → (𝒫 𝑗𝐴) ⊆ (𝒫 𝑗𝐵))
2 ssrexv 4012 . . . . . . 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 3163 . . . 4 (𝐴𝐵 → (∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦) → ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)))
65anim2d 613 . . 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 2733 . . . 4 𝑗 = 𝑗
87iscref 32482 . . 3 (𝑗 ∈ CovHasRef𝐴 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦)))
97iscref 32482 . . 3 (𝑗 ∈ CovHasRef𝐵 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)))
106, 8, 93imtr4g 296 . 2 (𝐴𝐵 → (𝑗 ∈ CovHasRef𝐴𝑗 ∈ CovHasRef𝐵))
1110ssrdv 3951 1 (𝐴𝐵 → CovHasRef𝐴 ⊆ CovHasRef𝐵)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 397   = wceq 1542  wcel 2107  wral 3061  wrex 3070  cin 3910  wss 3911  𝒫 cpw 4561   cuni 4866   class class class wbr 5106  Topctop 22258  Refcref 22869  CovHasRefccref 32480
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-ral 3062  df-rex 3071  df-rab 3407  df-v 3446  df-in 3918  df-ss 3928  df-pw 4563  df-uni 4867  df-cref 32481
This theorem is referenced by: (None)
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