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Theorem crefss 33883
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 4192 . . . . . . 7 (𝐴𝐵 → (𝒫 𝑗𝐴) ⊆ (𝒫 𝑗𝐵))
2 ssrexv 4000 . . . . . . 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 3147 . . . 4 (𝐴𝐵 → (∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦) → ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)))
65anim2d 612 . . 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 33878 . . 3 (𝑗 ∈ CovHasRef𝐴 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦)))
97iscref 33878 . . 3 (𝑗 ∈ CovHasRef𝐵 ↔ (𝑗 ∈ Top ∧ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)))
106, 8, 93imtr4g 296 . 2 (𝐴𝐵 → (𝑗 ∈ CovHasRef𝐴𝑗 ∈ CovHasRef𝐵))
1110ssrdv 3936 1 (𝐴𝐵 → CovHasRef𝐴 ⊆ CovHasRef𝐵)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1541  wcel 2113  wral 3048  wrex 3057  cin 3897  wss 3898  𝒫 cpw 4549   cuni 4858   class class class wbr 5093  Topctop 22809  Refcref 23418  CovHasRefccref 33876
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-ral 3049  df-rex 3058  df-rab 3397  df-v 3439  df-in 3905  df-ss 3915  df-pw 4551  df-uni 4859  df-cref 33877
This theorem is referenced by: (None)
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