Users' Mathboxes Mathbox for Thierry Arnoux < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  crefeq Structured version   Visualization version   GIF version

Theorem crefeq 31202
Description: Equality theorem for the "every open cover has an A refinement" predicate. (Contributed by Thierry Arnoux, 7-Jan-2020.)
Assertion
Ref Expression
crefeq (𝐴 = 𝐵 → CovHasRef𝐴 = CovHasRef𝐵)

Proof of Theorem crefeq
Dummy variables 𝑗 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 ineq2 4136 . . . . . 6 (𝐴 = 𝐵 → (𝒫 𝑗𝐴) = (𝒫 𝑗𝐵))
21rexeqdv 3368 . . . . 5 (𝐴 = 𝐵 → (∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦 ↔ ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦))
32imbi2d 344 . . . 4 (𝐴 = 𝐵 → (( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦) ↔ ( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)))
43ralbidv 3165 . . 3 (𝐴 = 𝐵 → (∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦) ↔ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)))
54rabbidv 3430 . 2 (𝐴 = 𝐵 → {𝑗 ∈ Top ∣ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦)} = {𝑗 ∈ Top ∣ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)})
6 df-cref 31200 . 2 CovHasRef𝐴 = {𝑗 ∈ Top ∣ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐴)𝑧Ref𝑦)}
7 df-cref 31200 . 2 CovHasRef𝐵 = {𝑗 ∈ Top ∣ ∀𝑦 ∈ 𝒫 𝑗( 𝑗 = 𝑦 → ∃𝑧 ∈ (𝒫 𝑗𝐵)𝑧Ref𝑦)}
85, 6, 73eqtr4g 2861 1 (𝐴 = 𝐵 → CovHasRef𝐴 = CovHasRef𝐵)
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1538  wral 3109  wrex 3110  {crab 3113  cin 3883  𝒫 cpw 4500   cuni 4803   class class class wbr 5033  Topctop 21502  Refcref 22111  CovHasRefccref 31199
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2114  ax-9 2122  ax-ext 2773
This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1541  df-ex 1782  df-sb 2070  df-clab 2780  df-cleq 2794  df-clel 2873  df-ral 3114  df-rex 3115  df-rab 3118  df-in 3891  df-cref 31200
This theorem is referenced by:  ispcmp  31214
  Copyright terms: Public domain W3C validator