Theorem crefeq 31109

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.

Step	Hyp	Ref	Expression
1		ineq2 4182	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝒫 𝑗 ∩ 𝐴) = (𝒫 𝑗 ∩ 𝐵))
2	1	rexeqdv 3416	. . . . 5 ⊢ (𝐴 = 𝐵 → (∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦 ↔ ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦))
3	2	imbi2d 343	. . . 4 ⊢ (𝐴 = 𝐵 → ((∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦) ↔ (∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)))
4	3	ralbidv 3197	. . 3 ⊢ (𝐴 = 𝐵 → (∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦) ↔ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)))
5	4	rabbidv 3480	. 2 ⊢ (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)})
6		df-cref 31107	. 2 ⊢ CovHasRef𝐴 = {𝑗 ∈ Top ∣ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦)}
7		df-cref 31107	. 2 ⊢ CovHasRef𝐵 = {𝑗 ∈ Top ∣ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)}
8	5, 6, 7	3eqtr4g 2881	1 ⊢ (𝐴 = 𝐵 → CovHasRef𝐴 = CovHasRef𝐵)

Syntax hints:   → wi 4   = wceq 1533  ∀wral 3138  ∃wrex 3139  {crab 3142   ∩ cin 3934  𝒫 cpw 4538  ∪ cuni 4837   class class class wbr 5065  Topctop 21500  Refcref 22109  CovHasRefccref 31106

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-tru 1536  df-ex 1777  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-ral 3143  df-rex 3144  df-rab 3147  df-in 3942  df-cref 31107

This theorem is referenced by:  ispcmp  31121