Theorem crefeq 32825

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 4207	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝒫 𝑗 ∩ 𝐴) = (𝒫 𝑗 ∩ 𝐵))
2	1	rexeqdv 3327	. . . . 5 ⊢ (𝐴 = 𝐵 → (∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦 ↔ ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦))
3	2	imbi2d 341	. . . 4 ⊢ (𝐴 = 𝐵 → ((∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦) ↔ (∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)))
4	3	ralbidv 3178	. . 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 3441	. 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 32823	. 2 ⊢ CovHasRef𝐴 = {𝑗 ∈ Top ∣ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐴)𝑧Ref𝑦)}
7		df-cref 32823	. 2 ⊢ CovHasRef𝐵 = {𝑗 ∈ Top ∣ ∀𝑦 ∈ 𝒫 𝑗(∪ 𝑗 = ∪ 𝑦 → ∃𝑧 ∈ (𝒫 𝑗 ∩ 𝐵)𝑧Ref𝑦)}
8	5, 6, 7	3eqtr4g 2798	1 ⊢ (𝐴 = 𝐵 → CovHasRef𝐴 = CovHasRef𝐵)

Syntax hints:   → wi 4   = wceq 1542  ∀wral 3062  ∃wrex 3071  {crab 3433   ∩ cin 3948  𝒫 cpw 4603  ∪ cuni 4909   class class class wbr 5149  Topctop 22395  Refcref 23006  CovHasRefccref 32822

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 3063  df-rex 3072  df-rab 3434  df-in 3956  df-cref 32823

This theorem is referenced by:  ispcmp  32837