Theorem opndisj 47623

Description: Two ways of saying that two open sets are disjoint, if 𝐽 is a topology and 𝑋 is an open set. (Contributed by Zhi Wang, 6-Sep-2024.)

Assertion

Ref	Expression
opndisj	⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅)))

Proof of Theorem opndisj

Step	Hyp	Ref	Expression
1		elpwg 4605	. . . 4 ⊢ (𝑌 ∈ 𝐽 → (𝑌 ∈ 𝒫 𝑍 ↔ 𝑌 ⊆ 𝑍))
2		sseq2 4008	. . . 4 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ⊆ 𝑍 ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
3	1, 2	sylan9bbr 510	. . 3 ⊢ ((𝑍 = (∪ 𝐽 ∖ 𝑋) ∧ 𝑌 ∈ 𝐽) → (𝑌 ∈ 𝒫 𝑍 ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
4	3	pm5.32da 578	. 2 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → ((𝑌 ∈ 𝐽 ∧ 𝑌 ∈ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))))
5		elin 3964	. 2 ⊢ (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ∈ 𝒫 𝑍))
6		elssuni 4941	. . . 4 ⊢ (𝑌 ∈ 𝐽 → 𝑌 ⊆ ∪ 𝐽)
7		incom 4201	. . . . . 6 ⊢ (𝑋 ∩ 𝑌) = (𝑌 ∩ 𝑋)
8	7	eqeq1i 2736	. . . . 5 ⊢ ((𝑋 ∩ 𝑌) = ∅ ↔ (𝑌 ∩ 𝑋) = ∅)
9		reldisj 4451	. . . . 5 ⊢ (𝑌 ⊆ ∪ 𝐽 → ((𝑌 ∩ 𝑋) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
10	8, 9	bitrid 283	. . . 4 ⊢ (𝑌 ⊆ ∪ 𝐽 → ((𝑋 ∩ 𝑌) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
11	6, 10	syl 17	. . 3 ⊢ (𝑌 ∈ 𝐽 → ((𝑋 ∩ 𝑌) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
12	11	pm5.32i 574	. 2 ⊢ ((𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
13	4, 5, 12	3bitr4g 314	1 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1540   ∈ wcel 2105   ∖ cdif 3945   ∩ cin 3947   ⊆ wss 3948  ∅c0 4322  𝒫 cpw 4602  ∪ cuni 4908

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 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1543  df-fal 1553  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-ral 3061  df-rab 3432  df-v 3475  df-dif 3951  df-in 3955  df-ss 3965  df-nul 4323  df-pw 4604  df-uni 4909

This theorem is referenced by:  clddisj  47624