Theorem opndisj 50222

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 4563	. . . 4 ⊢ (𝑌 ∈ 𝐽 → (𝑌 ∈ 𝒫 𝑍 ↔ 𝑌 ⊆ 𝑍))
2		sseq2 3939	. . . 4 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ⊆ 𝑍 ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
3	1, 2	sylan9bbr 520	. . 3 ⊢ ((𝑍 = (∪ 𝐽 ∖ 𝑋) ∧ 𝑌 ∈ 𝐽) → (𝑌 ∈ 𝒫 𝑍 ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
4	3	pm5.32da 590	. 2 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → ((𝑌 ∈ 𝐽 ∧ 𝑌 ∈ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))))
5		elin 4096	. 2 ⊢ (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ∈ 𝒫 𝑍))
6		elssuni 4902	. . . 4 ⊢ (𝑌 ∈ 𝐽 → 𝑌 ⊆ ∪ 𝐽)
7		incom 4109	. . . . . 6 ⊢ (𝑋 ∩ 𝑌) = (𝑌 ∩ 𝑋)
8	7	eqeq1i 2768	. . . . 5 ⊢ ((𝑋 ∩ 𝑌) = ∅ ↔ (𝑌 ∩ 𝑋) = ∅)
9		reldisj 4409	. . . . 5 ⊢ (𝑌 ⊆ ∪ 𝐽 → ((𝑌 ∩ 𝑋) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
10	8, 9	bitrid 286	. . . 4 ⊢ (𝑌 ⊆ ∪ 𝐽 → ((𝑋 ∩ 𝑌) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
11	6, 10	syl 18	. . 3 ⊢ (𝑌 ∈ 𝐽 → ((𝑋 ∩ 𝑌) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
12	11	pm5.32i 585	. 2 ⊢ ((𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))
13	4, 5, 12	3bitr4g 317	1 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2147   ⊆ wss 3900   ∖ cdif 4048   ∩ cin 4092  ∅c0 4281  𝒫 cpw 4560  ∪ cuni 4870

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-nul 4282  df-pw 4562  df-uni 4871

This theorem is used by:  clddisj  50223