Theorem iundifdifd 33156

Description: The intersection of a set is the complement of the union of the complements. (Contributed by Thierry Arnoux, 19-Dec-2016.)

Assertion

Ref	Expression
iundifdifd	⊢ (𝐴 ⊆ 𝒫 𝑂 → (𝐴 ≠ ∅ → ∩ 𝐴 = (𝑂 ∖ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥))))

Distinct variable groups:   𝑥,𝐴   𝑥,𝑂

Proof of Theorem iundifdifd

Step	Hyp	Ref	Expression
1		iundif2 5031	. . . . 5 ⊢ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥) = (𝑂 ∖ ∩ 𝑥 ∈ 𝐴 𝑥)
2		intiin 5017	. . . . . 6 ⊢ ∩ 𝐴 = ∩ 𝑥 ∈ 𝐴 𝑥
3	2	difeq2i 4060	. . . . 5 ⊢ (𝑂 ∖ ∩ 𝐴) = (𝑂 ∖ ∩ 𝑥 ∈ 𝐴 𝑥)
4	1, 3	eqtr4i 2786	. . . 4 ⊢ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥) = (𝑂 ∖ ∩ 𝐴)
5	4	difeq2i 4060	. . 3 ⊢ (𝑂 ∖ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥)) = (𝑂 ∖ (𝑂 ∖ ∩ 𝐴))
6		intssuni2 4932	. . . . 5 ⊢ ((𝐴 ⊆ 𝒫 𝑂 ∧ 𝐴 ≠ ∅) → ∩ 𝐴 ⊆ ∪ 𝒫 𝑂)
7		unipw 5417	. . . . 5 ⊢ ∪ 𝒫 𝑂 = 𝑂
8	6, 7	sseqtrdi 3948	. . . 4 ⊢ ((𝐴 ⊆ 𝒫 𝑂 ∧ 𝐴 ≠ ∅) → ∩ 𝐴 ⊆ 𝑂)
9		dfss4 4214	. . . 4 ⊢ (∩ 𝐴 ⊆ 𝑂 ↔ (𝑂 ∖ (𝑂 ∖ ∩ 𝐴)) = ∩ 𝐴)
10	8, 9	sylib 221	. . 3 ⊢ ((𝐴 ⊆ 𝒫 𝑂 ∧ 𝐴 ≠ ∅) → (𝑂 ∖ (𝑂 ∖ ∩ 𝐴)) = ∩ 𝐴)
11	5, 10	eqtr2id 2808	. 2 ⊢ ((𝐴 ⊆ 𝒫 𝑂 ∧ 𝐴 ≠ ∅) → ∩ 𝐴 = (𝑂 ∖ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥)))
12	11	ex 418	1 ⊢ (𝐴 ⊆ 𝒫 𝑂 → (𝐴 ≠ ∅ → ∩ 𝐴 = (𝑂 ∖ ∪ 𝑥 ∈ 𝐴 (𝑂 ∖ 𝑥))))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1569   ≠ wne 2955   ⊆ wss 3895   ∖ cdif 4045  ∅c0 4278  𝒫 cpw 4556  ∪ cuni 4866  ∩ cint 4906  ∪ ciun 4950  ∩ ciin 4951

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248  ax-pr 5390

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-pw 4558  df-sn 4584  df-pr 4586  df-uni 4867  df-int 4907  df-iun 4952  df-iin 4953

This theorem is used by:  sigaclci  34764