Theorem iinvdif 5037

Description: The indexed intersection of a complement. (Contributed by Gérard Lang, 5-Aug-2018.)

Assertion

Ref	Expression
iinvdif	⊢ ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵)

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝐵(𝑥)

Proof of Theorem iinvdif

Step	Hyp	Ref	Expression
1		dif0 4331	. . . 4 ⊢ (V ∖ ∅) = V
2		0iun 5020	. . . . 5 ⊢ ∪ 𝑥 ∈ ∅ 𝐵 = ∅
3	2	difeq2i 4077	. . . 4 ⊢ (V ∖ ∪ 𝑥 ∈ ∅ 𝐵) = (V ∖ ∅)
4		0iin 5021	. . . 4 ⊢ ∩ 𝑥 ∈ ∅ (V ∖ 𝐵) = V
5	1, 3, 4	3eqtr4ri 2796	. . 3 ⊢ ∩ 𝑥 ∈ ∅ (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ ∅ 𝐵)
6		iineq1 4967	. . 3 ⊢ (𝐴 = ∅ → ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = ∩ 𝑥 ∈ ∅ (V ∖ 𝐵))
7		iuneq1 4966	. . . 4 ⊢ (𝐴 = ∅ → ∪ 𝑥 ∈ 𝐴 𝐵 = ∪ 𝑥 ∈ ∅ 𝐵)
8	7	difeq2d 4080	. . 3 ⊢ (𝐴 = ∅ → (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵) = (V ∖ ∪ 𝑥 ∈ ∅ 𝐵))
9	5, 6, 8	3eqtr4a 2823	. 2 ⊢ (𝐴 = ∅ → ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵))
10		iindif2 5034	. 2 ⊢ (𝐴 ≠ ∅ → ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵))
11	9, 10	pm2.61ine 3040	1 ⊢ ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵)

Syntax hints:   = wceq 1560  Vcvv 3454   ∖ cdif 3901  ∅c0 4285  ∪ ciun 4949  ∩ ciin 4950

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-12 2212  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1563  df-fal 1573  df-ex 1800  df-nf 1804  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-ral 3077  df-rex 3087  df-rab 3415  df-v 3456  df-dif 3907  df-ss 3921  df-nul 4286  df-iun 4951  df-iin 4952

This theorem is referenced by: (None)