Theorem iinvdif 5032

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 5015	. . . . 5 ⊢ ∪ 𝑥 ∈ ∅ 𝐵 = ∅
3	2	difeq2i 4076	. . . 4 ⊢ (V ∖ ∪ 𝑥 ∈ ∅ 𝐵) = (V ∖ ∅)
4		0iin 5016	. . . 4 ⊢ ∩ 𝑥 ∈ ∅ (V ∖ 𝐵) = V
5	1, 3, 4	3eqtr4ri 2763	. . 3 ⊢ ∩ 𝑥 ∈ ∅ (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ ∅ 𝐵)
6		iineq1 4962	. . 3 ⊢ (𝐴 = ∅ → ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = ∩ 𝑥 ∈ ∅ (V ∖ 𝐵))
7		iuneq1 4961	. . . 4 ⊢ (𝐴 = ∅ → ∪ 𝑥 ∈ 𝐴 𝐵 = ∪ 𝑥 ∈ ∅ 𝐵)
8	7	difeq2d 4079	. . 3 ⊢ (𝐴 = ∅ → (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵) = (V ∖ ∪ 𝑥 ∈ ∅ 𝐵))
9	5, 6, 8	3eqtr4a 2790	. 2 ⊢ (𝐴 = ∅ → ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵))
10		iindif2 5029	. 2 ⊢ (𝐴 ≠ ∅ → ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵))
11	9, 10	pm2.61ine 3008	1 ⊢ ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵)

Syntax hints:   = wceq 1540  Vcvv 3438   ∖ cdif 3902  ∅c0 4286  ∪ ciun 4944  ∩ ciin 4945

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-12 2178  ax-ext 2701

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-ne 2926  df-ral 3045  df-rex 3054  df-rab 3397  df-v 3440  df-dif 3908  df-ss 3922  df-nul 4287  df-iun 4946  df-iin 4947

This theorem is referenced by: (None)