Theorem iinvdif 4869

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 4220	. . . 4 ⊢ (V ∖ ∅) = V
2		0iun 4853	. . . . 5 ⊢ ∪ 𝑥 ∈ ∅ 𝐵 = ∅
3	2	difeq2i 3988	. . . 4 ⊢ (V ∖ ∪ 𝑥 ∈ ∅ 𝐵) = (V ∖ ∅)
4		0iin 4854	. . . 4 ⊢ ∩ 𝑥 ∈ ∅ (V ∖ 𝐵) = V
5	1, 3, 4	3eqtr4ri 2813	. . 3 ⊢ ∩ 𝑥 ∈ ∅ (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ ∅ 𝐵)
6		iineq1 4809	. . 3 ⊢ (𝐴 = ∅ → ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = ∩ 𝑥 ∈ ∅ (V ∖ 𝐵))
7		iuneq1 4808	. . . 4 ⊢ (𝐴 = ∅ → ∪ 𝑥 ∈ 𝐴 𝐵 = ∪ 𝑥 ∈ ∅ 𝐵)
8	7	difeq2d 3991	. . 3 ⊢ (𝐴 = ∅ → (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵) = (V ∖ ∪ 𝑥 ∈ ∅ 𝐵))
9	5, 6, 8	3eqtr4a 2840	. 2 ⊢ (𝐴 = ∅ → ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵))
10		iindif2 4866	. 2 ⊢ (𝐴 ≠ ∅ → ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵))
11	9, 10	pm2.61ine 3051	1 ⊢ ∩ 𝑥 ∈ 𝐴 (V ∖ 𝐵) = (V ∖ ∪ 𝑥 ∈ 𝐴 𝐵)

Syntax hints:   = wceq 1507  Vcvv 3415   ∖ cdif 3828  ∅c0 4180  ∪ ciun 4793  ∩ ciin 4794

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-8 2052  ax-9 2059  ax-10 2079  ax-11 2093  ax-12 2106  ax-ext 2750

This theorem depends on definitions:  df-bi 199  df-an 388  df-or 834  df-tru 1510  df-ex 1743  df-nf 1747  df-sb 2016  df-clab 2759  df-cleq 2771  df-clel 2846  df-nfc 2918  df-ne 2968  df-ral 3093  df-rex 3094  df-rab 3097  df-v 3417  df-dif 3834  df-in 3838  df-ss 3845  df-nul 4181  df-iun 4795  df-iin 4796

This theorem is referenced by: (None)