Theorem difun1 4117

Description: A relationship involving double difference and union. (Contributed by NM, 29-Aug-2004.)

Assertion

Ref	Expression
difun1	⊢ (𝐴 ∖ (𝐵 ∪ 𝐶)) = ((𝐴 ∖ 𝐵) ∖ 𝐶)

Proof of Theorem difun1

Step	Hyp	Ref	Expression
1		inass 4048	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
2		invdif 4098	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
3	1, 2	eqtr3i 2851	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
4		undm 4115	. . . . 5 ⊢ (V ∖ (𝐵 ∪ 𝐶)) = ((V ∖ 𝐵) ∩ (V ∖ 𝐶))
5	4	ineq2i 4038	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
6		invdif 4098	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
7	5, 6	eqtr3i 2851	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
8	3, 7	eqtr3i 2851	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = (𝐴 ∖ (𝐵 ∪ 𝐶))
9		invdif 4098	. . 3 ⊢ (𝐴 ∩ (V ∖ 𝐵)) = (𝐴 ∖ 𝐵)
10	9	difeq1i 3951	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = ((𝐴 ∖ 𝐵) ∖ 𝐶)
11	8, 10	eqtr3i 2851	1 ⊢ (𝐴 ∖ (𝐵 ∪ 𝐶)) = ((𝐴 ∖ 𝐵) ∖ 𝐶)

Syntax hints:   = wceq 1658  Vcvv 3414   ∖ cdif 3795   ∪ cun 3796   ∩ cin 3797

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1896  ax-4 1910  ax-5 2011  ax-6 2077  ax-7 2114  ax-9 2175  ax-10 2194  ax-11 2209  ax-12 2222  ax-ext 2803

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 881  df-tru 1662  df-ex 1881  df-nf 1885  df-sb 2070  df-clab 2812  df-cleq 2818  df-clel 2821  df-nfc 2958  df-ral 3122  df-rab 3126  df-v 3416  df-dif 3801  df-un 3803  df-in 3805

This theorem is referenced by:  dif32  4120  difabs  4121  difpr  4552  infdiffi  8832  mreexexlem4d  16660  nulmbl2  23702  unmbl  23703  caragenuncllem  41520