Theorem difun1 4265

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 4194	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
2		invdif 4245	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
3	1, 2	eqtr3i 2755	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
4		undm 4263	. . . . 5 ⊢ (V ∖ (𝐵 ∪ 𝐶)) = ((V ∖ 𝐵) ∩ (V ∖ 𝐶))
5	4	ineq2i 4183	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
6		invdif 4245	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
7	5, 6	eqtr3i 2755	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
8	3, 7	eqtr3i 2755	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = (𝐴 ∖ (𝐵 ∪ 𝐶))
9		invdif 4245	. . 3 ⊢ (𝐴 ∩ (V ∖ 𝐵)) = (𝐴 ∖ 𝐵)
10	9	difeq1i 4088	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = ((𝐴 ∖ 𝐵) ∖ 𝐶)
11	8, 10	eqtr3i 2755	1 ⊢ (𝐴 ∖ (𝐵 ∪ 𝐶)) = ((𝐴 ∖ 𝐵) ∖ 𝐶)

Syntax hints:   = wceq 1540  Vcvv 3450   ∖ cdif 3914   ∪ cun 3915   ∩ cin 3916

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-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-rab 3409  df-v 3452  df-dif 3920  df-un 3922  df-in 3924

This theorem is referenced by:  dif32  4268  difabs  4269  difpr  4770  infdiffi  9618  mreexexlem4d  17615  nulmbl2  25444  unmbl  25445  caragenuncllem  46517