Theorem difun1 3197

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 3147	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
2		invdif 3179	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
3	1, 2	eqtr3i 2062	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
4		undm 3195	. . . . 5 ⊢ (V ∖ (𝐵 ∪ 𝐶)) = ((V ∖ 𝐵) ∩ (V ∖ 𝐶))
5	4	ineq2i 3135	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
6		invdif 3179	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
7	5, 6	eqtr3i 2062	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
8	3, 7	eqtr3i 2062	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = (𝐴 ∖ (𝐵 ∪ 𝐶))
9		invdif 3179	. . 3 ⊢ (𝐴 ∩ (V ∖ 𝐵)) = (𝐴 ∖ 𝐵)
10	9	difeq1i 3058	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = ((𝐴 ∖ 𝐵) ∖ 𝐶)
11	8, 10	eqtr3i 2062	1 ⊢ (𝐴 ∖ (𝐵 ∪ 𝐶)) = ((𝐴 ∖ 𝐵) ∖ 𝐶)

Syntax hints:   = wceq 1243  Vcvv 2557   ∖ cdif 2914   ∪ cun 2915   ∩ cin 2916

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-rab 2315  df-v 2559  df-dif 2920  df-un 2922  df-in 2924

This theorem is referenced by:  dif32  3200  difabs  3201  diffifi  6351