Theorem difun1 3397

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 3347	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
2		invdif 3379	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
3	1, 2	eqtr3i 2200	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
4		undm 3395	. . . . 5 ⊢ (V ∖ (𝐵 ∪ 𝐶)) = ((V ∖ 𝐵) ∩ (V ∖ 𝐶))
5	4	ineq2i 3335	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
6		invdif 3379	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
7	5, 6	eqtr3i 2200	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
8	3, 7	eqtr3i 2200	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = (𝐴 ∖ (𝐵 ∪ 𝐶))
9		invdif 3379	. . 3 ⊢ (𝐴 ∩ (V ∖ 𝐵)) = (𝐴 ∖ 𝐵)
10	9	difeq1i 3251	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = ((𝐴 ∖ 𝐵) ∖ 𝐶)
11	8, 10	eqtr3i 2200	1 ⊢ (𝐴 ∖ (𝐵 ∪ 𝐶)) = ((𝐴 ∖ 𝐵) ∖ 𝐶)

Syntax hints:   = wceq 1353  Vcvv 2739   ∖ cdif 3128   ∪ cun 3129   ∩ cin 3130

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-rab 2464  df-v 2741  df-dif 3133  df-un 3135  df-in 3137

This theorem is referenced by:  dif32  3400  difabs  3401  difpr  3736  diffifi  6896  difinfinf  7102