Theorem difun1 3515

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 3466	. . . 4 ⊢ ((A ∩ (V ∖ B)) ∩ (V ∖ C)) = (A ∩ ((V ∖ B) ∩ (V ∖ C)))
2		invdif 3497	. . . 4 ⊢ ((A ∩ (V ∖ B)) ∩ (V ∖ C)) = ((A ∩ (V ∖ B)) ∖ C)
3	1, 2	eqtr3i 2375	. . 3 ⊢ (A ∩ ((V ∖ B) ∩ (V ∖ C))) = ((A ∩ (V ∖ B)) ∖ C)
4		undm 3513	. . . . 5 ⊢ (V ∖ (B ∪ C)) = ((V ∖ B) ∩ (V ∖ C))
5	4	ineq2i 3455	. . . 4 ⊢ (A ∩ (V ∖ (B ∪ C))) = (A ∩ ((V ∖ B) ∩ (V ∖ C)))
6		invdif 3497	. . . 4 ⊢ (A ∩ (V ∖ (B ∪ C))) = (A ∖ (B ∪ C))
7	5, 6	eqtr3i 2375	. . 3 ⊢ (A ∩ ((V ∖ B) ∩ (V ∖ C))) = (A ∖ (B ∪ C))
8	3, 7	eqtr3i 2375	. 2 ⊢ ((A ∩ (V ∖ B)) ∖ C) = (A ∖ (B ∪ C))
9		invdif 3497	. . 3 ⊢ (A ∩ (V ∖ B)) = (A ∖ B)
10	9	difeq1i 3382	. 2 ⊢ ((A ∩ (V ∖ B)) ∖ C) = ((A ∖ B) ∖ C)
11	8, 10	eqtr3i 2375	1 ⊢ (A ∖ (B ∪ C)) = ((A ∖ B) ∖ C)

This proof depends on syntax axioms:   = wceq 1642  Vcvv 2860   ∖ cdif 3207   ∪ cun 3208   ∩ cin 3209

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-dif 3216

This theorem is used by:  dif32  3518  difabs  3519