Theorem difun1 4248

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 4176	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
2		invdif 4228	. . . 4 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∩ (V ∖ 𝐶)) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
3	1, 2	eqtr3i 2787	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶)
4		undm 4246	. . . . 5 ⊢ (V ∖ (𝐵 ∪ 𝐶)) = ((V ∖ 𝐵) ∩ (V ∖ 𝐶))
5	4	ineq2i 4166	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶)))
6		invdif 4228	. . . 4 ⊢ (𝐴 ∩ (V ∖ (𝐵 ∪ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
7	5, 6	eqtr3i 2787	. . 3 ⊢ (𝐴 ∩ ((V ∖ 𝐵) ∩ (V ∖ 𝐶))) = (𝐴 ∖ (𝐵 ∪ 𝐶))
8	3, 7	eqtr3i 2787	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = (𝐴 ∖ (𝐵 ∪ 𝐶))
9		invdif 4228	. . 3 ⊢ (𝐴 ∩ (V ∖ 𝐵)) = (𝐴 ∖ 𝐵)
10	9	difeq1i 4073	. 2 ⊢ ((𝐴 ∩ (V ∖ 𝐵)) ∖ 𝐶) = ((𝐴 ∖ 𝐵) ∖ 𝐶)
11	8, 10	eqtr3i 2787	1 ⊢ (𝐴 ∖ (𝐵 ∪ 𝐶)) = ((𝐴 ∖ 𝐵) ∖ 𝐶)

This proof depends on syntax axioms:   = wceq 1570  Vcvv 3453   ∖ cdif 3899   ∪ cun 3900   ∩ cin 3901

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909

This theorem is used by:  dif32  4251  difabs  4252  difpr  4769  infdiffi  9640  mreexexlem4d  17739  nulmbl2  25765  unmbl  25766  difelsiga  34632  caragenuncllem  47327