Theorem indif2 3366

Description: Bring an intersection in and out of a class difference. (Contributed by Jeff Hankins, 15-Jul-2009.)

Assertion

Ref	Expression
indif2	⊢ (𝐴 ∩ (𝐵 ∖ 𝐶)) = ((𝐴 ∩ 𝐵) ∖ 𝐶)

Proof of Theorem indif2

Step	Hyp	Ref	Expression
1		inass 3332	. 2 ⊢ ((𝐴 ∩ 𝐵) ∩ (V ∖ 𝐶)) = (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶)))
2		invdif 3364	. 2 ⊢ ((𝐴 ∩ 𝐵) ∩ (V ∖ 𝐶)) = ((𝐴 ∩ 𝐵) ∖ 𝐶)
3		invdif 3364	. . 3 ⊢ (𝐵 ∩ (V ∖ 𝐶)) = (𝐵 ∖ 𝐶)
4	3	ineq2i 3320	. 2 ⊢ (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶))) = (𝐴 ∩ (𝐵 ∖ 𝐶))
5	1, 2, 4	3eqtr3ri 2195	1 ⊢ (𝐴 ∩ (𝐵 ∖ 𝐶)) = ((𝐴 ∩ 𝐵) ∖ 𝐶)

Syntax hints:   = wceq 1343  Vcvv 2726   ∖ cdif 3113   ∩ cin 3115

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-v 2728  df-dif 3118  df-in 3122

This theorem is referenced by:  indif1  3367  indifcom  3368  difopn  12748