Theorem indif2 3181

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 3147	. 2 ⊢ ((𝐴 ∩ 𝐵) ∩ (V ∖ 𝐶)) = (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶)))
2		invdif 3179	. 2 ⊢ ((𝐴 ∩ 𝐵) ∩ (V ∖ 𝐶)) = ((𝐴 ∩ 𝐵) ∖ 𝐶)
3		invdif 3179	. . 3 ⊢ (𝐵 ∩ (V ∖ 𝐶)) = (𝐵 ∖ 𝐶)
4	3	ineq2i 3135	. 2 ⊢ (𝐴 ∩ (𝐵 ∩ (V ∖ 𝐶))) = (𝐴 ∩ (𝐵 ∖ 𝐶))
5	1, 2, 4	3eqtr3ri 2069	1 ⊢ (𝐴 ∩ (𝐵 ∖ 𝐶)) = ((𝐴 ∩ 𝐵) ∖ 𝐶)

Syntax hints:   = wceq 1243  Vcvv 2557   ∖ cdif 2914   ∩ 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-v 2559  df-dif 2920  df-in 2924

This theorem is referenced by:  indif1  3182  indifcom  3183