Theorem indif1 4232

Description: Bring an intersection in and out of a class difference. (Contributed by Mario Carneiro, 15-May-2015.)

Assertion

Ref	Expression
indif1	⊢ ((𝐴 ∖ 𝐶) ∩ 𝐵) = ((𝐴 ∩ 𝐵) ∖ 𝐶)

Proof of Theorem indif1

Step	Hyp	Ref	Expression
1		indif2 4231	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐵 ∩ 𝐴) ∖ 𝐶)
2		incom 4159	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐶) ∩ 𝐵)
3		incom 4159	. . 3 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
4	3	difeq1i 4072	. 2 ⊢ ((𝐵 ∩ 𝐴) ∖ 𝐶) = ((𝐴 ∩ 𝐵) ∖ 𝐶)
5	1, 2, 4	3eqtr3i 2765	1 ⊢ ((𝐴 ∖ 𝐶) ∩ 𝐵) = ((𝐴 ∩ 𝐵) ∖ 𝐶)

Syntax hints:   = wceq 1541   ∖ cdif 3896   ∩ cin 3898

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2713  df-cleq 2726  df-clel 2809  df-rab 3398  df-v 3440  df-dif 3902  df-in 3906

This theorem is referenced by:  resdifcom  5955  resdmdfsn  5988  hartogslem1  9445  fpwwe2  10552  leiso  14380  basdif0  22895  tgdif0  22934  kqdisj  23674  trufil  23852  difininv  32541  gtiso  32729  dfon4  36034  disjdifb  48997