Theorem indif1 4269

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 4268	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐵 ∩ 𝐴) ∖ 𝐶)
2		incom 4199	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐶) ∩ 𝐵)
3		incom 4199	. . 3 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
4	3	difeq1i 4116	. 2 ⊢ ((𝐵 ∩ 𝐴) ∖ 𝐶) = ((𝐴 ∩ 𝐵) ∖ 𝐶)
5	1, 2, 4	3eqtr3i 2769	1 ⊢ ((𝐴 ∖ 𝐶) ∩ 𝐵) = ((𝐴 ∩ 𝐵) ∖ 𝐶)

Syntax hints:   = wceq 1542   ∖ cdif 3943   ∩ cin 3945

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-rab 3434  df-v 3477  df-dif 3949  df-in 3953

This theorem is referenced by:  resdifcom  5997  resdmdfsn  6028  hartogslem1  9532  fpwwe2  10633  leiso  14415  basdif0  22437  tgdif0  22476  kqdisj  23217  trufil  23395  difininv  31732  gtiso  31899  dfon4  34802  disjdifb  47395