Theorem indif1 4160

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 4159	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐵 ∩ 𝐴) ∖ 𝐶)
2		incom 4089	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐶) ∩ 𝐵)
3		incom 4089	. . 3 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
4	3	difeq1i 4007	. 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 3838   ∩ cin 3840

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1916  ax-6 1974  ax-7 2019  ax-8 2115  ax-9 2123  ax-ext 2710

This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1545  df-ex 1787  df-sb 2074  df-clab 2717  df-cleq 2730  df-clel 2811  df-rab 3062  df-v 3399  df-dif 3844  df-in 3848

This theorem is referenced by:  resdifcom  5838  resdmdfsn  5869  hartogslem1  9072  fpwwe2  10136  leiso  13904  basdif0  21697  tgdif0  21736  kqdisj  22476  trufil  22654  difininv  30430  gtiso  30600  dfon4  33825  disjdifb  45671