Theorem indif1 4288

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 4287	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐵 ∩ 𝐴) ∖ 𝐶)
2		incom 4217	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐶) ∩ 𝐵)
3		incom 4217	. . 3 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
4	3	difeq1i 4132	. 2 ⊢ ((𝐵 ∩ 𝐴) ∖ 𝐶) = ((𝐴 ∩ 𝐵) ∖ 𝐶)
5	1, 2, 4	3eqtr3i 2771	1 ⊢ ((𝐴 ∖ 𝐶) ∩ 𝐵) = ((𝐴 ∩ 𝐵) ∖ 𝐶)

Syntax hints:   = wceq 1537   ∖ cdif 3960   ∩ cin 3962

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-dif 3966  df-in 3970

This theorem is referenced by:  resdifcom  6019  resdmdfsn  6051  hartogslem1  9580  fpwwe2  10681  leiso  14495  basdif0  22976  tgdif0  23015  kqdisj  23756  trufil  23934  difininv  32545  gtiso  32716  dfon4  35875  disjdifb  48658