Theorem indif1 4228

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 4227	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐵 ∩ 𝐴) ∖ 𝐶)
2		incom 4155	. 2 ⊢ (𝐵 ∩ (𝐴 ∖ 𝐶)) = ((𝐴 ∖ 𝐶) ∩ 𝐵)
3		incom 4155	. . 3 ⊢ (𝐵 ∩ 𝐴) = (𝐴 ∩ 𝐵)
4	3	difeq1i 4070	. 2 ⊢ ((𝐵 ∩ 𝐴) ∖ 𝐶) = ((𝐴 ∩ 𝐵) ∖ 𝐶)
5	1, 2, 4	3eqtr3i 2791	1 ⊢ ((𝐴 ∖ 𝐶) ∩ 𝐵) = ((𝐴 ∩ 𝐵) ∖ 𝐶)

This proof depends on syntax axioms:   = wceq 1570   ∖ cdif 3896   ∩ cin 3898

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-dif 3902  df-in 3906

This theorem is used by:  resdifcom  5991  resdmdfsn  6025  resdmdfsnOLD  6026  hartogslem1  9517  fpwwe2  10655  leiso  14527  basdif0  23181  tgdif0  23220  kqdisj  23961  trufil  24139  difininv  32995  gtiso  33176  dfon4  36473  disjdifb  49741