Theorem indif1 3500

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 3499	. 2 ⊢ (B ∩ (A ∖ C)) = ((B ∩ A) ∖ C)
2		incom 3449	. 2 ⊢ (B ∩ (A ∖ C)) = ((A ∖ C) ∩ B)
3		incom 3449	. . 3 ⊢ (B ∩ A) = (A ∩ B)
4	3	difeq1i 3382	. 2 ⊢ ((B ∩ A) ∖ C) = ((A ∩ B) ∖ C)
5	1, 2, 4	3eqtr3i 2381	1 ⊢ ((A ∖ C) ∩ B) = ((A ∩ B) ∖ C)

Syntax hints:   = wceq 1642   ∖ cdif 3207   ∩ cin 3209

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216

This theorem is referenced by: (None)