Theorem difxp2 6160

Description: Difference law for Cartesian product. (Contributed by Scott Fenton, 18-Feb-2013.) (Revised by Mario Carneiro, 26-Jun-2014.)

Assertion

Ref	Expression
difxp2	⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))

Proof of Theorem difxp2

Step	Hyp	Ref	Expression
1		difxp 6158	. 2 ⊢ ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶)))
2		difid 4356	. . . . 5 ⊢ (𝐴 ∖ 𝐴) = ∅
3	2	xpeq1i 5685	. . . 4 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = (∅ × 𝐵)
4		0xp 5758	. . . 4 ⊢ (∅ × 𝐵) = ∅
5	3, 4	eqtri 2759	. . 3 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = ∅
6	5	uneq1i 4144	. 2 ⊢ (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶)))
7		uncom 4138	. . 3 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅)
8		un0 4374	. . 3 ⊢ ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) = (𝐴 × (𝐵 ∖ 𝐶))
9	7, 8	eqtri 2759	. 2 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (𝐴 × (𝐵 ∖ 𝐶))
10	1, 6, 9	3eqtrri 2764	1 ⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))

Syntax hints:   = wceq 1540   ∖ cdif 3928   ∪ cun 3929  ∅c0 4313   × cxp 5657

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2708  ax-sep 5271  ax-nul 5281  ax-pr 5407

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810  df-ral 3053  df-rex 3062  df-rab 3421  df-v 3466  df-dif 3934  df-un 3936  df-ss 3948  df-nul 4314  df-if 4506  df-sn 4607  df-pr 4609  df-op 4613  df-opab 5187  df-xp 5665  df-rel 5666

This theorem is referenced by:  difxp2ss  32509  imadifxp  32587  sxbrsigalem2  34323