Theorem difxp1 6108

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

Assertion

Ref	Expression
difxp1	⊢ ((𝐴 ∖ 𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))

Proof of Theorem difxp1

Step	Hyp	Ref	Expression
1		difxp 6107	. 2 ⊢ ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶)))
2		difid 4322	. . . . 5 ⊢ (𝐶 ∖ 𝐶) = ∅
3	2	xpeq2i 5652	. . . 4 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = (𝐴 × ∅)
4		xp0 6101	. . . 4 ⊢ (𝐴 × ∅) = ∅
5	3, 4	eqtri 2765	. . 3 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = ∅
6	5	uneq2i 4112	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶))) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅)
7		un0 4342	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅) = ((𝐴 ∖ 𝐵) × 𝐶)
8	1, 6, 7	3eqtrri 2770	1 ⊢ ((𝐴 ∖ 𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))

Syntax hints:   = wceq 1541   ∖ cdif 3899   ∪ cun 3900  ∅c0 4274   × cxp 5623

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2708  ax-sep 5248  ax-nul 5255  ax-pr 5377

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 846  df-3an 1089  df-tru 1544  df-fal 1554  df-ex 1782  df-nf 1786  df-sb 2068  df-clab 2715  df-cleq 2729  df-clel 2815  df-ral 3063  df-rex 3072  df-rab 3405  df-v 3444  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4275  df-if 4479  df-sn 4579  df-pr 4581  df-op 4585  df-br 5098  df-opab 5160  df-xp 5631  df-rel 5632  df-cnv 5633

This theorem is referenced by:  resdifdi  6179  difxp1ss  31155  sxbrsigalem2  32551