Theorem difxp1 5804

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 5803	. 2 ⊢ ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶)))
2		difid 4180	. . . . 5 ⊢ (𝐶 ∖ 𝐶) = ∅
3	2	xpeq2i 5373	. . . 4 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = (𝐴 × ∅)
4		xp0 5797	. . . 4 ⊢ (𝐴 × ∅) = ∅
5	3, 4	eqtri 2849	. . 3 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = ∅
6	5	uneq2i 3993	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶))) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅)
7		un0 4194	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅) = ((𝐴 ∖ 𝐵) × 𝐶)
8	1, 6, 7	3eqtrri 2854	1 ⊢ ((𝐴 ∖ 𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))

Syntax hints:   = wceq 1656   ∖ cdif 3795   ∪ cun 3796  ∅c0 4146   × cxp 5344

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908  ax-5 2009  ax-6 2075  ax-7 2112  ax-9 2173  ax-10 2192  ax-11 2207  ax-12 2220  ax-13 2389  ax-ext 2803  ax-sep 5007  ax-nul 5015  ax-pr 5129

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 879  df-3an 1113  df-tru 1660  df-ex 1879  df-nf 1883  df-sb 2068  df-mo 2605  df-eu 2640  df-clab 2812  df-cleq 2818  df-clel 2821  df-nfc 2958  df-ral 3122  df-rex 3123  df-rab 3126  df-v 3416  df-dif 3801  df-un 3803  df-in 3805  df-ss 3812  df-nul 4147  df-if 4309  df-sn 4400  df-pr 4402  df-op 4406  df-br 4876  df-opab 4938  df-xp 5352  df-rel 5353  df-cnv 5354

This theorem is referenced by:  sxbrsigalem2  30889