Theorem difxp1 6057

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 6056	. 2 ⊢ ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶)))
2		difid 4301	. . . . 5 ⊢ (𝐶 ∖ 𝐶) = ∅
3	2	xpeq2i 5607	. . . 4 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = (𝐴 × ∅)
4		xp0 6050	. . . 4 ⊢ (𝐴 × ∅) = ∅
5	3, 4	eqtri 2766	. . 3 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = ∅
6	5	uneq2i 4090	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶))) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅)
7		un0 4321	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅) = ((𝐴 ∖ 𝐵) × 𝐶)
8	1, 6, 7	3eqtrri 2771	1 ⊢ ((𝐴 ∖ 𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))

Syntax hints:   = wceq 1539   ∖ cdif 3880   ∪ cun 3881  ∅c0 4253   × cxp 5578

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-br 5071  df-opab 5133  df-xp 5586  df-rel 5587  df-cnv 5588

This theorem is referenced by:  resdifdi  6128  difxp1ss  30771  sxbrsigalem2  32153