Theorem difxp1 6068

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 6067	. 2 ⊢ ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶)))
2		difid 4304	. . . . 5 ⊢ (𝐶 ∖ 𝐶) = ∅
3	2	xpeq2i 5616	. . . 4 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = (𝐴 × ∅)
4		xp0 6061	. . . 4 ⊢ (𝐴 × ∅) = ∅
5	3, 4	eqtri 2766	. . 3 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = ∅
6	5	uneq2i 4094	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶))) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅)
7		un0 4324	. 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 3884   ∪ cun 3885  ∅c0 4256   × cxp 5587

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  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 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070  df-rab 3073  df-v 3434  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568  df-br 5075  df-opab 5137  df-xp 5595  df-rel 5596  df-cnv 5597

This theorem is referenced by:  resdifdi  6139  difxp1ss  30870  sxbrsigalem2  32253