Theorem difxp1 6025

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 6024	. 2 ⊢ ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶)))
2		difid 4333	. . . . 5 ⊢ (𝐶 ∖ 𝐶) = ∅
3	2	xpeq2i 5585	. . . 4 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = (𝐴 × ∅)
4		xp0 6018	. . . 4 ⊢ (𝐴 × ∅) = ∅
5	3, 4	eqtri 2847	. . 3 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = ∅
6	5	uneq2i 4139	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶))) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅)
7		un0 4347	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅) = ((𝐴 ∖ 𝐵) × 𝐶)
8	1, 6, 7	3eqtrri 2852	1 ⊢ ((𝐴 ∖ 𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))

Syntax hints:   = wceq 1536   ∖ cdif 3936   ∪ cun 3937  ∅c0 4294   × cxp 5556

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 1969  ax-7 2014  ax-8 2115  ax-9 2123  ax-10 2144  ax-11 2160  ax-12 2176  ax-ext 2796  ax-sep 5206  ax-nul 5213  ax-pr 5333

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1539  df-ex 1780  df-nf 1784  df-sb 2069  df-mo 2621  df-eu 2653  df-clab 2803  df-cleq 2817  df-clel 2896  df-nfc 2966  df-ral 3146  df-rex 3147  df-rab 3150  df-v 3499  df-dif 3942  df-un 3944  df-in 3946  df-ss 3955  df-nul 4295  df-if 4471  df-sn 4571  df-pr 4573  df-op 4577  df-br 5070  df-opab 5132  df-xp 5564  df-rel 5565  df-cnv 5566

This theorem is referenced by:  difxp1ss  30285  sxbrsigalem2  31548