Theorem difxp2 6120

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

Assertion

Ref	Expression
difxp2	⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))

Proof of Theorem difxp2

Step	Hyp	Ref	Expression
1		difxp 6118	. 2 ⊢ ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶)))
2		difid 4325	. . . . 5 ⊢ (𝐴 ∖ 𝐴) = ∅
3	2	xpeq1i 5647	. . . 4 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = (∅ × 𝐵)
4		0xp 5720	. . . 4 ⊢ (∅ × 𝐵) = ∅
5	3, 4	eqtri 2756	. . 3 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = ∅
6	5	uneq1i 4113	. 2 ⊢ (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶)))
7		uncom 4107	. . 3 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅)
8		un0 4343	. . 3 ⊢ ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) = (𝐴 × (𝐵 ∖ 𝐶))
9	7, 8	eqtri 2756	. 2 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (𝐴 × (𝐵 ∖ 𝐶))
10	1, 6, 9	3eqtrri 2761	1 ⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶))

Syntax hints:   = wceq 1541   ∖ cdif 3895   ∪ cun 3896  ∅c0 4282   × cxp 5619

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705  ax-sep 5238  ax-nul 5248  ax-pr 5374

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-ral 3049  df-rex 3058  df-rab 3397  df-v 3439  df-dif 3901  df-un 3903  df-ss 3915  df-nul 4283  df-if 4477  df-sn 4578  df-pr 4580  df-op 4584  df-opab 5158  df-xp 5627  df-rel 5628

This theorem is referenced by:  difxp2ss  32507  imadifxp  32585  sxbrsigalem2  34322