Theorem difxp1 6163

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 6162	. 2 ⊢ ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶)) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶)))
2		difid 4366	. . . . 5 ⊢ (𝐶 ∖ 𝐶) = ∅
3	2	xpeq2i 5699	. . . 4 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = (𝐴 × ∅)
4		xp0 6156	. . . 4 ⊢ (𝐴 × ∅) = ∅
5	3, 4	eqtri 2755	. . 3 ⊢ (𝐴 × (𝐶 ∖ 𝐶)) = ∅
6	5	uneq2i 4156	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ (𝐴 × (𝐶 ∖ 𝐶))) = (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅)
7		un0 4386	. 2 ⊢ (((𝐴 ∖ 𝐵) × 𝐶) ∪ ∅) = ((𝐴 ∖ 𝐵) × 𝐶)
8	1, 6, 7	3eqtrri 2760	1 ⊢ ((𝐴 ∖ 𝐵) × 𝐶) = ((𝐴 × 𝐶) ∖ (𝐵 × 𝐶))

Syntax hints:   = wceq 1534   ∖ cdif 3941   ∪ cun 3942  ∅c0 4318   × cxp 5670

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2164  ax-ext 2698  ax-sep 5293  ax-nul 5300  ax-pr 5423

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-3an 1087  df-tru 1537  df-fal 1547  df-ex 1775  df-nf 1779  df-sb 2061  df-clab 2705  df-cleq 2719  df-clel 2805  df-ral 3057  df-rex 3066  df-rab 3428  df-v 3471  df-dif 3947  df-un 3949  df-in 3951  df-ss 3961  df-nul 4319  df-if 4525  df-sn 4625  df-pr 4627  df-op 4631  df-br 5143  df-opab 5205  df-xp 5678  df-rel 5679  df-cnv 5680

This theorem is referenced by:  resdifdi  6234  difxp1ss  32312  sxbrsigalem2  33896