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Mirrors > Home > MPE Home > Th. List > difxp2 | Structured version Visualization version GIF version |
Description: Difference law for Cartesian product. (Contributed by Scott Fenton, 18-Feb-2013.) (Revised by Mario Carneiro, 26-Jun-2014.) |
Ref | Expression |
---|---|
difxp2 | ⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | difxp 6078 | . 2 ⊢ ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) | |
2 | difid 4310 | . . . . 5 ⊢ (𝐴 ∖ 𝐴) = ∅ | |
3 | 2 | xpeq1i 5622 | . . . 4 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = (∅ × 𝐵) |
4 | 0xp 5692 | . . . 4 ⊢ (∅ × 𝐵) = ∅ | |
5 | 3, 4 | eqtri 2764 | . . 3 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = ∅ |
6 | 5 | uneq1i 4099 | . 2 ⊢ (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) |
7 | uncom 4093 | . . 3 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) | |
8 | un0 4330 | . . 3 ⊢ ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) = (𝐴 × (𝐵 ∖ 𝐶)) | |
9 | 7, 8 | eqtri 2764 | . 2 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (𝐴 × (𝐵 ∖ 𝐶)) |
10 | 1, 6, 9 | 3eqtrri 2769 | 1 ⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1539 ∖ cdif 3889 ∪ cun 3890 ∅c0 4262 × cxp 5594 |
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 1911 ax-6 1969 ax-7 2009 ax-8 2106 ax-9 2114 ax-ext 2707 ax-sep 5232 ax-nul 5239 ax-pr 5361 |
This theorem depends on definitions: df-bi 206 df-an 398 df-or 846 df-3an 1089 df-tru 1542 df-fal 1552 df-ex 1780 df-sb 2066 df-clab 2714 df-cleq 2728 df-clel 2814 df-ral 3063 df-rex 3072 df-rab 3287 df-v 3439 df-dif 3895 df-un 3897 df-in 3899 df-ss 3909 df-nul 4263 df-if 4466 df-sn 4566 df-pr 4568 df-op 4572 df-opab 5144 df-xp 5602 df-rel 5603 |
This theorem is referenced by: difxp2ss 30912 imadifxp 30981 sxbrsigalem2 32294 |
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