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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 6089 | . 2 ⊢ ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) | |
2 | difid 4314 | . . . . 5 ⊢ (𝐴 ∖ 𝐴) = ∅ | |
3 | 2 | xpeq1i 5633 | . . . 4 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = (∅ × 𝐵) |
4 | 0xp 5703 | . . . 4 ⊢ (∅ × 𝐵) = ∅ | |
5 | 3, 4 | eqtri 2764 | . . 3 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = ∅ |
6 | 5 | uneq1i 4103 | . 2 ⊢ (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) |
7 | uncom 4097 | . . 3 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) | |
8 | un0 4334 | . . 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 1540 ∖ cdif 3893 ∪ cun 3894 ∅c0 4266 × cxp 5605 |
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 1912 ax-6 1970 ax-7 2010 ax-8 2107 ax-9 2115 ax-ext 2707 ax-sep 5237 ax-nul 5244 ax-pr 5366 |
This theorem depends on definitions: df-bi 206 df-an 397 df-or 845 df-3an 1088 df-tru 1543 df-fal 1553 df-ex 1781 df-sb 2067 df-clab 2714 df-cleq 2728 df-clel 2814 df-ral 3062 df-rex 3071 df-rab 3404 df-v 3442 df-dif 3899 df-un 3901 df-in 3903 df-ss 3913 df-nul 4267 df-if 4471 df-sn 4571 df-pr 4573 df-op 4577 df-opab 5149 df-xp 5613 df-rel 5614 |
This theorem is referenced by: difxp2ss 31002 imadifxp 31071 sxbrsigalem2 32389 |
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