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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 6027 | . 2 ⊢ ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) | |
2 | difid 4285 | . . . . 5 ⊢ (𝐴 ∖ 𝐴) = ∅ | |
3 | 2 | xpeq1i 5577 | . . . 4 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = (∅ × 𝐵) |
4 | 0xp 5646 | . . . 4 ⊢ (∅ × 𝐵) = ∅ | |
5 | 3, 4 | eqtri 2765 | . . 3 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = ∅ |
6 | 5 | uneq1i 4073 | . 2 ⊢ (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) |
7 | uncom 4067 | . . 3 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) | |
8 | un0 4305 | . . 3 ⊢ ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) = (𝐴 × (𝐵 ∖ 𝐶)) | |
9 | 7, 8 | eqtri 2765 | . 2 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (𝐴 × (𝐵 ∖ 𝐶)) |
10 | 1, 6, 9 | 3eqtrri 2770 | 1 ⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1543 ∖ cdif 3863 ∪ cun 3864 ∅c0 4237 × cxp 5549 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1803 ax-4 1817 ax-5 1918 ax-6 1976 ax-7 2016 ax-8 2112 ax-9 2120 ax-12 2175 ax-ext 2708 ax-sep 5192 ax-nul 5199 ax-pr 5322 |
This theorem depends on definitions: df-bi 210 df-an 400 df-or 848 df-3an 1091 df-tru 1546 df-fal 1556 df-ex 1788 df-sb 2071 df-clab 2715 df-cleq 2729 df-clel 2816 df-ral 3066 df-rex 3067 df-rab 3070 df-v 3410 df-dif 3869 df-un 3871 df-in 3873 df-ss 3883 df-nul 4238 df-if 4440 df-sn 4542 df-pr 4544 df-op 4548 df-opab 5116 df-xp 5557 df-rel 5558 |
This theorem is referenced by: difxp2ss 30590 imadifxp 30659 sxbrsigalem2 31965 |
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