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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 6175 | . 2 ⊢ ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) | |
2 | difid 4375 | . . . . 5 ⊢ (𝐴 ∖ 𝐴) = ∅ | |
3 | 2 | xpeq1i 5708 | . . . 4 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = (∅ × 𝐵) |
4 | 0xp 5780 | . . . 4 ⊢ (∅ × 𝐵) = ∅ | |
5 | 3, 4 | eqtri 2754 | . . 3 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = ∅ |
6 | 5 | uneq1i 4159 | . 2 ⊢ (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) |
7 | uncom 4153 | . . 3 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) | |
8 | un0 4395 | . . 3 ⊢ ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) = (𝐴 × (𝐵 ∖ 𝐶)) | |
9 | 7, 8 | eqtri 2754 | . 2 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (𝐴 × (𝐵 ∖ 𝐶)) |
10 | 1, 6, 9 | 3eqtrri 2759 | 1 ⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1534 ∖ cdif 3944 ∪ cun 3945 ∅c0 4325 × cxp 5680 |
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-ext 2697 ax-sep 5304 ax-nul 5311 ax-pr 5433 |
This theorem depends on definitions: df-bi 206 df-an 395 df-or 846 df-3an 1086 df-tru 1537 df-fal 1547 df-ex 1775 df-sb 2061 df-clab 2704 df-cleq 2718 df-clel 2803 df-ral 3052 df-rex 3061 df-rab 3420 df-v 3464 df-dif 3950 df-un 3952 df-ss 3964 df-nul 4326 df-if 4534 df-sn 4634 df-pr 4636 df-op 4640 df-opab 5216 df-xp 5688 df-rel 5689 |
This theorem is referenced by: difxp2ss 32450 imadifxp 32521 sxbrsigalem2 34120 |
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