| Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
| 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 6137 | . 2 ⊢ ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) = (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) | |
| 2 | difid 4339 | . . . . 5 ⊢ (𝐴 ∖ 𝐴) = ∅ | |
| 3 | 2 | xpeq1i 5664 | . . . 4 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = (∅ × 𝐵) |
| 4 | 0xp 5737 | . . . 4 ⊢ (∅ × 𝐵) = ∅ | |
| 5 | 3, 4 | eqtri 2752 | . . 3 ⊢ ((𝐴 ∖ 𝐴) × 𝐵) = ∅ |
| 6 | 5 | uneq1i 4127 | . 2 ⊢ (((𝐴 ∖ 𝐴) × 𝐵) ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) |
| 7 | uncom 4121 | . . 3 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) | |
| 8 | un0 4357 | . . 3 ⊢ ((𝐴 × (𝐵 ∖ 𝐶)) ∪ ∅) = (𝐴 × (𝐵 ∖ 𝐶)) | |
| 9 | 7, 8 | eqtri 2752 | . 2 ⊢ (∅ ∪ (𝐴 × (𝐵 ∖ 𝐶))) = (𝐴 × (𝐵 ∖ 𝐶)) |
| 10 | 1, 6, 9 | 3eqtrri 2757 | 1 ⊢ (𝐴 × (𝐵 ∖ 𝐶)) = ((𝐴 × 𝐵) ∖ (𝐴 × 𝐶)) |
| Colors of variables: wff setvar class |
| Syntax hints: = wceq 1540 ∖ cdif 3911 ∪ cun 3912 ∅c0 4296 × cxp 5636 |
| 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 1910 ax-6 1967 ax-7 2008 ax-8 2111 ax-9 2119 ax-ext 2701 ax-sep 5251 ax-nul 5261 ax-pr 5387 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3an 1088 df-tru 1543 df-fal 1553 df-ex 1780 df-sb 2066 df-clab 2708 df-cleq 2721 df-clel 2803 df-ral 3045 df-rex 3054 df-rab 3406 df-v 3449 df-dif 3917 df-un 3919 df-ss 3931 df-nul 4297 df-if 4489 df-sn 4590 df-pr 4592 df-op 4596 df-opab 5170 df-xp 5644 df-rel 5645 |
| This theorem is referenced by: difxp2ss 32452 imadifxp 32530 sxbrsigalem2 34277 |
| Copyright terms: Public domain | W3C validator |