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Mirrors > Home > MPE Home > Th. List > Mathboxes > difuncomp | Structured version Visualization version GIF version |
Description: Express a class difference using unions and class complements. (Contributed by Thierry Arnoux, 21-Jun-2020.) |
Ref | Expression |
---|---|
difuncomp | ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | incom 3767 | . . . 4 ⊢ (𝐶 ∩ 𝐴) = (𝐴 ∩ 𝐶) | |
2 | sseqin2 3779 | . . . . 5 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∩ 𝐴) = 𝐴) | |
3 | 2 | biimpi 205 | . . . 4 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∩ 𝐴) = 𝐴) |
4 | 1, 3 | syl5reqr 2659 | . . 3 ⊢ (𝐴 ⊆ 𝐶 → 𝐴 = (𝐴 ∩ 𝐶)) |
5 | 4 | difeq1d 3689 | . 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = ((𝐴 ∩ 𝐶) ∖ 𝐵)) |
6 | difundi 3838 | . . . 4 ⊢ (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵)) | |
7 | dfss4 3820 | . . . . . 6 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴) | |
8 | 7 | biimpi 205 | . . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ (𝐶 ∖ 𝐴)) = 𝐴) |
9 | 8 | ineq1d 3775 | . . . 4 ⊢ (𝐴 ⊆ 𝐶 → ((𝐶 ∖ (𝐶 ∖ 𝐴)) ∩ (𝐶 ∖ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵))) |
10 | 6, 9 | syl5eq 2656 | . . 3 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = (𝐴 ∩ (𝐶 ∖ 𝐵))) |
11 | indif2 3829 | . . 3 ⊢ (𝐴 ∩ (𝐶 ∖ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵) | |
12 | 10, 11 | syl6eq 2660 | . 2 ⊢ (𝐴 ⊆ 𝐶 → (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵)) = ((𝐴 ∩ 𝐶) ∖ 𝐵)) |
13 | 5, 12 | eqtr4d 2647 | 1 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = (𝐶 ∖ ((𝐶 ∖ 𝐴) ∪ 𝐵))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 = wceq 1475 ∖ cdif 3537 ∪ cun 3538 ∩ cin 3539 ⊆ wss 3540 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ral 2901 df-rab 2905 df-v 3175 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 |
This theorem is referenced by: ldgenpisyslem1 29553 |
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