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Mirrors > Home > MPE Home > Th. List > disjssun | Structured version Visualization version GIF version |
Description: Subset relation for disjoint classes. (Contributed by NM, 25-Oct-2005.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) |
Ref | Expression |
---|---|
disjssun | ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | uneq2 3761 | . . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = ((𝐴 ∩ 𝐶) ∪ ∅)) | |
2 | indi 3873 | . . . . 5 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) | |
3 | 2 | equncomi 3759 | . . . 4 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) |
4 | un0 3967 | . . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∪ ∅) = (𝐴 ∩ 𝐶) | |
5 | 4 | eqcomi 2631 | . . . 4 ⊢ (𝐴 ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∪ ∅) |
6 | 1, 3, 5 | 3eqtr4g 2681 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ (𝐵 ∪ 𝐶)) = (𝐴 ∩ 𝐶)) |
7 | 6 | eqeq1d 2624 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴 ↔ (𝐴 ∩ 𝐶) = 𝐴)) |
8 | df-ss 3588 | . 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴) | |
9 | df-ss 3588 | . 2 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐴 ∩ 𝐶) = 𝐴) | |
10 | 7, 8, 9 | 3bitr4g 303 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶)) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 196 = wceq 1483 ∪ cun 3572 ∩ cin 3573 ⊆ wss 3574 ∅c0 3915 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1722 ax-4 1737 ax-5 1839 ax-6 1888 ax-7 1935 ax-9 1999 ax-10 2019 ax-11 2034 ax-12 2047 ax-13 2246 ax-ext 2602 |
This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-tru 1486 df-ex 1705 df-nf 1710 df-sb 1881 df-clab 2609 df-cleq 2615 df-clel 2618 df-nfc 2753 df-v 3202 df-dif 3577 df-un 3579 df-in 3581 df-ss 3588 df-nul 3916 |
This theorem is referenced by: hashbclem 13236 alexsubALTlem2 21852 iccntr 22624 reconnlem1 22629 dvne0 23774 |
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