Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
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 4130 | . . . 4 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) = ((𝐴 ∩ 𝐶) ∪ ∅)) | |
2 | indi 4247 | . . . . 5 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)) | |
3 | 2 | equncomi 4128 | . . . 4 ⊢ (𝐴 ∩ (𝐵 ∪ 𝐶)) = ((𝐴 ∩ 𝐶) ∪ (𝐴 ∩ 𝐵)) |
4 | un0 4341 | . . . . 5 ⊢ ((𝐴 ∩ 𝐶) ∪ ∅) = (𝐴 ∩ 𝐶) | |
5 | 4 | eqcomi 2827 | . . . 4 ⊢ (𝐴 ∩ 𝐶) = ((𝐴 ∩ 𝐶) ∪ ∅) |
6 | 1, 3, 5 | 3eqtr4g 2878 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ∩ (𝐵 ∪ 𝐶)) = (𝐴 ∩ 𝐶)) |
7 | 6 | eqeq1d 2820 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ → ((𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴 ↔ (𝐴 ∩ 𝐶) = 𝐴)) |
8 | df-ss 3949 | . 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∩ (𝐵 ∪ 𝐶)) = 𝐴) | |
9 | df-ss 3949 | . 2 ⊢ (𝐴 ⊆ 𝐶 ↔ (𝐴 ∩ 𝐶) = 𝐴) | |
10 | 7, 8, 9 | 3bitr4g 315 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ → (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ 𝐶)) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 207 = wceq 1528 ∪ cun 3931 ∩ cin 3932 ⊆ wss 3933 ∅c0 4288 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1787 ax-4 1801 ax-5 1902 ax-6 1961 ax-7 2006 ax-8 2107 ax-9 2115 ax-10 2136 ax-11 2151 ax-12 2167 ax-ext 2790 |
This theorem depends on definitions: df-bi 208 df-an 397 df-or 842 df-tru 1531 df-ex 1772 df-nf 1776 df-sb 2061 df-clab 2797 df-cleq 2811 df-clel 2890 df-nfc 2960 df-v 3494 df-dif 3936 df-un 3938 df-in 3940 df-ss 3949 df-nul 4289 |
This theorem is referenced by: hashbclem 13798 alexsubALTlem2 22584 iccntr 23356 reconnlem1 23361 dvne0 24535 |
Copyright terms: Public domain | W3C validator |