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| Mirrors > Home > MPE Home > Th. List > ordequn | Structured version Visualization version GIF version | ||
| Description: The maximum (i.e. union) of two ordinals is either one or the other. Similar to Exercise 14 of [TakeutiZaring] p. 40. (Contributed by NM, 28-Nov-2003.) |
| Ref | Expression |
|---|---|
| ordequn | ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | ordtri2or2 6462 | . . 3 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) | |
| 2 | 1 | orcomd 884 | . 2 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶)) |
| 3 | ssequn2 4142 | . . . 4 ⊢ (𝐶 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵) | |
| 4 | eqeq1 2767 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵)) | |
| 5 | 3, 4 | bitr4id 293 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐶 ⊆ 𝐵 ↔ 𝐴 = 𝐵)) |
| 6 | ssequn1 4139 | . . . 4 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶) | |
| 7 | eqeq1 2767 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶)) | |
| 8 | 6, 7 | bitr4id 293 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐵 ⊆ 𝐶 ↔ 𝐴 = 𝐶)) |
| 9 | 5, 8 | orbi12d 931 | . 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → ((𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶) ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
| 10 | 2, 9 | syl5ibcom 248 | 1 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 ∧ wa 400 ∨ wo 860 = wceq 1570 ∪ cun 3903 ⊆ wss 3905 Ord word 6359 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1825 ax-4 1839 ax-5 1940 ax-6 1997 ax-7 2038 ax-8 2145 ax-9 2153 ax-ext 2735 ax-sep 5257 ax-pr 5404 |
| This theorem depends on definitions: df-bi 210 df-an 401 df-or 861 df-3or 1104 df-3an 1105 df-tru 1573 df-fal 1583 df-ex 1810 df-sb 2097 df-clab 2742 df-cleq 2755 df-clel 2838 df-ne 2959 df-ral 3080 df-rex 3090 df-rab 3417 df-v 3457 df-dif 3908 df-un 3910 df-in 3912 df-ss 3922 df-pss 3925 df-nul 4287 df-if 4488 df-pw 4564 df-sn 4590 df-pr 4592 df-op 4596 df-uni 4873 df-br 5110 df-opab 5174 df-tr 5219 df-eprel 5561 df-po 5569 df-so 5570 df-fr 5614 df-we 5616 df-ord 6363 |
| This theorem is referenced by: ordun 6467 inar1 10755 |
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