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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 6470 | . . 3 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) | |
| 2 | 1 | orcomd 885 | . 2 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶)) |
| 3 | ssequn2 4189 | . . . 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 4186 | . . . 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 932 | . 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 |
| This proof depends on syntax axioms: → wi 4 ∧ wa 401 ∨ wo 861 = wceq 1569 ⊆ wss 3899 ∪ cun 4151 Ord word 6367 |
| This proof depends on axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1827 ax-4 1841 ax-5 1944 ax-6 2001 ax-7 2042 ax-8 2148 ax-9 2156 ax-ext 2735 ax-sep 5251 ax-pr 5394 |
| This proof depends on definitions: df-bi 210 df-an 402 df-or 862 df-3or 1104 df-3an 1105 df-tru 1572 df-fal 1582 df-ex 1812 df-sb 2101 df-clab 2742 df-cleq 2755 df-clel 2838 df-ne 2959 df-ral 3080 df-rex 3090 df-rab 3416 df-v 3455 df-ss 3900 df-pss 4016 df-dif 4049 df-in 4093 df-un 4153 df-nul 4281 df-if 4485 df-pw 4561 df-sn 4588 df-pr 4590 df-op 4838 df-uni 4870 df-br 5106 df-opab 5170 df-tr 5215 df-eprel 5554 df-po 5562 df-so 5563 df-fr 5607 df-we 5609 df-ord 6371 |
| This theorem is used by: ordun 6475 inar1 10870 |
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