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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 6418 | . . 3 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) | |
| 2 | 1 | orcomd 871 | . 2 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶)) |
| 3 | ssequn2 4141 | . . . 4 ⊢ (𝐶 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵) | |
| 4 | eqeq1 2740 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵)) | |
| 5 | 3, 4 | bitr4id 290 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐶 ⊆ 𝐵 ↔ 𝐴 = 𝐵)) |
| 6 | ssequn1 4138 | . . . 4 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶) | |
| 7 | eqeq1 2740 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶)) | |
| 8 | 6, 7 | bitr4id 290 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐵 ⊆ 𝐶 ↔ 𝐴 = 𝐶)) |
| 9 | 5, 8 | orbi12d 918 | . 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → ((𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶) ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
| 10 | 2, 9 | syl5ibcom 245 | 1 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 ∧ wa 395 ∨ wo 847 = wceq 1541 ∪ cun 3899 ⊆ wss 3901 Ord word 6316 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1968 ax-7 2009 ax-8 2115 ax-9 2123 ax-ext 2708 ax-sep 5241 ax-nul 5251 ax-pr 5377 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3or 1087 df-3an 1088 df-tru 1544 df-fal 1554 df-ex 1781 df-sb 2068 df-clab 2715 df-cleq 2728 df-clel 2811 df-ne 2933 df-ral 3052 df-rex 3061 df-rab 3400 df-v 3442 df-dif 3904 df-un 3906 df-in 3908 df-ss 3918 df-pss 3921 df-nul 4286 df-if 4480 df-pw 4556 df-sn 4581 df-pr 4583 df-op 4587 df-uni 4864 df-br 5099 df-opab 5161 df-tr 5206 df-eprel 5524 df-po 5532 df-so 5533 df-fr 5577 df-we 5579 df-ord 6320 |
| This theorem is referenced by: ordun 6423 inar1 10686 |
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