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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 6494 | . . 3 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) | |
2 | 1 | orcomd 870 | . 2 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶)) |
3 | ssequn2 4212 | . . . 4 ⊢ (𝐶 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵) | |
4 | eqeq1 2744 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵)) | |
5 | 3, 4 | bitr4id 290 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐶 ⊆ 𝐵 ↔ 𝐴 = 𝐵)) |
6 | ssequn1 4209 | . . . 4 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶) | |
7 | eqeq1 2744 | . . . 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 917 | . 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 846 = wceq 1537 ∪ cun 3974 ⊆ wss 3976 Ord word 6394 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1793 ax-4 1807 ax-5 1909 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-ext 2711 ax-sep 5317 ax-nul 5324 ax-pr 5447 |
This theorem depends on definitions: df-bi 207 df-an 396 df-or 847 df-3or 1088 df-3an 1089 df-tru 1540 df-fal 1550 df-ex 1778 df-sb 2065 df-clab 2718 df-cleq 2732 df-clel 2819 df-ne 2947 df-ral 3068 df-rex 3077 df-rab 3444 df-v 3490 df-dif 3979 df-un 3981 df-in 3983 df-ss 3993 df-pss 3996 df-nul 4353 df-if 4549 df-pw 4624 df-sn 4649 df-pr 4651 df-op 4655 df-uni 4932 df-br 5167 df-opab 5229 df-tr 5284 df-eprel 5599 df-po 5607 df-so 5608 df-fr 5652 df-we 5654 df-ord 6398 |
This theorem is referenced by: ordun 6499 inar1 10844 |
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