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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 6467 | . . 3 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) | |
2 | 1 | orcomd 869 | . 2 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶)) |
3 | ssequn2 4181 | . . . 4 ⊢ (𝐶 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵) | |
4 | eqeq1 2730 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵)) | |
5 | 3, 4 | bitr4id 289 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐶 ⊆ 𝐵 ↔ 𝐴 = 𝐵)) |
6 | ssequn1 4178 | . . . 4 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶) | |
7 | eqeq1 2730 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶)) | |
8 | 6, 7 | bitr4id 289 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐵 ⊆ 𝐶 ↔ 𝐴 = 𝐶)) |
9 | 5, 8 | orbi12d 916 | . 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → ((𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶) ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
10 | 2, 9 | syl5ibcom 244 | 1 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ∧ wa 394 ∨ wo 845 = wceq 1534 ∪ cun 3944 ⊆ wss 3946 Ord word 6367 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1790 ax-4 1804 ax-5 1906 ax-6 1964 ax-7 2004 ax-8 2101 ax-9 2109 ax-ext 2697 ax-sep 5296 ax-nul 5303 ax-pr 5425 |
This theorem depends on definitions: df-bi 206 df-an 395 df-or 846 df-3or 1085 df-3an 1086 df-tru 1537 df-fal 1547 df-ex 1775 df-sb 2061 df-clab 2704 df-cleq 2718 df-clel 2803 df-ne 2931 df-ral 3052 df-rex 3061 df-rab 3420 df-v 3464 df-dif 3949 df-un 3951 df-in 3953 df-ss 3963 df-pss 3966 df-nul 4323 df-if 4524 df-pw 4599 df-sn 4624 df-pr 4626 df-op 4630 df-uni 4906 df-br 5146 df-opab 5208 df-tr 5263 df-eprel 5578 df-po 5586 df-so 5587 df-fr 5629 df-we 5631 df-ord 6371 |
This theorem is referenced by: ordun 6472 inar1 10809 |
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