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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 6485 | . . 3 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) | |
2 | 1 | orcomd 871 | . 2 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶)) |
3 | ssequn2 4199 | . . . 4 ⊢ (𝐶 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵) | |
4 | eqeq1 2739 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵)) | |
5 | 3, 4 | bitr4id 290 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐶 ⊆ 𝐵 ↔ 𝐴 = 𝐵)) |
6 | ssequn1 4196 | . . . 4 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶) | |
7 | eqeq1 2739 | . . . 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 1537 ∪ cun 3961 ⊆ wss 3963 Ord word 6385 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1908 ax-6 1965 ax-7 2005 ax-8 2108 ax-9 2116 ax-ext 2706 ax-sep 5302 ax-nul 5312 ax-pr 5438 |
This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3or 1087 df-3an 1088 df-tru 1540 df-fal 1550 df-ex 1777 df-sb 2063 df-clab 2713 df-cleq 2727 df-clel 2814 df-ne 2939 df-ral 3060 df-rex 3069 df-rab 3434 df-v 3480 df-dif 3966 df-un 3968 df-in 3970 df-ss 3980 df-pss 3983 df-nul 4340 df-if 4532 df-pw 4607 df-sn 4632 df-pr 4634 df-op 4638 df-uni 4913 df-br 5149 df-opab 5211 df-tr 5266 df-eprel 5589 df-po 5597 df-so 5598 df-fr 5641 df-we 5643 df-ord 6389 |
This theorem is referenced by: ordun 6490 inar1 10813 |
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