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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 6483 | . . 3 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) | |
| 2 | 1 | orcomd 872 | . 2 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶)) | 
| 3 | ssequn2 4189 | . . . 4 ⊢ (𝐶 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵) | |
| 4 | eqeq1 2741 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵)) | |
| 5 | 3, 4 | bitr4id 290 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐶 ⊆ 𝐵 ↔ 𝐴 = 𝐵)) | 
| 6 | ssequn1 4186 | . . . 4 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶) | |
| 7 | eqeq1 2741 | . . . 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 919 | . 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 848 = wceq 1540 ∪ cun 3949 ⊆ wss 3951 Ord word 6383 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-ext 2708 ax-sep 5296 ax-nul 5306 ax-pr 5432 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 849 df-3or 1088 df-3an 1089 df-tru 1543 df-fal 1553 df-ex 1780 df-sb 2065 df-clab 2715 df-cleq 2729 df-clel 2816 df-ne 2941 df-ral 3062 df-rex 3071 df-rab 3437 df-v 3482 df-dif 3954 df-un 3956 df-in 3958 df-ss 3968 df-pss 3971 df-nul 4334 df-if 4526 df-pw 4602 df-sn 4627 df-pr 4629 df-op 4633 df-uni 4908 df-br 5144 df-opab 5206 df-tr 5260 df-eprel 5584 df-po 5592 df-so 5593 df-fr 5637 df-we 5639 df-ord 6387 | 
| This theorem is referenced by: ordun 6488 inar1 10815 | 
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