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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 6280 | . . 3 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) | |
2 | 1 | orcomd 865 | . 2 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶)) |
3 | eqeq1 2822 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵)) | |
4 | ssequn2 4156 | . . . 4 ⊢ (𝐶 ⊆ 𝐵 ↔ (𝐵 ∪ 𝐶) = 𝐵) | |
5 | 3, 4 | syl6rbbr 291 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐶 ⊆ 𝐵 ↔ 𝐴 = 𝐵)) |
6 | eqeq1 2822 | . . . 4 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶)) | |
7 | ssequn1 4153 | . . . 4 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐵 ∪ 𝐶) = 𝐶) | |
8 | 6, 7 | syl6rbbr 291 | . . 3 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → (𝐵 ⊆ 𝐶 ↔ 𝐴 = 𝐶)) |
9 | 5, 8 | orbi12d 912 | . 2 ⊢ (𝐴 = (𝐵 ∪ 𝐶) → ((𝐶 ⊆ 𝐵 ∨ 𝐵 ⊆ 𝐶) ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
10 | 2, 9 | syl5ibcom 246 | 1 ⊢ ((Ord 𝐵 ∧ Ord 𝐶) → (𝐴 = (𝐵 ∪ 𝐶) → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ∧ wa 396 ∨ wo 841 = wceq 1528 ∪ cun 3931 ⊆ wss 3933 Ord word 6183 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1787 ax-4 1801 ax-5 1902 ax-6 1961 ax-7 2006 ax-8 2107 ax-9 2115 ax-10 2136 ax-11 2151 ax-12 2167 ax-ext 2790 ax-sep 5194 ax-nul 5201 ax-pr 5320 |
This theorem depends on definitions: df-bi 208 df-an 397 df-or 842 df-3or 1080 df-3an 1081 df-tru 1531 df-ex 1772 df-nf 1776 df-sb 2061 df-mo 2615 df-eu 2647 df-clab 2797 df-cleq 2811 df-clel 2890 df-nfc 2960 df-ne 3014 df-ral 3140 df-rex 3141 df-rab 3144 df-v 3494 df-sbc 3770 df-dif 3936 df-un 3938 df-in 3940 df-ss 3949 df-pss 3951 df-nul 4289 df-if 4464 df-sn 4558 df-pr 4560 df-op 4564 df-uni 4831 df-br 5058 df-opab 5120 df-tr 5164 df-eprel 5458 df-po 5467 df-so 5468 df-fr 5507 df-we 5509 df-ord 6187 |
This theorem is referenced by: ordun 6285 inar1 10185 |
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