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| Mirrors > Home > MPE Home > Th. List > ecopover | Structured version Visualization version GIF version | ||
| Description: Assuming that operation 𝐹 is commutative (second hypothesis), closed (third hypothesis), associative (fourth hypothesis), and has the cancellation property (fifth hypothesis), show that the relation ∼, specified by the first hypothesis, is an equivalence relation. (Contributed by NM, 16-Feb-1996.) (Revised by Mario Carneiro, 12-Aug-2015.) (Proof shortened by AV, 1-May-2021.) |
| Ref | Expression |
|---|---|
| ecopopr.1 | ⊢ ∼ = {〈𝑥, 𝑦〉 ∣ ((𝑥 ∈ (𝑆 × 𝑆) ∧ 𝑦 ∈ (𝑆 × 𝑆)) ∧ ∃𝑧∃𝑤∃𝑣∃𝑢((𝑥 = 〈𝑧, 𝑤〉 ∧ 𝑦 = 〈𝑣, 𝑢〉) ∧ (𝑧 + 𝑢) = (𝑤 + 𝑣)))} |
| ecopopr.com | ⊢ (𝑥 + 𝑦) = (𝑦 + 𝑥) |
| ecopopr.cl | ⊢ ((𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆) → (𝑥 + 𝑦) ∈ 𝑆) |
| ecopopr.ass | ⊢ ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧)) |
| ecopopr.can | ⊢ ((𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆) → ((𝑥 + 𝑦) = (𝑥 + 𝑧) → 𝑦 = 𝑧)) |
| Ref | Expression |
|---|---|
| ecopover | ⊢ ∼ Er (𝑆 × 𝑆) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | ecopopr.1 | . . 3 ⊢ ∼ = {〈𝑥, 𝑦〉 ∣ ((𝑥 ∈ (𝑆 × 𝑆) ∧ 𝑦 ∈ (𝑆 × 𝑆)) ∧ ∃𝑧∃𝑤∃𝑣∃𝑢((𝑥 = 〈𝑧, 𝑤〉 ∧ 𝑦 = 〈𝑣, 𝑢〉) ∧ (𝑧 + 𝑢) = (𝑤 + 𝑣)))} | |
| 2 | 1 | relopabiv 5770 | . 2 ⊢ Rel ∼ |
| 3 | ecopopr.com | . . 3 ⊢ (𝑥 + 𝑦) = (𝑦 + 𝑥) | |
| 4 | 1, 3 | ecopovsym 8763 | . 2 ⊢ (𝑓 ∼ 𝑔 → 𝑔 ∼ 𝑓) |
| 5 | ecopopr.cl | . . 3 ⊢ ((𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆) → (𝑥 + 𝑦) ∈ 𝑆) | |
| 6 | ecopopr.ass | . . 3 ⊢ ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧)) | |
| 7 | ecopopr.can | . . 3 ⊢ ((𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆) → ((𝑥 + 𝑦) = (𝑥 + 𝑧) → 𝑦 = 𝑧)) | |
| 8 | 1, 3, 5, 6, 7 | ecopovtrn 8764 | . 2 ⊢ ((𝑓 ∼ 𝑔 ∧ 𝑔 ∼ ℎ) → 𝑓 ∼ ℎ) |
| 9 | vex 3436 | . . . . . . . . 9 ⊢ 𝑔 ∈ V | |
| 10 | vex 3436 | . . . . . . . . 9 ⊢ ℎ ∈ V | |
| 11 | 9, 10, 3 | caovcom 7560 | . . . . . . . 8 ⊢ (𝑔 + ℎ) = (ℎ + 𝑔) |
| 12 | 1 | ecopoveq 8762 | . . . . . . . 8 ⊢ (((𝑔 ∈ 𝑆 ∧ ℎ ∈ 𝑆) ∧ (𝑔 ∈ 𝑆 ∧ ℎ ∈ 𝑆)) → (〈𝑔, ℎ〉 ∼ 〈𝑔, ℎ〉 ↔ (𝑔 + ℎ) = (ℎ + 𝑔))) |
| 13 | 11, 12 | mpbiri 259 | . . . . . . 7 ⊢ (((𝑔 ∈ 𝑆 ∧ ℎ ∈ 𝑆) ∧ (𝑔 ∈ 𝑆 ∧ ℎ ∈ 𝑆)) → 〈𝑔, ℎ〉 ∼ 〈𝑔, ℎ〉) |
| 14 | 13 | anidms 571 | . . . . . 6 ⊢ ((𝑔 ∈ 𝑆 ∧ ℎ ∈ 𝑆) → 〈𝑔, ℎ〉 ∼ 〈𝑔, ℎ〉) |
| 15 | 14 | rgen2 3180 | . . . . 5 ⊢ ∀𝑔 ∈ 𝑆 ∀ℎ ∈ 𝑆 〈𝑔, ℎ〉 ∼ 〈𝑔, ℎ〉 |
| 16 | breq12 5084 | . . . . . . 7 ⊢ ((𝑓 = 〈𝑔, ℎ〉 ∧ 𝑓 = 〈𝑔, ℎ〉) → (𝑓 ∼ 𝑓 ↔ 〈𝑔, ℎ〉 ∼ 〈𝑔, ℎ〉)) | |
| 17 | 16 | anidms 571 | . . . . . 6 ⊢ (𝑓 = 〈𝑔, ℎ〉 → (𝑓 ∼ 𝑓 ↔ 〈𝑔, ℎ〉 ∼ 〈𝑔, ℎ〉)) |
| 18 | 17 | ralxp 5790 | . . . . 5 ⊢ (∀𝑓 ∈ (𝑆 × 𝑆)𝑓 ∼ 𝑓 ↔ ∀𝑔 ∈ 𝑆 ∀ℎ ∈ 𝑆 〈𝑔, ℎ〉 ∼ 〈𝑔, ℎ〉) |
| 19 | 15, 18 | mpbir 232 | . . . 4 ⊢ ∀𝑓 ∈ (𝑆 × 𝑆)𝑓 ∼ 𝑓 |
| 20 | 19 | rspec 3231 | . . 3 ⊢ (𝑓 ∈ (𝑆 × 𝑆) → 𝑓 ∼ 𝑓) |
| 21 | opabssxp 5717 | . . . . . 6 ⊢ {〈𝑥, 𝑦〉 ∣ ((𝑥 ∈ (𝑆 × 𝑆) ∧ 𝑦 ∈ (𝑆 × 𝑆)) ∧ ∃𝑧∃𝑤∃𝑣∃𝑢((𝑥 = 〈𝑧, 𝑤〉 ∧ 𝑦 = 〈𝑣, 𝑢〉) ∧ (𝑧 + 𝑢) = (𝑤 + 𝑣)))} ⊆ ((𝑆 × 𝑆) × (𝑆 × 𝑆)) | |
| 22 | 1, 21 | eqsstri 3968 | . . . . 5 ⊢ ∼ ⊆ ((𝑆 × 𝑆) × (𝑆 × 𝑆)) |
| 23 | 22 | ssbri 5124 | . . . 4 ⊢ (𝑓 ∼ 𝑓 → 𝑓((𝑆 × 𝑆) × (𝑆 × 𝑆))𝑓) |
| 24 | brxp 5674 | . . . . 5 ⊢ (𝑓((𝑆 × 𝑆) × (𝑆 × 𝑆))𝑓 ↔ (𝑓 ∈ (𝑆 × 𝑆) ∧ 𝑓 ∈ (𝑆 × 𝑆))) | |
| 25 | 24 | simplbi 497 | . . . 4 ⊢ (𝑓((𝑆 × 𝑆) × (𝑆 × 𝑆))𝑓 → 𝑓 ∈ (𝑆 × 𝑆)) |
| 26 | 23, 25 | syl 17 | . . 3 ⊢ (𝑓 ∼ 𝑓 → 𝑓 ∈ (𝑆 × 𝑆)) |
| 27 | 20, 26 | impbii 210 | . 2 ⊢ (𝑓 ∈ (𝑆 × 𝑆) ↔ 𝑓 ∼ 𝑓) |
| 28 | 2, 4, 8, 27 | iseri 8668 | 1 ⊢ ∼ Er (𝑆 × 𝑆) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 ↔ wb 207 ∧ wa 396 = wceq 1547 ∃wex 1786 ∈ wcel 2119 ∀wral 3054 〈cop 4568 class class class wbr 5079 {copab 5141 × cxp 5623 (class class class)co 7363 Er wer 8637 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1802 ax-4 1816 ax-5 1917 ax-6 1974 ax-7 2015 ax-8 2121 ax-9 2129 ax-10 2152 ax-11 2168 ax-12 2189 ax-ext 2712 ax-sep 5225 ax-nul 5235 ax-pr 5369 |
| This theorem depends on definitions: df-bi 208 df-an 397 df-or 854 df-3an 1094 df-tru 1550 df-fal 1560 df-ex 1787 df-nf 1791 df-sb 2074 df-clab 2719 df-cleq 2732 df-clel 2815 df-nfc 2889 df-ne 2936 df-ral 3055 df-rex 3065 df-rab 3393 df-v 3434 df-sbc 3731 df-csb 3839 df-dif 3893 df-un 3895 df-in 3897 df-ss 3907 df-nul 4269 df-if 4462 df-sn 4563 df-pr 4565 df-op 4569 df-uni 4846 df-iun 4930 df-br 5080 df-opab 5142 df-xp 5631 df-rel 5632 df-cnv 5633 df-co 5634 df-dm 5635 df-iota 6448 df-fv 6500 df-ov 7366 df-er 8640 |
| This theorem is referenced by: enqer 10842 enrer 10984 |
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