Proof of Theorem 4atex2-0bOLDN
Step | Hyp | Ref
| Expression |
1 | | simp1 1134 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) ∧ (𝑃 ≠ 𝑄 ∧ 𝑇 = (0.‘𝐾) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → (𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻)) |
2 | | simp21 1204 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) ∧ (𝑃 ≠ 𝑄 ∧ 𝑇 = (0.‘𝐾) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) |
3 | | simp22 1205 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) ∧ (𝑃 ≠ 𝑄 ∧ 𝑇 = (0.‘𝐾) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊)) |
4 | | simp32 1208 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) ∧ (𝑃 ≠ 𝑄 ∧ 𝑇 = (0.‘𝐾) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → 𝑇 = (0.‘𝐾)) |
5 | | simp31 1207 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) ∧ (𝑃 ≠ 𝑄 ∧ 𝑇 = (0.‘𝐾) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → 𝑃 ≠ 𝑄) |
6 | | simp23 1206 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) ∧ (𝑃 ≠ 𝑄 ∧ 𝑇 = (0.‘𝐾) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) |
7 | | simp33 1209 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) ∧ (𝑃 ≠ 𝑄 ∧ 𝑇 = (0.‘𝐾) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟))) |
8 | | 4that.l |
. . . 4
⊢ ≤ =
(le‘𝐾) |
9 | | 4that.j |
. . . 4
⊢ ∨ =
(join‘𝐾) |
10 | | 4that.a |
. . . 4
⊢ 𝐴 = (Atoms‘𝐾) |
11 | | 4that.h |
. . . 4
⊢ 𝐻 = (LHyp‘𝐾) |
12 | 8, 9, 10, 11 | 4atex2-0aOLDN 38019 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ 𝑇 = (0.‘𝐾)) ∧ (𝑃 ≠ 𝑄 ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → ∃𝑧 ∈ 𝐴 (¬ 𝑧 ≤ 𝑊 ∧ (𝑇 ∨ 𝑧) = (𝑆 ∨ 𝑧))) |
13 | 1, 2, 3, 4, 5, 6, 7, 12 | syl133anc 1391 |
. 2
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) ∧ (𝑃 ≠ 𝑄 ∧ 𝑇 = (0.‘𝐾) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → ∃𝑧 ∈ 𝐴 (¬ 𝑧 ≤ 𝑊 ∧ (𝑇 ∨ 𝑧) = (𝑆 ∨ 𝑧))) |
14 | | eqcom 2745 |
. . . 4
⊢ ((𝑆 ∨ 𝑧) = (𝑇 ∨ 𝑧) ↔ (𝑇 ∨ 𝑧) = (𝑆 ∨ 𝑧)) |
15 | 14 | anbi2i 622 |
. . 3
⊢ ((¬
𝑧 ≤ 𝑊 ∧ (𝑆 ∨ 𝑧) = (𝑇 ∨ 𝑧)) ↔ (¬ 𝑧 ≤ 𝑊 ∧ (𝑇 ∨ 𝑧) = (𝑆 ∨ 𝑧))) |
16 | 15 | rexbii 3177 |
. 2
⊢
(∃𝑧 ∈
𝐴 (¬ 𝑧 ≤ 𝑊 ∧ (𝑆 ∨ 𝑧) = (𝑇 ∨ 𝑧)) ↔ ∃𝑧 ∈ 𝐴 (¬ 𝑧 ≤ 𝑊 ∧ (𝑇 ∨ 𝑧) = (𝑆 ∨ 𝑧))) |
17 | 13, 16 | sylibr 233 |
1
⊢ (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑄 ∈ 𝐴 ∧ ¬ 𝑄 ≤ 𝑊) ∧ (𝑆 ∈ 𝐴 ∧ ¬ 𝑆 ≤ 𝑊)) ∧ (𝑃 ≠ 𝑄 ∧ 𝑇 = (0.‘𝐾) ∧ ∃𝑟 ∈ 𝐴 (¬ 𝑟 ≤ 𝑊 ∧ (𝑃 ∨ 𝑟) = (𝑄 ∨ 𝑟)))) → ∃𝑧 ∈ 𝐴 (¬ 𝑧 ≤ 𝑊 ∧ (𝑆 ∨ 𝑧) = (𝑇 ∨ 𝑧))) |