Proof of Theorem atcvrj1
Step | Hyp | Ref
| Expression |
1 | | simp3l 1200 |
. . . 4
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → 𝑃 ≠ 𝑅) |
2 | | hlatl 37374 |
. . . . . 6
⊢ (𝐾 ∈ HL → 𝐾 ∈ AtLat) |
3 | 2 | 3ad2ant1 1132 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → 𝐾 ∈ AtLat) |
4 | | simp21 1205 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → 𝑃 ∈ 𝐴) |
5 | | simp23 1207 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → 𝑅 ∈ 𝐴) |
6 | | eqid 2738 |
. . . . . 6
⊢
(meet‘𝐾) =
(meet‘𝐾) |
7 | | eqid 2738 |
. . . . . 6
⊢
(0.‘𝐾) =
(0.‘𝐾) |
8 | | atcvrj1x.a |
. . . . . 6
⊢ 𝐴 = (Atoms‘𝐾) |
9 | 6, 7, 8 | atnem0 37332 |
. . . . 5
⊢ ((𝐾 ∈ AtLat ∧ 𝑃 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) → (𝑃 ≠ 𝑅 ↔ (𝑃(meet‘𝐾)𝑅) = (0.‘𝐾))) |
10 | 3, 4, 5, 9 | syl3anc 1370 |
. . . 4
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → (𝑃 ≠ 𝑅 ↔ (𝑃(meet‘𝐾)𝑅) = (0.‘𝐾))) |
11 | 1, 10 | mpbid 231 |
. . 3
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → (𝑃(meet‘𝐾)𝑅) = (0.‘𝐾)) |
12 | | simp1 1135 |
. . . 4
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → 𝐾 ∈ HL) |
13 | | eqid 2738 |
. . . . . 6
⊢
(Base‘𝐾) =
(Base‘𝐾) |
14 | 13, 8 | atbase 37303 |
. . . . 5
⊢ (𝑃 ∈ 𝐴 → 𝑃 ∈ (Base‘𝐾)) |
15 | 4, 14 | syl 17 |
. . . 4
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → 𝑃 ∈ (Base‘𝐾)) |
16 | | atcvrj1x.j |
. . . . 5
⊢ ∨ =
(join‘𝐾) |
17 | | atcvrj1x.c |
. . . . 5
⊢ 𝐶 = ( ⋖ ‘𝐾) |
18 | 13, 16, 6, 7, 17, 8 | cvrp 37430 |
. . . 4
⊢ ((𝐾 ∈ HL ∧ 𝑃 ∈ (Base‘𝐾) ∧ 𝑅 ∈ 𝐴) → ((𝑃(meet‘𝐾)𝑅) = (0.‘𝐾) ↔ 𝑃𝐶(𝑃 ∨ 𝑅))) |
19 | 12, 15, 5, 18 | syl3anc 1370 |
. . 3
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → ((𝑃(meet‘𝐾)𝑅) = (0.‘𝐾) ↔ 𝑃𝐶(𝑃 ∨ 𝑅))) |
20 | 11, 19 | mpbid 231 |
. 2
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → 𝑃𝐶(𝑃 ∨ 𝑅)) |
21 | | simp3r 1201 |
. . 3
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → 𝑃 ≤ (𝑄 ∨ 𝑅)) |
22 | | atcvrj1x.l |
. . . . 5
⊢ ≤ =
(le‘𝐾) |
23 | 22, 16, 8 | hlatexchb2 37408 |
. . . 4
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ 𝑃 ≠ 𝑅) → (𝑃 ≤ (𝑄 ∨ 𝑅) ↔ (𝑃 ∨ 𝑅) = (𝑄 ∨ 𝑅))) |
24 | 23 | 3adant3r 1180 |
. . 3
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → (𝑃 ≤ (𝑄 ∨ 𝑅) ↔ (𝑃 ∨ 𝑅) = (𝑄 ∨ 𝑅))) |
25 | 21, 24 | mpbid 231 |
. 2
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → (𝑃 ∨ 𝑅) = (𝑄 ∨ 𝑅)) |
26 | 20, 25 | breqtrd 5100 |
1
⊢ ((𝐾 ∈ HL ∧ (𝑃 ∈ 𝐴 ∧ 𝑄 ∈ 𝐴 ∧ 𝑅 ∈ 𝐴) ∧ (𝑃 ≠ 𝑅 ∧ 𝑃 ≤ (𝑄 ∨ 𝑅))) → 𝑃𝐶(𝑄 ∨ 𝑅)) |