Proof of Theorem cvrval2
Step | Hyp | Ref
| Expression |
1 | | cvrletr.b |
. . 3
⊢ 𝐵 = (Base‘𝐾) |
2 | | cvrletr.s |
. . 3
⊢ < =
(lt‘𝐾) |
3 | | cvrletr.c |
. . 3
⊢ 𝐶 = ( ⋖ ‘𝐾) |
4 | 1, 2, 3 | cvrval 37283 |
. 2
⊢ ((𝐾 ∈ 𝐴 ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋𝐶𝑌 ↔ (𝑋 < 𝑌 ∧ ¬ ∃𝑧 ∈ 𝐵 (𝑋 < 𝑧 ∧ 𝑧 < 𝑌)))) |
5 | | iman 402 |
. . . . . . . 8
⊢ (((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) → 𝑧 = 𝑌) ↔ ¬ ((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) ∧ ¬ 𝑧 = 𝑌)) |
6 | | df-ne 2944 |
. . . . . . . . 9
⊢ (𝑧 ≠ 𝑌 ↔ ¬ 𝑧 = 𝑌) |
7 | 6 | anbi2i 623 |
. . . . . . . 8
⊢ (((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) ∧ 𝑧 ≠ 𝑌) ↔ ((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) ∧ ¬ 𝑧 = 𝑌)) |
8 | 5, 7 | xchbinxr 335 |
. . . . . . 7
⊢ (((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) → 𝑧 = 𝑌) ↔ ¬ ((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) ∧ 𝑧 ≠ 𝑌)) |
9 | | anass 469 |
. . . . . . . . 9
⊢ (((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) ∧ 𝑧 ≠ 𝑌) ↔ (𝑋 < 𝑧 ∧ (𝑧 ≤ 𝑌 ∧ 𝑧 ≠ 𝑌))) |
10 | | cvrletr.l |
. . . . . . . . . . . . 13
⊢ ≤ =
(le‘𝐾) |
11 | 10, 2 | pltval 18050 |
. . . . . . . . . . . 12
⊢ ((𝐾 ∈ 𝐴 ∧ 𝑧 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑧 < 𝑌 ↔ (𝑧 ≤ 𝑌 ∧ 𝑧 ≠ 𝑌))) |
12 | 11 | 3com23 1125 |
. . . . . . . . . . 11
⊢ ((𝐾 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵 ∧ 𝑧 ∈ 𝐵) → (𝑧 < 𝑌 ↔ (𝑧 ≤ 𝑌 ∧ 𝑧 ≠ 𝑌))) |
13 | 12 | 3expa 1117 |
. . . . . . . . . 10
⊢ (((𝐾 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) ∧ 𝑧 ∈ 𝐵) → (𝑧 < 𝑌 ↔ (𝑧 ≤ 𝑌 ∧ 𝑧 ≠ 𝑌))) |
14 | 13 | anbi2d 629 |
. . . . . . . . 9
⊢ (((𝐾 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) ∧ 𝑧 ∈ 𝐵) → ((𝑋 < 𝑧 ∧ 𝑧 < 𝑌) ↔ (𝑋 < 𝑧 ∧ (𝑧 ≤ 𝑌 ∧ 𝑧 ≠ 𝑌)))) |
15 | 9, 14 | bitr4id 290 |
. . . . . . . 8
⊢ (((𝐾 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) ∧ 𝑧 ∈ 𝐵) → (((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) ∧ 𝑧 ≠ 𝑌) ↔ (𝑋 < 𝑧 ∧ 𝑧 < 𝑌))) |
16 | 15 | notbid 318 |
. . . . . . 7
⊢ (((𝐾 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) ∧ 𝑧 ∈ 𝐵) → (¬ ((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) ∧ 𝑧 ≠ 𝑌) ↔ ¬ (𝑋 < 𝑧 ∧ 𝑧 < 𝑌))) |
17 | 8, 16 | syl5bb 283 |
. . . . . 6
⊢ (((𝐾 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) ∧ 𝑧 ∈ 𝐵) → (((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) → 𝑧 = 𝑌) ↔ ¬ (𝑋 < 𝑧 ∧ 𝑧 < 𝑌))) |
18 | 17 | ralbidva 3111 |
. . . . 5
⊢ ((𝐾 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) → (∀𝑧 ∈ 𝐵 ((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) → 𝑧 = 𝑌) ↔ ∀𝑧 ∈ 𝐵 ¬ (𝑋 < 𝑧 ∧ 𝑧 < 𝑌))) |
19 | | ralnex 3167 |
. . . . 5
⊢
(∀𝑧 ∈
𝐵 ¬ (𝑋 < 𝑧 ∧ 𝑧 < 𝑌) ↔ ¬ ∃𝑧 ∈ 𝐵 (𝑋 < 𝑧 ∧ 𝑧 < 𝑌)) |
20 | 18, 19 | bitrdi 287 |
. . . 4
⊢ ((𝐾 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) → (∀𝑧 ∈ 𝐵 ((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) → 𝑧 = 𝑌) ↔ ¬ ∃𝑧 ∈ 𝐵 (𝑋 < 𝑧 ∧ 𝑧 < 𝑌))) |
21 | 20 | anbi2d 629 |
. . 3
⊢ ((𝐾 ∈ 𝐴 ∧ 𝑌 ∈ 𝐵) → ((𝑋 < 𝑌 ∧ ∀𝑧 ∈ 𝐵 ((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) → 𝑧 = 𝑌)) ↔ (𝑋 < 𝑌 ∧ ¬ ∃𝑧 ∈ 𝐵 (𝑋 < 𝑧 ∧ 𝑧 < 𝑌)))) |
22 | 21 | 3adant2 1130 |
. 2
⊢ ((𝐾 ∈ 𝐴 ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑋 < 𝑌 ∧ ∀𝑧 ∈ 𝐵 ((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) → 𝑧 = 𝑌)) ↔ (𝑋 < 𝑌 ∧ ¬ ∃𝑧 ∈ 𝐵 (𝑋 < 𝑧 ∧ 𝑧 < 𝑌)))) |
23 | 4, 22 | bitr4d 281 |
1
⊢ ((𝐾 ∈ 𝐴 ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋𝐶𝑌 ↔ (𝑋 < 𝑌 ∧ ∀𝑧 ∈ 𝐵 ((𝑋 < 𝑧 ∧ 𝑧 ≤ 𝑌) → 𝑧 = 𝑌)))) |