Proof of Theorem tridc
Step | Hyp | Ref
| Expression |
1 | | simpr 109 |
. . . 4
⊢ ((𝜑 ∧ 𝐵𝑅𝐶) → 𝐵𝑅𝐶) |
2 | 1 | orcd 728 |
. . 3
⊢ ((𝜑 ∧ 𝐵𝑅𝐶) → (𝐵𝑅𝐶 ∨ ¬ 𝐵𝑅𝐶)) |
3 | | df-dc 830 |
. . 3
⊢
(DECID 𝐵𝑅𝐶 ↔ (𝐵𝑅𝐶 ∨ ¬ 𝐵𝑅𝐶)) |
4 | 2, 3 | sylibr 133 |
. 2
⊢ ((𝜑 ∧ 𝐵𝑅𝐶) → DECID 𝐵𝑅𝐶) |
5 | | tridc.po |
. . . . . . 7
⊢ (𝜑 → 𝑅 Po 𝐴) |
6 | | tridc.c |
. . . . . . 7
⊢ (𝜑 → 𝐶 ∈ 𝐴) |
7 | | poirr 4292 |
. . . . . . 7
⊢ ((𝑅 Po 𝐴 ∧ 𝐶 ∈ 𝐴) → ¬ 𝐶𝑅𝐶) |
8 | 5, 6, 7 | syl2anc 409 |
. . . . . 6
⊢ (𝜑 → ¬ 𝐶𝑅𝐶) |
9 | 8 | adantr 274 |
. . . . 5
⊢ ((𝜑 ∧ 𝐵 = 𝐶) → ¬ 𝐶𝑅𝐶) |
10 | | simpr 109 |
. . . . . 6
⊢ ((𝜑 ∧ 𝐵 = 𝐶) → 𝐵 = 𝐶) |
11 | 10 | breq1d 3999 |
. . . . 5
⊢ ((𝜑 ∧ 𝐵 = 𝐶) → (𝐵𝑅𝐶 ↔ 𝐶𝑅𝐶)) |
12 | 9, 11 | mtbird 668 |
. . . 4
⊢ ((𝜑 ∧ 𝐵 = 𝐶) → ¬ 𝐵𝑅𝐶) |
13 | 12 | olcd 729 |
. . 3
⊢ ((𝜑 ∧ 𝐵 = 𝐶) → (𝐵𝑅𝐶 ∨ ¬ 𝐵𝑅𝐶)) |
14 | 13, 3 | sylibr 133 |
. 2
⊢ ((𝜑 ∧ 𝐵 = 𝐶) → DECID 𝐵𝑅𝐶) |
15 | | tridc.b |
. . . . . . 7
⊢ (𝜑 → 𝐵 ∈ 𝐴) |
16 | | po2nr 4294 |
. . . . . . 7
⊢ ((𝑅 Po 𝐴 ∧ (𝐶 ∈ 𝐴 ∧ 𝐵 ∈ 𝐴)) → ¬ (𝐶𝑅𝐵 ∧ 𝐵𝑅𝐶)) |
17 | 5, 6, 15, 16 | syl12anc 1231 |
. . . . . 6
⊢ (𝜑 → ¬ (𝐶𝑅𝐵 ∧ 𝐵𝑅𝐶)) |
18 | 17 | adantr 274 |
. . . . 5
⊢ ((𝜑 ∧ 𝐶𝑅𝐵) → ¬ (𝐶𝑅𝐵 ∧ 𝐵𝑅𝐶)) |
19 | | simplr 525 |
. . . . . 6
⊢ (((𝜑 ∧ 𝐶𝑅𝐵) ∧ 𝐵𝑅𝐶) → 𝐶𝑅𝐵) |
20 | | simpr 109 |
. . . . . 6
⊢ (((𝜑 ∧ 𝐶𝑅𝐵) ∧ 𝐵𝑅𝐶) → 𝐵𝑅𝐶) |
21 | 19, 20 | jca 304 |
. . . . 5
⊢ (((𝜑 ∧ 𝐶𝑅𝐵) ∧ 𝐵𝑅𝐶) → (𝐶𝑅𝐵 ∧ 𝐵𝑅𝐶)) |
22 | 18, 21 | mtand 660 |
. . . 4
⊢ ((𝜑 ∧ 𝐶𝑅𝐵) → ¬ 𝐵𝑅𝐶) |
23 | 22 | olcd 729 |
. . 3
⊢ ((𝜑 ∧ 𝐶𝑅𝐵) → (𝐵𝑅𝐶 ∨ ¬ 𝐵𝑅𝐶)) |
24 | 23, 3 | sylibr 133 |
. 2
⊢ ((𝜑 ∧ 𝐶𝑅𝐵) → DECID 𝐵𝑅𝐶) |
25 | | tridc.tri |
. . 3
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) |
26 | | breq1 3992 |
. . . . 5
⊢ (𝑥 = 𝐵 → (𝑥𝑅𝑦 ↔ 𝐵𝑅𝑦)) |
27 | | eqeq1 2177 |
. . . . 5
⊢ (𝑥 = 𝐵 → (𝑥 = 𝑦 ↔ 𝐵 = 𝑦)) |
28 | | breq2 3993 |
. . . . 5
⊢ (𝑥 = 𝐵 → (𝑦𝑅𝑥 ↔ 𝑦𝑅𝐵)) |
29 | 26, 27, 28 | 3orbi123d 1306 |
. . . 4
⊢ (𝑥 = 𝐵 → ((𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥) ↔ (𝐵𝑅𝑦 ∨ 𝐵 = 𝑦 ∨ 𝑦𝑅𝐵))) |
30 | | breq2 3993 |
. . . . 5
⊢ (𝑦 = 𝐶 → (𝐵𝑅𝑦 ↔ 𝐵𝑅𝐶)) |
31 | | eqeq2 2180 |
. . . . 5
⊢ (𝑦 = 𝐶 → (𝐵 = 𝑦 ↔ 𝐵 = 𝐶)) |
32 | | breq1 3992 |
. . . . 5
⊢ (𝑦 = 𝐶 → (𝑦𝑅𝐵 ↔ 𝐶𝑅𝐵)) |
33 | 30, 31, 32 | 3orbi123d 1306 |
. . . 4
⊢ (𝑦 = 𝐶 → ((𝐵𝑅𝑦 ∨ 𝐵 = 𝑦 ∨ 𝑦𝑅𝐵) ↔ (𝐵𝑅𝐶 ∨ 𝐵 = 𝐶 ∨ 𝐶𝑅𝐵))) |
34 | 29, 33 | rspc2va 2848 |
. . 3
⊢ (((𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) → (𝐵𝑅𝐶 ∨ 𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)) |
35 | 15, 6, 25, 34 | syl21anc 1232 |
. 2
⊢ (𝜑 → (𝐵𝑅𝐶 ∨ 𝐵 = 𝐶 ∨ 𝐶𝑅𝐵)) |
36 | 4, 14, 24, 35 | mpjao3dan 1302 |
1
⊢ (𝜑 → DECID 𝐵𝑅𝐶) |