Proof of Theorem pmod2iN
Step | Hyp | Ref
| Expression |
1 | | incom 4135 |
. . . . . 6
⊢ (𝑋 ∩ 𝑌) = (𝑌 ∩ 𝑋) |
2 | 1 | oveq1i 7285 |
. . . . 5
⊢ ((𝑋 ∩ 𝑌) + 𝑍) = ((𝑌 ∩ 𝑋) + 𝑍) |
3 | | hllat 37377 |
. . . . . . 7
⊢ (𝐾 ∈ HL → 𝐾 ∈ Lat) |
4 | 3 | 3ad2ant1 1132 |
. . . . . 6
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → 𝐾 ∈ Lat) |
5 | | simp22 1206 |
. . . . . . 7
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → 𝑌 ⊆ 𝐴) |
6 | | ssinss1 4171 |
. . . . . . 7
⊢ (𝑌 ⊆ 𝐴 → (𝑌 ∩ 𝑋) ⊆ 𝐴) |
7 | 5, 6 | syl 17 |
. . . . . 6
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → (𝑌 ∩ 𝑋) ⊆ 𝐴) |
8 | | simp23 1207 |
. . . . . 6
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → 𝑍 ⊆ 𝐴) |
9 | | pmod.a |
. . . . . . 7
⊢ 𝐴 = (Atoms‘𝐾) |
10 | | pmod.p |
. . . . . . 7
⊢ + =
(+𝑃‘𝐾) |
11 | 9, 10 | paddcom 37827 |
. . . . . 6
⊢ ((𝐾 ∈ Lat ∧ (𝑌 ∩ 𝑋) ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) → ((𝑌 ∩ 𝑋) + 𝑍) = (𝑍 + (𝑌 ∩ 𝑋))) |
12 | 4, 7, 8, 11 | syl3anc 1370 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → ((𝑌 ∩ 𝑋) + 𝑍) = (𝑍 + (𝑌 ∩ 𝑋))) |
13 | 2, 12 | eqtrid 2790 |
. . . 4
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → ((𝑋 ∩ 𝑌) + 𝑍) = (𝑍 + (𝑌 ∩ 𝑋))) |
14 | | simp21 1205 |
. . . . . 6
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → 𝑋 ∈ 𝑆) |
15 | 8, 5, 14 | 3jca 1127 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → (𝑍 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ∈ 𝑆)) |
16 | | pmod.s |
. . . . . . 7
⊢ 𝑆 = (PSubSp‘𝐾) |
17 | 9, 16, 10 | pmod1i 37862 |
. . . . . 6
⊢ ((𝐾 ∈ HL ∧ (𝑍 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ∈ 𝑆)) → (𝑍 ⊆ 𝑋 → ((𝑍 + 𝑌) ∩ 𝑋) = (𝑍 + (𝑌 ∩ 𝑋)))) |
18 | 17 | 3impia 1116 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ (𝑍 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑋 ∈ 𝑆) ∧ 𝑍 ⊆ 𝑋) → ((𝑍 + 𝑌) ∩ 𝑋) = (𝑍 + (𝑌 ∩ 𝑋))) |
19 | 15, 18 | syld3an2 1410 |
. . . 4
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → ((𝑍 + 𝑌) ∩ 𝑋) = (𝑍 + (𝑌 ∩ 𝑋))) |
20 | 9, 10 | paddcom 37827 |
. . . . . 6
⊢ ((𝐾 ∈ Lat ∧ 𝑍 ⊆ 𝐴 ∧ 𝑌 ⊆ 𝐴) → (𝑍 + 𝑌) = (𝑌 + 𝑍)) |
21 | 4, 8, 5, 20 | syl3anc 1370 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → (𝑍 + 𝑌) = (𝑌 + 𝑍)) |
22 | 21 | ineq1d 4145 |
. . . 4
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → ((𝑍 + 𝑌) ∩ 𝑋) = ((𝑌 + 𝑍) ∩ 𝑋)) |
23 | 13, 19, 22 | 3eqtr2d 2784 |
. . 3
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → ((𝑋 ∩ 𝑌) + 𝑍) = ((𝑌 + 𝑍) ∩ 𝑋)) |
24 | | incom 4135 |
. . 3
⊢ ((𝑌 + 𝑍) ∩ 𝑋) = (𝑋 ∩ (𝑌 + 𝑍)) |
25 | 23, 24 | eqtrdi 2794 |
. 2
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴) ∧ 𝑍 ⊆ 𝑋) → ((𝑋 ∩ 𝑌) + 𝑍) = (𝑋 ∩ (𝑌 + 𝑍))) |
26 | 25 | 3expia 1120 |
1
⊢ ((𝐾 ∈ HL ∧ (𝑋 ∈ 𝑆 ∧ 𝑌 ⊆ 𝐴 ∧ 𝑍 ⊆ 𝐴)) → (𝑍 ⊆ 𝑋 → ((𝑋 ∩ 𝑌) + 𝑍) = (𝑋 ∩ (𝑌 + 𝑍)))) |