Proof of Theorem ntrin
| Step | Hyp | Ref
| Expression |
| 1 | | inss1 4219 |
. . . . 5
⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴 |
| 2 | | clscld.1 |
. . . . . 6
⊢ 𝑋 = ∪
𝐽 |
| 3 | 2 | ntrss 23028 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ (𝐴 ∩ 𝐵) ⊆ 𝐴) → ((int‘𝐽)‘(𝐴 ∩ 𝐵)) ⊆ ((int‘𝐽)‘𝐴)) |
| 4 | 1, 3 | mp3an3 1451 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋) → ((int‘𝐽)‘(𝐴 ∩ 𝐵)) ⊆ ((int‘𝐽)‘𝐴)) |
| 5 | 4 | 3adant3 1132 |
. . 3
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝐴 ∩ 𝐵)) ⊆ ((int‘𝐽)‘𝐴)) |
| 6 | | inss2 4220 |
. . . . 5
⊢ (𝐴 ∩ 𝐵) ⊆ 𝐵 |
| 7 | 2 | ntrss 23028 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝑋 ∧ (𝐴 ∩ 𝐵) ⊆ 𝐵) → ((int‘𝐽)‘(𝐴 ∩ 𝐵)) ⊆ ((int‘𝐽)‘𝐵)) |
| 8 | 6, 7 | mp3an3 1451 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝐴 ∩ 𝐵)) ⊆ ((int‘𝐽)‘𝐵)) |
| 9 | 8 | 3adant2 1131 |
. . 3
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝐴 ∩ 𝐵)) ⊆ ((int‘𝐽)‘𝐵)) |
| 10 | 5, 9 | ssind 4223 |
. 2
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝐴 ∩ 𝐵)) ⊆ (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵))) |
| 11 | | simp1 1136 |
. . 3
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → 𝐽 ∈ Top) |
| 12 | | ssinss1 4228 |
. . . 4
⊢ (𝐴 ⊆ 𝑋 → (𝐴 ∩ 𝐵) ⊆ 𝑋) |
| 13 | 12 | 3ad2ant2 1134 |
. . 3
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (𝐴 ∩ 𝐵) ⊆ 𝑋) |
| 14 | 2 | ntropn 23022 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋) → ((int‘𝐽)‘𝐴) ∈ 𝐽) |
| 15 | 14 | 3adant3 1132 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘𝐴) ∈ 𝐽) |
| 16 | 2 | ntropn 23022 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘𝐵) ∈ 𝐽) |
| 17 | 16 | 3adant2 1131 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘𝐵) ∈ 𝐽) |
| 18 | | inopn 22872 |
. . . 4
⊢ ((𝐽 ∈ Top ∧
((int‘𝐽)‘𝐴) ∈ 𝐽 ∧ ((int‘𝐽)‘𝐵) ∈ 𝐽) → (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ∈ 𝐽) |
| 19 | 11, 15, 17, 18 | syl3anc 1372 |
. . 3
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ∈ 𝐽) |
| 20 | | inss1 4219 |
. . . . 5
⊢
(((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ⊆ ((int‘𝐽)‘𝐴) |
| 21 | 2 | ntrss2 23030 |
. . . . . 6
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋) → ((int‘𝐽)‘𝐴) ⊆ 𝐴) |
| 22 | 21 | 3adant3 1132 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘𝐴) ⊆ 𝐴) |
| 23 | 20, 22 | sstrid 3977 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ⊆ 𝐴) |
| 24 | | inss2 4220 |
. . . . 5
⊢
(((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ⊆ ((int‘𝐽)‘𝐵) |
| 25 | 2 | ntrss2 23030 |
. . . . . 6
⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘𝐵) ⊆ 𝐵) |
| 26 | 25 | 3adant2 1131 |
. . . . 5
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘𝐵) ⊆ 𝐵) |
| 27 | 24, 26 | sstrid 3977 |
. . . 4
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ⊆ 𝐵) |
| 28 | 23, 27 | ssind 4223 |
. . 3
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ⊆ (𝐴 ∩ 𝐵)) |
| 29 | 2 | ssntr 23031 |
. . 3
⊢ (((𝐽 ∈ Top ∧ (𝐴 ∩ 𝐵) ⊆ 𝑋) ∧ ((((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ∈ 𝐽 ∧ (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ⊆ (𝐴 ∩ 𝐵))) → (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ⊆ ((int‘𝐽)‘(𝐴 ∩ 𝐵))) |
| 30 | 11, 13, 19, 28, 29 | syl22anc 838 |
. 2
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵)) ⊆ ((int‘𝐽)‘(𝐴 ∩ 𝐵))) |
| 31 | 10, 30 | eqssd 3983 |
1
⊢ ((𝐽 ∈ Top ∧ 𝐴 ⊆ 𝑋 ∧ 𝐵 ⊆ 𝑋) → ((int‘𝐽)‘(𝐴 ∩ 𝐵)) = (((int‘𝐽)‘𝐴) ∩ ((int‘𝐽)‘𝐵))) |