Proof of Theorem pexmidN
Step | Hyp | Ref
| Expression |
1 | | simpll 763 |
. . . . 5
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → 𝐾 ∈ HL) |
2 | | simplr 765 |
. . . . 5
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → 𝑋 ⊆ 𝐴) |
3 | | pexmid.a |
. . . . . . 7
⊢ 𝐴 = (Atoms‘𝐾) |
4 | | pexmid.o |
. . . . . . 7
⊢ ⊥ =
(⊥𝑃‘𝐾) |
5 | 3, 4 | polssatN 37849 |
. . . . . 6
⊢ ((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) → ( ⊥ ‘𝑋) ⊆ 𝐴) |
6 | 5 | adantr 480 |
. . . . 5
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → ( ⊥ ‘𝑋) ⊆ 𝐴) |
7 | | pexmid.p |
. . . . . 6
⊢ + =
(+𝑃‘𝐾) |
8 | 3, 7, 4 | poldmj1N 37869 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴 ∧ ( ⊥ ‘𝑋) ⊆ 𝐴) → ( ⊥ ‘(𝑋 + ( ⊥ ‘𝑋))) = (( ⊥ ‘𝑋) ∩ ( ⊥ ‘( ⊥
‘𝑋)))) |
9 | 1, 2, 6, 8 | syl3anc 1369 |
. . . 4
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → ( ⊥ ‘(𝑋 + ( ⊥ ‘𝑋))) = (( ⊥ ‘𝑋) ∩ ( ⊥ ‘( ⊥
‘𝑋)))) |
10 | 3, 4 | pnonsingN 37874 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ ( ⊥
‘𝑋) ⊆ 𝐴) → (( ⊥ ‘𝑋) ∩ ( ⊥ ‘( ⊥
‘𝑋))) =
∅) |
11 | 1, 6, 10 | syl2anc 583 |
. . . 4
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → (( ⊥ ‘𝑋) ∩ ( ⊥ ‘( ⊥
‘𝑋))) =
∅) |
12 | 9, 11 | eqtrd 2778 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → ( ⊥ ‘(𝑋 + ( ⊥ ‘𝑋))) = ∅) |
13 | 12 | fveq2d 6760 |
. 2
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → ( ⊥ ‘( ⊥
‘(𝑋 + ( ⊥
‘𝑋)))) = ( ⊥
‘∅)) |
14 | | simpr 484 |
. . . . 5
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) |
15 | | eqid 2738 |
. . . . . . 7
⊢
(PSubCl‘𝐾) =
(PSubCl‘𝐾) |
16 | 3, 4, 15 | ispsubclN 37878 |
. . . . . 6
⊢ (𝐾 ∈ HL → (𝑋 ∈ (PSubCl‘𝐾) ↔ (𝑋 ⊆ 𝐴 ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋))) |
17 | 16 | ad2antrr 722 |
. . . . 5
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → (𝑋 ∈ (PSubCl‘𝐾) ↔ (𝑋 ⊆ 𝐴 ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋))) |
18 | 2, 14, 17 | mpbir2and 709 |
. . . 4
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → 𝑋 ∈ (PSubCl‘𝐾)) |
19 | 3, 4, 15 | polsubclN 37893 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) → ( ⊥ ‘𝑋) ∈ (PSubCl‘𝐾)) |
20 | 19 | adantr 480 |
. . . 4
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → ( ⊥ ‘𝑋) ∈ (PSubCl‘𝐾)) |
21 | 3, 4 | 2polssN 37856 |
. . . . 5
⊢ ((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) → 𝑋 ⊆ ( ⊥ ‘( ⊥
‘𝑋))) |
22 | 21 | adantr 480 |
. . . 4
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → 𝑋 ⊆ ( ⊥ ‘( ⊥
‘𝑋))) |
23 | 7, 4, 15 | osumclN 37908 |
. . . 4
⊢ (((𝐾 ∈ HL ∧ 𝑋 ∈ (PSubCl‘𝐾) ∧ ( ⊥ ‘𝑋) ∈ (PSubCl‘𝐾)) ∧ 𝑋 ⊆ ( ⊥ ‘( ⊥
‘𝑋))) → (𝑋 + ( ⊥ ‘𝑋)) ∈ (PSubCl‘𝐾)) |
24 | 1, 18, 20, 22, 23 | syl31anc 1371 |
. . 3
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → (𝑋 + ( ⊥ ‘𝑋)) ∈ (PSubCl‘𝐾)) |
25 | 4, 15 | psubcli2N 37880 |
. . 3
⊢ ((𝐾 ∈ HL ∧ (𝑋 + ( ⊥ ‘𝑋)) ∈ (PSubCl‘𝐾)) → ( ⊥ ‘( ⊥
‘(𝑋 + ( ⊥
‘𝑋)))) = (𝑋 + ( ⊥ ‘𝑋))) |
26 | 1, 24, 25 | syl2anc 583 |
. 2
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → ( ⊥ ‘( ⊥
‘(𝑋 + ( ⊥
‘𝑋)))) = (𝑋 + ( ⊥ ‘𝑋))) |
27 | 3, 4 | pol0N 37850 |
. . 3
⊢ (𝐾 ∈ HL → ( ⊥
‘∅) = 𝐴) |
28 | 27 | ad2antrr 722 |
. 2
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → ( ⊥ ‘∅) =
𝐴) |
29 | 13, 26, 28 | 3eqtr3d 2786 |
1
⊢ (((𝐾 ∈ HL ∧ 𝑋 ⊆ 𝐴) ∧ ( ⊥ ‘( ⊥
‘𝑋)) = 𝑋) → (𝑋 + ( ⊥ ‘𝑋)) = 𝐴) |