Metamath Proof Explorer


Theorem mdsl2bi

Description: If the modular pair property holds in a sublattice, it holds in the whole lattice. Lemma 1.4 of MaedaMaeda p. 2. (Contributed by NM, 24-Dec-2006) (New usage is discouraged.)

Ref Expression
Hypotheses mdsl.1 ⊢ A ∈ C ℋ
mdsl.2 ⊢ B ∈ C ℋ
Assertion mdsl2bi ⊢ A 𝑀 ℋ B ↔ ∀ x ∈ C ℋ A ∩ B ⊆ x ∧ x ⊆ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B

Proof

Step Hyp Ref Expression
1 mdsl.1 ⊢ A ∈ C ℋ
2 mdsl.2 ⊢ B ∈ C ℋ
3 1 2 mdsl2i ⊢ A 𝑀 ℋ B ↔ ∀ x ∈ C ℋ A ∩ B ⊆ x ∧ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B
4 1 2 chincli ⊢ A ∩ B ∈ C ℋ
5 inss1 ⊢ A ∩ B ⊆ A
6 chlej2 ⊢ A ∩ B ∈ C ℋ ∧ A ∈ C ℋ ∧ x ∈ C ℋ ∧ A ∩ B ⊆ A → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A
7 5 6 mpan2 ⊢ A ∩ B ∈ C ℋ ∧ A ∈ C ℋ ∧ x ∈ C ℋ → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A
8 4 1 7 mp3an12 ⊢ x ∈ C ℋ → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A
9 8 adantr ⊢ x ∈ C ℋ ∧ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A
10 simpr ⊢ x ∈ C ℋ ∧ x ⊆ B → x ⊆ B
11 inss2 ⊢ A ∩ B ⊆ B
12 10 11 jctir ⊢ x ∈ C ℋ ∧ x ⊆ B → x ⊆ B ∧ A ∩ B ⊆ B
13 chlub ⊢ x ∈ C ℋ ∧ A ∩ B ∈ C ℋ ∧ B ∈ C ℋ → x ⊆ B ∧ A ∩ B ⊆ B ↔ x ∨ ℋ A ∩ B ⊆ B
14 4 2 13 mp3an23 ⊢ x ∈ C ℋ → x ⊆ B ∧ A ∩ B ⊆ B ↔ x ∨ ℋ A ∩ B ⊆ B
15 14 adantr ⊢ x ∈ C ℋ ∧ x ⊆ B → x ⊆ B ∧ A ∩ B ⊆ B ↔ x ∨ ℋ A ∩ B ⊆ B
16 12 15 mpbid ⊢ x ∈ C ℋ ∧ x ⊆ B → x ∨ ℋ A ∩ B ⊆ B
17 9 16 ssind ⊢ x ∈ C ℋ ∧ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B
18 17 biantrud ⊢ x ∈ C ℋ ∧ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B ↔ x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B ∧ x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B
19 eqss ⊢ x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B ↔ x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B ∧ x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B
20 18 19 bitr4di ⊢ x ∈ C ℋ ∧ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B ↔ x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
21 20 ex ⊢ x ∈ C ℋ → x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B ↔ x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
22 21 adantld ⊢ x ∈ C ℋ → A ∩ B ⊆ x ∧ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B ↔ x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
23 22 pm5.74d ⊢ x ∈ C ℋ → A ∩ B ⊆ x ∧ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B ↔ A ∩ B ⊆ x ∧ x ⊆ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
24 23 ralbiia ⊢ ∀ x ∈ C ℋ A ∩ B ⊆ x ∧ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B ↔ ∀ x ∈ C ℋ A ∩ B ⊆ x ∧ x ⊆ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
25 3 24 bitri ⊢ A 𝑀 ℋ B ↔ ∀ x ∈ C ℋ A ∩ B ⊆ x ∧ x ⊆ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B