Metamath Proof Explorer


Theorem cvmdi

Description: The covering property implies the modular pair property. Lemma 7.5.1 of MaedaMaeda p. 31. (Contributed by NM, 16-Jun-2004) (New usage is discouraged.)

Ref Expression
Hypotheses mdsl.1 ⊢ A ∈ C ℋ
mdsl.2 ⊢ B ∈ C ℋ
Assertion cvmdi ⊢ A ∩ B ⋖ ℋ B → A 𝑀 ℋ B

Proof

Step Hyp Ref Expression
1 mdsl.1 ⊢ A ∈ C ℋ
2 mdsl.2 ⊢ B ∈ C ℋ
3 anass ⊢ A ∩ B ⊆ x ∧ x ⊆ A ∨ ℋ B ∧ x ⊆ B ↔ A ∩ B ⊆ x ∧ x ⊆ A ∨ ℋ B ∧ x ⊆ B
4 2 1 chub2i ⊢ B ⊆ A ∨ ℋ B
5 sstr ⊢ x ⊆ B ∧ B ⊆ A ∨ ℋ B → x ⊆ A ∨ ℋ B
6 4 5 mpan2 ⊢ x ⊆ B → x ⊆ A ∨ ℋ B
7 6 pm4.71ri ⊢ x ⊆ B ↔ x ⊆ A ∨ ℋ B ∧ x ⊆ B
8 7 anbi2i ⊢ A ∩ B ⊆ x ∧ x ⊆ B ↔ A ∩ B ⊆ x ∧ x ⊆ A ∨ ℋ B ∧ x ⊆ B
9 3 8 bitr4i ⊢ A ∩ B ⊆ x ∧ x ⊆ A ∨ ℋ B ∧ x ⊆ B ↔ A ∩ B ⊆ x ∧ x ⊆ B
10 1 2 chincli ⊢ A ∩ B ∈ C ℋ
11 cvnbtwn4 ⊢ A ∩ B ∈ C ℋ ∧ B ∈ C ℋ ∧ x ∈ C ℋ → A ∩ B ⋖ ℋ B → A ∩ B ⊆ x ∧ x ⊆ B → x = A ∩ B ∨ x = B
12 10 2 11 mp3an12 ⊢ x ∈ C ℋ → A ∩ B ⋖ ℋ B → A ∩ B ⊆ x ∧ x ⊆ B → x = A ∩ B ∨ x = B
13 12 impcom ⊢ A ∩ B ⋖ ℋ B ∧ x ∈ C ℋ → A ∩ B ⊆ x ∧ x ⊆ B → x = A ∩ B ∨ x = B
14 10 1 chjcomi ⊢ A ∩ B ∨ ℋ A = A ∨ ℋ A ∩ B
15 1 2 chabs1i ⊢ A ∨ ℋ A ∩ B = A
16 14 15 eqtri ⊢ A ∩ B ∨ ℋ A = A
17 16 ineq1i ⊢ A ∩ B ∨ ℋ A ∩ B = A ∩ B
18 10 chjidmi ⊢ A ∩ B ∨ ℋ A ∩ B = A ∩ B
19 17 18 eqtr4i ⊢ A ∩ B ∨ ℋ A ∩ B = A ∩ B ∨ ℋ A ∩ B
20 oveq1 ⊢ x = A ∩ B → x ∨ ℋ A = A ∩ B ∨ ℋ A
21 20 ineq1d ⊢ x = A ∩ B → x ∨ ℋ A ∩ B = A ∩ B ∨ ℋ A ∩ B
22 oveq1 ⊢ x = A ∩ B → x ∨ ℋ A ∩ B = A ∩ B ∨ ℋ A ∩ B
23 19 21 22 3eqtr4a ⊢ x = A ∩ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
24 incom ⊢ B ∨ ℋ A ∩ B = B ∩ B ∨ ℋ A
25 2 1 chabs2i ⊢ B ∩ B ∨ ℋ A = B
26 2 1 chabs1i ⊢ B ∨ ℋ B ∩ A = B
27 incom ⊢ B ∩ A = A ∩ B
28 27 oveq2i ⊢ B ∨ ℋ B ∩ A = B ∨ ℋ A ∩ B
29 25 26 28 3eqtr2i ⊢ B ∩ B ∨ ℋ A = B ∨ ℋ A ∩ B
30 24 29 eqtri ⊢ B ∨ ℋ A ∩ B = B ∨ ℋ A ∩ B
31 oveq1 ⊢ x = B → x ∨ ℋ A = B ∨ ℋ A
32 31 ineq1d ⊢ x = B → x ∨ ℋ A ∩ B = B ∨ ℋ A ∩ B
33 oveq1 ⊢ x = B → x ∨ ℋ A ∩ B = B ∨ ℋ A ∩ B
34 30 32 33 3eqtr4a ⊢ x = B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
35 23 34 jaoi ⊢ x = A ∩ B ∨ x = B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
36 13 35 syl6 ⊢ A ∩ B ⋖ ℋ B ∧ x ∈ C ℋ → A ∩ B ⊆ x ∧ x ⊆ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
37 9 36 biimtrid ⊢ A ∩ B ⋖ ℋ B ∧ x ∈ C ℋ → A ∩ B ⊆ x ∧ x ⊆ A ∨ ℋ B ∧ x ⊆ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
38 37 exp4b ⊢ A ∩ B ⋖ ℋ B → x ∈ C ℋ → A ∩ B ⊆ x ∧ x ⊆ A ∨ ℋ B → x ⊆ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
39 38 ralrimiv ⊢ A ∩ B ⋖ ℋ B → ∀ x ∈ C ℋ A ∩ B ⊆ x ∧ x ⊆ A ∨ ℋ B → x ⊆ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B
40 1 2 mdsl1i ⊢ ∀ x ∈ C ℋ A ∩ B ⊆ x ∧ x ⊆ A ∨ ℋ B → x ⊆ B → x ∨ ℋ A ∩ B = x ∨ ℋ A ∩ B ↔ A 𝑀 ℋ B
41 39 40 sylib ⊢ A ∩ B ⋖ ℋ B → A 𝑀 ℋ B