Metamath Proof Explorer


Theorem mdsl0

Description: A sublattice condition that transfers the modular pair property. Exercise 12 of Kalmbach p. 103. Also Lemma 1.5.3 of MaedaMaeda p. 2. (Contributed by NM, 22-Jun-2004) (New usage is discouraged.)

Ref Expression
Assertion mdsl0 ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ → C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ ∧ A 𝑀 ℋ B → C 𝑀 ℋ D

Proof

Step Hyp Ref Expression
1 sstr2 ⊢ x ⊆ D → D ⊆ B → x ⊆ B
2 1 com12 ⊢ D ⊆ B → x ⊆ D → x ⊆ B
3 2 ad2antlr ⊢ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ → x ⊆ D → x ⊆ B
4 3 ad2antlr ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ ∧ x ∈ C ℋ → x ⊆ D → x ⊆ B
5 chlej2 ⊢ C ∈ C ℋ ∧ A ∈ C ℋ ∧ x ∈ C ℋ ∧ C ⊆ A → x ∨ ℋ C ⊆ x ∨ ℋ A
6 ss2in ⊢ x ∨ ℋ C ⊆ x ∨ ℋ A ∧ D ⊆ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
7 6 ex ⊢ x ∨ ℋ C ⊆ x ∨ ℋ A → D ⊆ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
8 5 7 syl ⊢ C ∈ C ℋ ∧ A ∈ C ℋ ∧ x ∈ C ℋ ∧ C ⊆ A → D ⊆ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
9 8 ex ⊢ C ∈ C ℋ ∧ A ∈ C ℋ ∧ x ∈ C ℋ → C ⊆ A → D ⊆ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
10 9 3expia ⊢ C ∈ C ℋ ∧ A ∈ C ℋ → x ∈ C ℋ → C ⊆ A → D ⊆ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
11 10 ancoms ⊢ A ∈ C ℋ ∧ C ∈ C ℋ → x ∈ C ℋ → C ⊆ A → D ⊆ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
12 11 ad2ant2r ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ → x ∈ C ℋ → C ⊆ A → D ⊆ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
13 12 imp43 ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
14 13 adantrr ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
15 oveq2 ⊢ A ∩ B = 0 ℋ → x ∨ ℋ A ∩ B = x ∨ ℋ 0 ℋ
16 chj0 ⊢ x ∈ C ℋ → x ∨ ℋ 0 ℋ = x
17 15 16 sylan9eqr ⊢ x ∈ C ℋ ∧ A ∩ B = 0 ℋ → x ∨ ℋ A ∩ B = x
18 17 adantl ⊢ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ ∧ A ∩ B = 0 ℋ → x ∨ ℋ A ∩ B = x
19 chincl ⊢ C ∈ C ℋ ∧ D ∈ C ℋ → C ∩ D ∈ C ℋ
20 chub1 ⊢ x ∈ C ℋ ∧ C ∩ D ∈ C ℋ → x ⊆ x ∨ ℋ C ∩ D
21 20 ancoms ⊢ C ∩ D ∈ C ℋ ∧ x ∈ C ℋ → x ⊆ x ∨ ℋ C ∩ D
22 19 21 sylan ⊢ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ → x ⊆ x ∨ ℋ C ∩ D
23 22 adantrr ⊢ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ ∧ A ∩ B = 0 ℋ → x ⊆ x ∨ ℋ C ∩ D
24 18 23 eqsstrd ⊢ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ ∧ A ∩ B = 0 ℋ → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ C ∩ D
25 24 adantll ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ ∧ A ∩ B = 0 ℋ → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ C ∩ D
26 25 anassrs ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ ∧ A ∩ B = 0 ℋ → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ C ∩ D
27 26 adantrl ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ C ∩ D
28 sstr2 ⊢ x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B
29 sstr2 ⊢ x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ C ∩ D → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ C ∩ D
30 28 29 syl6 ⊢ x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ C ∩ D → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ C ∩ D
31 30 com23 ⊢ x ∨ ℋ C ∩ D ⊆ x ∨ ℋ A ∩ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ C ∩ D → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ C ∩ D
32 14 27 31 sylc ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ x ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ C ∩ D
33 32 an32s ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ ∧ x ∈ C ℋ → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ C ∩ D
34 4 33 imim12d ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ ∧ x ∈ C ℋ → x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B → x ⊆ D → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ C ∩ D
35 34 ralimdva ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ → ∀ x ∈ C ℋ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B → ∀ x ∈ C ℋ x ⊆ D → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ C ∩ D
36 mdbr2 ⊢ A ∈ C ℋ ∧ B ∈ C ℋ → A 𝑀 ℋ B ↔ ∀ x ∈ C ℋ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B
37 36 ad2antrr ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ → A 𝑀 ℋ B ↔ ∀ x ∈ C ℋ x ⊆ B → x ∨ ℋ A ∩ B ⊆ x ∨ ℋ A ∩ B
38 mdbr2 ⊢ C ∈ C ℋ ∧ D ∈ C ℋ → C 𝑀 ℋ D ↔ ∀ x ∈ C ℋ x ⊆ D → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ C ∩ D
39 38 ad2antlr ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ → C 𝑀 ℋ D ↔ ∀ x ∈ C ℋ x ⊆ D → x ∨ ℋ C ∩ D ⊆ x ∨ ℋ C ∩ D
40 35 37 39 3imtr4d ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ ∧ C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ → A 𝑀 ℋ B → C 𝑀 ℋ D
41 40 expimpd ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ ∧ D ∈ C ℋ → C ⊆ A ∧ D ⊆ B ∧ A ∩ B = 0 ℋ ∧ A 𝑀 ℋ B → C 𝑀 ℋ D