Metamath Proof Explorer


Theorem shlub

Description: Hilbert lattice join is the least upper bound (among Hilbert lattice elements) of two subspaces. (Contributed by NM, 15-Jun-2004) (New usage is discouraged.)

Ref Expression
Assertion shlub ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ⊆ C ∧ B ⊆ C ↔ A ∨ ℋ B ⊆ C

Proof

Step Hyp Ref Expression
1 unss ⊢ A ⊆ C ∧ B ⊆ C ↔ A ∪ B ⊆ C
2 simp1 ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ∈ S ℋ
3 shss ⊢ A ∈ S ℋ → A ⊆ ℋ
4 2 3 syl ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ⊆ ℋ
5 simp2 ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → B ∈ S ℋ
6 shss ⊢ B ∈ S ℋ → B ⊆ ℋ
7 5 6 syl ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → B ⊆ ℋ
8 4 7 unssd ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ∪ B ⊆ ℋ
9 chss ⊢ C ∈ C ℋ → C ⊆ ℋ
10 9 3ad2ant3 ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → C ⊆ ℋ
11 occon2 ⊢ A ∪ B ⊆ ℋ ∧ C ⊆ ℋ → A ∪ B ⊆ C → ⊥ ⁡ ⊥ ⁡ A ∪ B ⊆ ⊥ ⁡ ⊥ ⁡ C
12 8 10 11 syl2anc ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ∪ B ⊆ C → ⊥ ⁡ ⊥ ⁡ A ∪ B ⊆ ⊥ ⁡ ⊥ ⁡ C
13 1 12 biimtrid ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ⊆ C ∧ B ⊆ C → ⊥ ⁡ ⊥ ⁡ A ∪ B ⊆ ⊥ ⁡ ⊥ ⁡ C
14 shjval ⊢ A ∈ S ℋ ∧ B ∈ S ℋ → A ∨ ℋ B = ⊥ ⁡ ⊥ ⁡ A ∪ B
15 2 5 14 syl2anc ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ∨ ℋ B = ⊥ ⁡ ⊥ ⁡ A ∪ B
16 ococ ⊢ C ∈ C ℋ → ⊥ ⁡ ⊥ ⁡ C = C
17 16 3ad2ant3 ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → ⊥ ⁡ ⊥ ⁡ C = C
18 17 eqcomd ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → C = ⊥ ⁡ ⊥ ⁡ C
19 15 18 sseq12d ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ∨ ℋ B ⊆ C ↔ ⊥ ⁡ ⊥ ⁡ A ∪ B ⊆ ⊥ ⁡ ⊥ ⁡ C
20 13 19 sylibrd ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ⊆ C ∧ B ⊆ C → A ∨ ℋ B ⊆ C
21 shub1 ⊢ A ∈ S ℋ ∧ B ∈ S ℋ → A ⊆ A ∨ ℋ B
22 2 5 21 syl2anc ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ⊆ A ∨ ℋ B
23 sstr ⊢ A ⊆ A ∨ ℋ B ∧ A ∨ ℋ B ⊆ C → A ⊆ C
24 22 23 sylan ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ ∧ A ∨ ℋ B ⊆ C → A ⊆ C
25 shub2 ⊢ B ∈ S ℋ ∧ A ∈ S ℋ → B ⊆ A ∨ ℋ B
26 5 2 25 syl2anc ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → B ⊆ A ∨ ℋ B
27 sstr ⊢ B ⊆ A ∨ ℋ B ∧ A ∨ ℋ B ⊆ C → B ⊆ C
28 26 27 sylan ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ ∧ A ∨ ℋ B ⊆ C → B ⊆ C
29 24 28 jca ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ ∧ A ∨ ℋ B ⊆ C → A ⊆ C ∧ B ⊆ C
30 29 ex ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ∨ ℋ B ⊆ C → A ⊆ C ∧ B ⊆ C
31 20 30 impbid ⊢ A ∈ S ℋ ∧ B ∈ S ℋ ∧ C ∈ C ℋ → A ⊆ C ∧ B ⊆ C ↔ A ∨ ℋ B ⊆ C