Metamath Proof Explorer


Theorem atcvati

Description: A nonzero Hilbert lattice element less than the join of two atoms is an atom. (Contributed by NM, 28-Jun-2004) (New usage is discouraged.)

Ref Expression
Hypothesis atoml.1 ⊢ A ∈ C ℋ
Assertion atcvati ⊢ B ∈ HAtoms ∧ C ∈ HAtoms → A ≠ 0 ℋ ∧ A ⊂ B ∨ ℋ C → A ∈ HAtoms

Proof

Step Hyp Ref Expression
1 atoml.1 ⊢ A ∈ C ℋ
2 1 atcvatlem ⊢ B ∈ HAtoms ∧ C ∈ HAtoms ∧ A ≠ 0 ℋ ∧ A ⊂ B ∨ ℋ C → ¬ B ⊆ A → A ∈ HAtoms
3 atelch ⊢ C ∈ HAtoms → C ∈ C ℋ
4 atelch ⊢ B ∈ HAtoms → B ∈ C ℋ
5 chjcom ⊢ C ∈ C ℋ ∧ B ∈ C ℋ → C ∨ ℋ B = B ∨ ℋ C
6 3 4 5 syl2an ⊢ C ∈ HAtoms ∧ B ∈ HAtoms → C ∨ ℋ B = B ∨ ℋ C
7 6 psseq2d ⊢ C ∈ HAtoms ∧ B ∈ HAtoms → A ⊂ C ∨ ℋ B ↔ A ⊂ B ∨ ℋ C
8 7 anbi2d ⊢ C ∈ HAtoms ∧ B ∈ HAtoms → A ≠ 0 ℋ ∧ A ⊂ C ∨ ℋ B ↔ A ≠ 0 ℋ ∧ A ⊂ B ∨ ℋ C
9 1 atcvatlem ⊢ C ∈ HAtoms ∧ B ∈ HAtoms ∧ A ≠ 0 ℋ ∧ A ⊂ C ∨ ℋ B → ¬ C ⊆ A → A ∈ HAtoms
10 9 ex ⊢ C ∈ HAtoms ∧ B ∈ HAtoms → A ≠ 0 ℋ ∧ A ⊂ C ∨ ℋ B → ¬ C ⊆ A → A ∈ HAtoms
11 8 10 sylbird ⊢ C ∈ HAtoms ∧ B ∈ HAtoms → A ≠ 0 ℋ ∧ A ⊂ B ∨ ℋ C → ¬ C ⊆ A → A ∈ HAtoms
12 11 ancoms ⊢ B ∈ HAtoms ∧ C ∈ HAtoms → A ≠ 0 ℋ ∧ A ⊂ B ∨ ℋ C → ¬ C ⊆ A → A ∈ HAtoms
13 12 imp ⊢ B ∈ HAtoms ∧ C ∈ HAtoms ∧ A ≠ 0 ℋ ∧ A ⊂ B ∨ ℋ C → ¬ C ⊆ A → A ∈ HAtoms
14 chlub ⊢ B ∈ C ℋ ∧ C ∈ C ℋ ∧ A ∈ C ℋ → B ⊆ A ∧ C ⊆ A ↔ B ∨ ℋ C ⊆ A
15 14 3comr ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ → B ⊆ A ∧ C ⊆ A ↔ B ∨ ℋ C ⊆ A
16 ssnpss ⊢ B ∨ ℋ C ⊆ A → ¬ A ⊂ B ∨ ℋ C
17 15 16 biimtrdi ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ → B ⊆ A ∧ C ⊆ A → ¬ A ⊂ B ∨ ℋ C
18 17 con2d ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ → A ⊂ B ∨ ℋ C → ¬ B ⊆ A ∧ C ⊆ A
19 ianor ⊢ ¬ B ⊆ A ∧ C ⊆ A ↔ ¬ B ⊆ A ∨ ¬ C ⊆ A
20 18 19 imbitrdi ⊢ A ∈ C ℋ ∧ B ∈ C ℋ ∧ C ∈ C ℋ → A ⊂ B ∨ ℋ C → ¬ B ⊆ A ∨ ¬ C ⊆ A
21 1 20 mp3an1 ⊢ B ∈ C ℋ ∧ C ∈ C ℋ → A ⊂ B ∨ ℋ C → ¬ B ⊆ A ∨ ¬ C ⊆ A
22 4 3 21 syl2an ⊢ B ∈ HAtoms ∧ C ∈ HAtoms → A ⊂ B ∨ ℋ C → ¬ B ⊆ A ∨ ¬ C ⊆ A
23 22 imp ⊢ B ∈ HAtoms ∧ C ∈ HAtoms ∧ A ⊂ B ∨ ℋ C → ¬ B ⊆ A ∨ ¬ C ⊆ A
24 23 adantrl ⊢ B ∈ HAtoms ∧ C ∈ HAtoms ∧ A ≠ 0 ℋ ∧ A ⊂ B ∨ ℋ C → ¬ B ⊆ A ∨ ¬ C ⊆ A
25 2 13 24 mpjaod ⊢ B ∈ HAtoms ∧ C ∈ HAtoms ∧ A ≠ 0 ℋ ∧ A ⊂ B ∨ ℋ C → A ∈ HAtoms
26 25 ex ⊢ B ∈ HAtoms ∧ C ∈ HAtoms → A ≠ 0 ℋ ∧ A ⊂ B ∨ ℋ C → A ∈ HAtoms