Metamath Proof Explorer


Theorem atcvrj1

Description: Condition for an atom to be covered by the join of two others. (Contributed by NM, 7-Feb-2012)

Ref Expression
Hypotheses atcvrj1x.l ⊢ ≤ ˙ = ≤ K
atcvrj1x.j ⊢ ∨ ˙ = join ⁡ K
atcvrj1x.c ⊢ C = ⋖ K
atcvrj1x.a ⊢ A = Atoms ⁡ K
Assertion atcvrj1 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P C Q ∨ ˙ R

Proof

Step Hyp Ref Expression
1 atcvrj1x.l ⊢ ≤ ˙ = ≤ K
2 atcvrj1x.j ⊢ ∨ ˙ = join ⁡ K
3 atcvrj1x.c ⊢ C = ⋖ K
4 atcvrj1x.a ⊢ A = Atoms ⁡ K
5 simp3l ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P ≠ R
6 hlatl ⊢ K ∈ HL → K ∈ AtLat
7 6 3ad2ant1 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → K ∈ AtLat
8 simp21 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P ∈ A
9 simp23 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → R ∈ A
10 eqid ⊢ meet ⁡ K = meet ⁡ K
11 eqid ⊢ 0. ⁡ K = 0. ⁡ K
12 10 11 4 atnem0 ⊢ K ∈ AtLat ∧ P ∈ A ∧ R ∈ A → P ≠ R ↔ P meet ⁡ K R = 0. ⁡ K
13 7 8 9 12 syl3anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P ≠ R ↔ P meet ⁡ K R = 0. ⁡ K
14 5 13 mpbid ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P meet ⁡ K R = 0. ⁡ K
15 simp1 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → K ∈ HL
16 eqid ⊢ Base K = Base K
17 16 4 atbase ⊢ P ∈ A → P ∈ Base K
18 8 17 syl ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P ∈ Base K
19 16 2 10 11 3 4 cvrp ⊢ K ∈ HL ∧ P ∈ Base K ∧ R ∈ A → P meet ⁡ K R = 0. ⁡ K ↔ P C P ∨ ˙ R
20 15 18 9 19 syl3anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P meet ⁡ K R = 0. ⁡ K ↔ P C P ∨ ˙ R
21 14 20 mpbid ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P C P ∨ ˙ R
22 simp3r ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P ≤ ˙ Q ∨ ˙ R
23 1 2 4 hlatexchb2 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R → P ≤ ˙ Q ∨ ˙ R ↔ P ∨ ˙ R = Q ∨ ˙ R
24 23 3adant3r ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P ≤ ˙ Q ∨ ˙ R ↔ P ∨ ˙ R = Q ∨ ˙ R
25 22 24 mpbid ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P ∨ ˙ R = Q ∨ ˙ R
26 21 25 breqtrd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ R ∈ A ∧ P ≠ R ∧ P ≤ ˙ Q ∨ ˙ R → P C Q ∨ ˙ R