Metamath Proof Explorer


Theorem llni2

Description: The join of two different atoms is a lattice line. (Contributed by NM, 26-Jun-2012)

Ref Expression
Hypotheses llni2.j ⊢ ∨ ˙ = join ⁡ K
llni2.a ⊢ A = Atoms ⁡ K
llni2.n ⊢ N = LLines ⁡ K
Assertion llni2 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → P ∨ ˙ Q ∈ N

Proof

Step Hyp Ref Expression
1 llni2.j ⊢ ∨ ˙ = join ⁡ K
2 llni2.a ⊢ A = Atoms ⁡ K
3 llni2.n ⊢ N = LLines ⁡ K
4 simpl2 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → P ∈ A
5 simpl3 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → Q ∈ A
6 simpr ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → P ≠ Q
7 eqidd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → P ∨ ˙ Q = P ∨ ˙ Q
8 neeq1 ⊢ r = P → r ≠ s ↔ P ≠ s
9 oveq1 ⊢ r = P → r ∨ ˙ s = P ∨ ˙ s
10 9 eqeq2d ⊢ r = P → P ∨ ˙ Q = r ∨ ˙ s ↔ P ∨ ˙ Q = P ∨ ˙ s
11 8 10 anbi12d ⊢ r = P → r ≠ s ∧ P ∨ ˙ Q = r ∨ ˙ s ↔ P ≠ s ∧ P ∨ ˙ Q = P ∨ ˙ s
12 neeq2 ⊢ s = Q → P ≠ s ↔ P ≠ Q
13 oveq2 ⊢ s = Q → P ∨ ˙ s = P ∨ ˙ Q
14 13 eqeq2d ⊢ s = Q → P ∨ ˙ Q = P ∨ ˙ s ↔ P ∨ ˙ Q = P ∨ ˙ Q
15 12 14 anbi12d ⊢ s = Q → P ≠ s ∧ P ∨ ˙ Q = P ∨ ˙ s ↔ P ≠ Q ∧ P ∨ ˙ Q = P ∨ ˙ Q
16 11 15 rspc2ev ⊢ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ P ∨ ˙ Q = P ∨ ˙ Q → ∃ r ∈ A ∃ s ∈ A r ≠ s ∧ P ∨ ˙ Q = r ∨ ˙ s
17 4 5 6 7 16 syl112anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → ∃ r ∈ A ∃ s ∈ A r ≠ s ∧ P ∨ ˙ Q = r ∨ ˙ s
18 simpl1 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → K ∈ HL
19 eqid ⊢ Base K = Base K
20 19 1 2 hlatjcl ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A → P ∨ ˙ Q ∈ Base K
21 20 adantr ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → P ∨ ˙ Q ∈ Base K
22 19 1 2 3 islln3 ⊢ K ∈ HL ∧ P ∨ ˙ Q ∈ Base K → P ∨ ˙ Q ∈ N ↔ ∃ r ∈ A ∃ s ∈ A r ≠ s ∧ P ∨ ˙ Q = r ∨ ˙ s
23 18 21 22 syl2anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → P ∨ ˙ Q ∈ N ↔ ∃ r ∈ A ∃ s ∈ A r ≠ s ∧ P ∨ ˙ Q = r ∨ ˙ s
24 17 23 mpbird ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q → P ∨ ˙ Q ∈ N