Metamath Proof Explorer


Theorem llnexatN

Description: Given an atom on a line, there is another atom whose join equals the line. (Contributed by NM, 26-Jun-2012) (New usage is discouraged.)

Ref Expression
Hypotheses llnexat.l ⊢ ≤ ˙ = ≤ K
llnexat.j ⊢ ∨ ˙ = join ⁡ K
llnexat.a ⊢ A = Atoms ⁡ K
llnexat.n ⊢ N = LLines ⁡ K
Assertion llnexatN ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → ∃ q ∈ A P ≠ q ∧ X = P ∨ ˙ q

Proof

Step Hyp Ref Expression
1 llnexat.l ⊢ ≤ ˙ = ≤ K
2 llnexat.j ⊢ ∨ ˙ = join ⁡ K
3 llnexat.a ⊢ A = Atoms ⁡ K
4 llnexat.n ⊢ N = LLines ⁡ K
5 simp1 ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A → K ∈ HL
6 simp3 ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A → P ∈ A
7 simp2 ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A → X ∈ N
8 5 6 7 3jca ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A → K ∈ HL ∧ P ∈ A ∧ X ∈ N
9 eqid ⊢ ⋖ K = ⋖ K
10 1 9 3 4 atcvrlln2 ⊢ K ∈ HL ∧ P ∈ A ∧ X ∈ N ∧ P ≤ ˙ X → P ⋖ K X
11 8 10 sylan ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → P ⋖ K X
12 simpl1 ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → K ∈ HL
13 simpl3 ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → P ∈ A
14 eqid ⊢ Base K = Base K
15 14 3 atbase ⊢ P ∈ A → P ∈ Base K
16 13 15 syl ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → P ∈ Base K
17 simpl2 ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → X ∈ N
18 14 4 llnbase ⊢ X ∈ N → X ∈ Base K
19 17 18 syl ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → X ∈ Base K
20 14 1 2 9 3 cvrval3 ⊢ K ∈ HL ∧ P ∈ Base K ∧ X ∈ Base K → P ⋖ K X ↔ ∃ q ∈ A ¬ q ≤ ˙ P ∧ P ∨ ˙ q = X
21 12 16 19 20 syl3anc ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → P ⋖ K X ↔ ∃ q ∈ A ¬ q ≤ ˙ P ∧ P ∨ ˙ q = X
22 simpll1 ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X ∧ q ∈ A → K ∈ HL
23 hlatl ⊢ K ∈ HL → K ∈ AtLat
24 22 23 syl ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X ∧ q ∈ A → K ∈ AtLat
25 simpr ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X ∧ q ∈ A → q ∈ A
26 simpll3 ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X ∧ q ∈ A → P ∈ A
27 1 3 atncmp ⊢ K ∈ AtLat ∧ q ∈ A ∧ P ∈ A → ¬ q ≤ ˙ P ↔ q ≠ P
28 24 25 26 27 syl3anc ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X ∧ q ∈ A → ¬ q ≤ ˙ P ↔ q ≠ P
29 28 anbi1d ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X ∧ q ∈ A → ¬ q ≤ ˙ P ∧ P ∨ ˙ q = X ↔ q ≠ P ∧ P ∨ ˙ q = X
30 necom ⊢ q ≠ P ↔ P ≠ q
31 eqcom ⊢ P ∨ ˙ q = X ↔ X = P ∨ ˙ q
32 30 31 anbi12i ⊢ q ≠ P ∧ P ∨ ˙ q = X ↔ P ≠ q ∧ X = P ∨ ˙ q
33 29 32 bitrdi ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X ∧ q ∈ A → ¬ q ≤ ˙ P ∧ P ∨ ˙ q = X ↔ P ≠ q ∧ X = P ∨ ˙ q
34 33 rexbidva ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → ∃ q ∈ A ¬ q ≤ ˙ P ∧ P ∨ ˙ q = X ↔ ∃ q ∈ A P ≠ q ∧ X = P ∨ ˙ q
35 21 34 bitrd ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → P ⋖ K X ↔ ∃ q ∈ A P ≠ q ∧ X = P ∨ ˙ q
36 11 35 mpbid ⊢ K ∈ HL ∧ X ∈ N ∧ P ∈ A ∧ P ≤ ˙ X → ∃ q ∈ A P ≠ q ∧ X = P ∨ ˙ q