Metamath Proof Explorer


Theorem lnjatN

Description: Given an atom in a line, there is another atom which when joined equals the line. (Contributed by NM, 30-Apr-2012) (New usage is discouraged.)

Ref Expression
Hypotheses lnjat.b ⊢ B = Base K
lnjat.l ⊢ ≤ ˙ = ≤ K
lnjat.j ⊢ ∨ ˙ = join ⁡ K
lnjat.a ⊢ A = Atoms ⁡ K
lnjat.n ⊢ N = Lines ⁡ K
lnjat.m ⊢ M = pmap ⁡ K
Assertion lnjatN ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X → ∃ q ∈ A q ≠ P ∧ X = P ∨ ˙ q

Proof

Step Hyp Ref Expression
1 lnjat.b ⊢ B = Base K
2 lnjat.l ⊢ ≤ ˙ = ≤ K
3 lnjat.j ⊢ ∨ ˙ = join ⁡ K
4 lnjat.a ⊢ A = Atoms ⁡ K
5 lnjat.n ⊢ N = Lines ⁡ K
6 lnjat.m ⊢ M = pmap ⁡ K
7 simpl1 ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X → K ∈ HL
8 simpl2 ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X → X ∈ B
9 simprl ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X → M ⁡ X ∈ N
10 1 2 4 5 6 lnatexN ⊢ K ∈ HL ∧ X ∈ B ∧ M ⁡ X ∈ N → ∃ q ∈ A q ≠ P ∧ q ≤ ˙ X
11 7 8 9 10 syl3anc ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X → ∃ q ∈ A q ≠ P ∧ q ≤ ˙ X
12 simp3l ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → q ≠ P
13 simp1l1 ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → K ∈ HL
14 simp1l2 ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → X ∈ B
15 simp1rl ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → M ⁡ X ∈ N
16 simp1l3 ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → P ∈ A
17 simp2 ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → q ∈ A
18 12 necomd ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → P ≠ q
19 simp1rr ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → P ≤ ˙ X
20 simp3r ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → q ≤ ˙ X
21 1 2 3 4 5 6 lneq2at ⊢ K ∈ HL ∧ X ∈ B ∧ M ⁡ X ∈ N ∧ P ∈ A ∧ q ∈ A ∧ P ≠ q ∧ P ≤ ˙ X ∧ q ≤ ˙ X → X = P ∨ ˙ q
22 13 14 15 16 17 18 19 20 21 syl332anc ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → X = P ∨ ˙ q
23 12 22 jca ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X ∧ q ∈ A ∧ q ≠ P ∧ q ≤ ˙ X → q ≠ P ∧ X = P ∨ ˙ q
24 23 3exp ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X → q ∈ A → q ≠ P ∧ q ≤ ˙ X → q ≠ P ∧ X = P ∨ ˙ q
25 24 reximdvai ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X → ∃ q ∈ A q ≠ P ∧ q ≤ ˙ X → ∃ q ∈ A q ≠ P ∧ X = P ∨ ˙ q
26 11 25 mpd ⊢ K ∈ HL ∧ X ∈ B ∧ P ∈ A ∧ M ⁡ X ∈ N ∧ P ≤ ˙ X → ∃ q ∈ A q ≠ P ∧ X = P ∨ ˙ q