Metamath Proof Explorer


Theorem lnnat

Description: A line (the join of two distinct atoms) is not an atom. (Contributed by NM, 14-Jun-2012)

Ref Expression
Hypotheses lnnat.j ˙ = join K
lnnat.a A = Atoms K
Assertion lnnat K HL P A Q A P Q ¬ P ˙ Q A

Proof

Step Hyp Ref Expression
1 lnnat.j ˙ = join K
2 lnnat.a A = Atoms K
3 simpl1 K HL P A Q A P Q K HL
4 simpl2 K HL P A Q A P Q P A
5 eqid 0. K = 0. K
6 eqid K = K
7 5 6 2 atcvr0 K HL P A 0. K K P
8 3 4 7 syl2anc K HL P A Q A P Q 0. K K P
9 1 6 2 atcvr1 K HL P A Q A P Q P K P ˙ Q
10 9 biimpa K HL P A Q A P Q P K P ˙ Q
11 hlop K HL K OP
12 eqid Base K = Base K
13 12 5 op0cl K OP 0. K Base K
14 3 11 13 3syl K HL P A Q A P Q 0. K Base K
15 12 2 atbase P A P Base K
16 4 15 syl K HL P A Q A P Q P Base K
17 3 hllatd K HL P A Q A P Q K Lat
18 simpl3 K HL P A Q A P Q Q A
19 12 2 atbase Q A Q Base K
20 18 19 syl K HL P A Q A P Q Q Base K
21 12 1 latjcl K Lat P Base K Q Base K P ˙ Q Base K
22 17 16 20 21 syl3anc K HL P A Q A P Q P ˙ Q Base K
23 12 6 cvrntr K HL 0. K Base K P Base K P ˙ Q Base K 0. K K P P K P ˙ Q ¬ 0. K K P ˙ Q
24 3 14 16 22 23 syl13anc K HL P A Q A P Q 0. K K P P K P ˙ Q ¬ 0. K K P ˙ Q
25 8 10 24 mp2and K HL P A Q A P Q ¬ 0. K K P ˙ Q
26 simpll1 K HL P A Q A P Q P ˙ Q A K HL
27 5 6 2 atcvr0 K HL P ˙ Q A 0. K K P ˙ Q
28 26 27 sylancom K HL P A Q A P Q P ˙ Q A 0. K K P ˙ Q
29 25 28 mtand K HL P A Q A P Q ¬ P ˙ Q A
30 29 ex K HL P A Q A P Q ¬ P ˙ Q A
31 1 2 hlatjidm K HL P A P ˙ P = P
32 31 3adant3 K HL P A Q A P ˙ P = P
33 simp2 K HL P A Q A P A
34 32 33 eqeltrd K HL P A Q A P ˙ P A
35 oveq2 P = Q P ˙ P = P ˙ Q
36 35 eleq1d P = Q P ˙ P A P ˙ Q A
37 34 36 syl5ibcom K HL P A Q A P = Q P ˙ Q A
38 37 necon3bd K HL P A Q A ¬ P ˙ Q A P Q
39 30 38 impbid K HL P A Q A P Q ¬ P ˙ Q A