Metamath Proof Explorer


Theorem dalemcea

Description: Lemma for dath . Frequently-used utility lemma. Here we show that C must be an atom. This is an assumption in most presentations of Desargues's theorem; instead, we assume only the C is a lattice element, in order to make later substitutions for C easier. (Contributed by NM, 23-Sep-2012)

Ref Expression
Hypotheses dalema.ph φ K HL C Base K P A Q A R A S A T A U A Y O Z O ¬ C ˙ P ˙ Q ¬ C ˙ Q ˙ R ¬ C ˙ R ˙ P ¬ C ˙ S ˙ T ¬ C ˙ T ˙ U ¬ C ˙ U ˙ S C ˙ P ˙ S C ˙ Q ˙ T C ˙ R ˙ U
dalemc.l ˙ = K
dalemc.j ˙ = join K
dalemc.a A = Atoms K
dalem1.o O = LPlanes K
dalem1.y Y = P ˙ Q ˙ R
Assertion dalemcea φ C A

Proof

Step Hyp Ref Expression
1 dalema.ph φ K HL C Base K P A Q A R A S A T A U A Y O Z O ¬ C ˙ P ˙ Q ¬ C ˙ Q ˙ R ¬ C ˙ R ˙ P ¬ C ˙ S ˙ T ¬ C ˙ T ˙ U ¬ C ˙ U ˙ S C ˙ P ˙ S C ˙ Q ˙ T C ˙ R ˙ U
2 dalemc.l ˙ = K
3 dalemc.j ˙ = join K
4 dalemc.a A = Atoms K
5 dalem1.o O = LPlanes K
6 dalem1.y Y = P ˙ Q ˙ R
7 1 dalemkeop φ K OP
8 1 4 dalemceb φ C Base K
9 1 dalemkehl φ K HL
10 1 2 3 4 5 6 dalempjsen φ P ˙ S LLines K
11 1 dalemqea φ Q A
12 1 dalemtea φ T A
13 1 2 3 4 5 6 dalemqnet φ Q T
14 eqid LLines K = LLines K
15 3 4 14 llni2 K HL Q A T A Q T Q ˙ T LLines K
16 9 11 12 13 15 syl31anc φ Q ˙ T LLines K
17 1 2 3 4 5 6 dalem1 φ P ˙ S Q ˙ T
18 1 dalem-clpjq φ ¬ C ˙ P ˙ Q
19 1 3 4 dalempjqeb φ P ˙ Q Base K
20 eqid Base K = Base K
21 eqid 0. K = 0. K
22 20 2 21 op0le K OP P ˙ Q Base K 0. K ˙ P ˙ Q
23 7 19 22 syl2anc φ 0. K ˙ P ˙ Q
24 breq1 C = 0. K C ˙ P ˙ Q 0. K ˙ P ˙ Q
25 23 24 syl5ibrcom φ C = 0. K C ˙ P ˙ Q
26 25 necon3bd φ ¬ C ˙ P ˙ Q C 0. K
27 18 26 mpd φ C 0. K
28 eqid < K = < K
29 20 28 21 opltn0 K OP C Base K 0. K < K C C 0. K
30 7 8 29 syl2anc φ 0. K < K C C 0. K
31 27 30 mpbird φ 0. K < K C
32 1 dalemclpjs φ C ˙ P ˙ S
33 1 dalemclqjt φ C ˙ Q ˙ T
34 1 dalemkelat φ K Lat
35 1 dalempea φ P A
36 1 dalemsea φ S A
37 20 3 4 hlatjcl K HL P A S A P ˙ S Base K
38 9 35 36 37 syl3anc φ P ˙ S Base K
39 20 3 4 hlatjcl K HL Q A T A Q ˙ T Base K
40 9 11 12 39 syl3anc φ Q ˙ T Base K
41 eqid meet K = meet K
42 20 2 41 latlem12 K Lat C Base K P ˙ S Base K Q ˙ T Base K C ˙ P ˙ S C ˙ Q ˙ T C ˙ P ˙ S meet K Q ˙ T
43 34 8 38 40 42 syl13anc φ C ˙ P ˙ S C ˙ Q ˙ T C ˙ P ˙ S meet K Q ˙ T
44 32 33 43 mpbi2and φ C ˙ P ˙ S meet K Q ˙ T
45 opposet K OP K Poset
46 7 45 syl φ K Poset
47 20 21 op0cl K OP 0. K Base K
48 7 47 syl φ 0. K Base K
49 20 41 latmcl K Lat P ˙ S Base K Q ˙ T Base K P ˙ S meet K Q ˙ T Base K
50 34 38 40 49 syl3anc φ P ˙ S meet K Q ˙ T Base K
51 20 2 28 pltletr K Poset 0. K Base K C Base K P ˙ S meet K Q ˙ T Base K 0. K < K C C ˙ P ˙ S meet K Q ˙ T 0. K < K P ˙ S meet K Q ˙ T
52 46 48 8 50 51 syl13anc φ 0. K < K C C ˙ P ˙ S meet K Q ˙ T 0. K < K P ˙ S meet K Q ˙ T
53 31 44 52 mp2and φ 0. K < K P ˙ S meet K Q ˙ T
54 20 28 21 opltn0 K OP P ˙ S meet K Q ˙ T Base K 0. K < K P ˙ S meet K Q ˙ T P ˙ S meet K Q ˙ T 0. K
55 7 50 54 syl2anc φ 0. K < K P ˙ S meet K Q ˙ T P ˙ S meet K Q ˙ T 0. K
56 53 55 mpbid φ P ˙ S meet K Q ˙ T 0. K
57 41 21 4 14 2llnmat K HL P ˙ S LLines K Q ˙ T LLines K P ˙ S Q ˙ T P ˙ S meet K Q ˙ T 0. K P ˙ S meet K Q ˙ T A
58 9 10 16 17 56 57 syl32anc φ P ˙ S meet K Q ˙ T A
59 20 2 21 4 leat2 K OP C Base K P ˙ S meet K Q ˙ T A C 0. K C ˙ P ˙ S meet K Q ˙ T C = P ˙ S meet K Q ˙ T
60 7 8 58 27 44 59 syl32anc φ C = P ˙ S meet K Q ˙ T
61 60 58 eqeltrd φ C A