Metamath Proof Explorer


Theorem tglineneq

Description: Given three non-colinear points, build two different lines. (Contributed by Thierry Arnoux, 6-Aug-2019)

Ref Expression
Hypotheses tglineintmo.p ⊢ P = Base G
tglineintmo.i ⊢ I = Itv ⁡ G
tglineintmo.l ⊢ L = Line 𝒢 ⁡ G
tglineintmo.g ⊢ φ → G ∈ 𝒢 Tarski
tglineinteq.a ⊢ φ → A ∈ P
tglineinteq.b ⊢ φ → B ∈ P
tglineinteq.c ⊢ φ → C ∈ P
tglineinteq.d ⊢ φ → D ∈ P
tglineinteq.e ⊢ φ → ¬ A ∈ B L C ∨ B = C
Assertion tglineneq ⊢ φ → A L B ≠ C L D

Proof

Step Hyp Ref Expression
1 tglineintmo.p ⊢ P = Base G
2 tglineintmo.i ⊢ I = Itv ⁡ G
3 tglineintmo.l ⊢ L = Line 𝒢 ⁡ G
4 tglineintmo.g ⊢ φ → G ∈ 𝒢 Tarski
5 tglineinteq.a ⊢ φ → A ∈ P
6 tglineinteq.b ⊢ φ → B ∈ P
7 tglineinteq.c ⊢ φ → C ∈ P
8 tglineinteq.d ⊢ φ → D ∈ P
9 tglineinteq.e ⊢ φ → ¬ A ∈ B L C ∨ B = C
10 1 2 3 4 5 6 7 9 ncolne1 ⊢ φ → A ≠ B
11 1 2 3 4 5 6 10 tglinerflx1 ⊢ φ → A ∈ A L B
12 simplr ⊢ φ ∧ C = D ∧ A ∈ C L D → C = D
13 4 adantr ⊢ φ ∧ A ∈ C L D → G ∈ 𝒢 Tarski
14 7 adantr ⊢ φ ∧ A ∈ C L D → C ∈ P
15 8 adantr ⊢ φ ∧ A ∈ C L D → D ∈ P
16 simpr ⊢ φ ∧ A ∈ C L D → A ∈ C L D
17 1 3 2 13 14 15 16 tglngne ⊢ φ ∧ A ∈ C L D → C ≠ D
18 17 adantlr ⊢ φ ∧ C = D ∧ A ∈ C L D → C ≠ D
19 18 neneqd ⊢ φ ∧ C = D ∧ A ∈ C L D → ¬ C = D
20 12 19 pm2.65da ⊢ φ ∧ C = D → ¬ A ∈ C L D
21 nelne1 ⊢ A ∈ A L B ∧ ¬ A ∈ C L D → A L B ≠ C L D
22 11 20 21 syl2an2r ⊢ φ ∧ C = D → A L B ≠ C L D
23 4 ad2antrr ⊢ φ ∧ C ≠ D ∧ A L B = C L D → G ∈ 𝒢 Tarski
24 6 ad2antrr ⊢ φ ∧ C ≠ D ∧ A L B = C L D → B ∈ P
25 7 ad2antrr ⊢ φ ∧ C ≠ D ∧ A L B = C L D → C ∈ P
26 5 ad2antrr ⊢ φ ∧ C ≠ D ∧ A L B = C L D → A ∈ P
27 pm2.46 ⊢ ¬ A ∈ B L C ∨ B = C → ¬ B = C
28 9 27 syl ⊢ φ → ¬ B = C
29 28 neqned ⊢ φ → B ≠ C
30 29 ad2antrr ⊢ φ ∧ C ≠ D ∧ A L B = C L D → B ≠ C
31 8 ad2antrr ⊢ φ ∧ C ≠ D ∧ A L B = C L D → D ∈ P
32 simplr ⊢ φ ∧ C ≠ D ∧ A L B = C L D → C ≠ D
33 1 2 3 23 25 31 32 tglinerflx1 ⊢ φ ∧ C ≠ D ∧ A L B = C L D → C ∈ C L D
34 simpr ⊢ φ ∧ C ≠ D ∧ A L B = C L D → A L B = C L D
35 33 34 eleqtrrd ⊢ φ ∧ C ≠ D ∧ A L B = C L D → C ∈ A L B
36 1 3 2 23 26 24 35 tglngne ⊢ φ ∧ C ≠ D ∧ A L B = C L D → A ≠ B
37 1 2 3 23 24 25 26 30 35 36 lnrot1 ⊢ φ ∧ C ≠ D ∧ A L B = C L D → A ∈ B L C
38 37 orcd ⊢ φ ∧ C ≠ D ∧ A L B = C L D → A ∈ B L C ∨ B = C
39 9 ad2antrr ⊢ φ ∧ C ≠ D ∧ A L B = C L D → ¬ A ∈ B L C ∨ B = C
40 38 39 pm2.65da ⊢ φ ∧ C ≠ D → ¬ A L B = C L D
41 40 neqned ⊢ φ ∧ C ≠ D → A L B ≠ C L D
42 22 41 pm2.61dane ⊢ φ → A L B ≠ C L D