Metamath Proof Explorer


Theorem mirln2

Description: If a point and its mirror point are both on the same line, so is the center of the point inversion. (Contributed by Thierry Arnoux, 3-Mar-2020)

Ref Expression
Hypotheses mirval.p ⊢ P = Base G
mirval.d ⊢ - ˙ = dist ⁡ G
mirval.i ⊢ I = Itv ⁡ G
mirval.l ⊢ L = Line 𝒢 ⁡ G
mirval.s ⊢ S = pInv 𝒢 ⁡ G
mirval.g ⊢ φ → G ∈ 𝒢 Tarski
mirln2.m ⊢ M = S ⁡ A
mirln2.d ⊢ φ → D ∈ ran ⁡ L
mirln2.a ⊢ φ → A ∈ P
mirln2.1 ⊢ φ → B ∈ D
mirln2.2 ⊢ φ → M ⁡ B ∈ D
Assertion mirln2 ⊢ φ → A ∈ D

Proof

Step Hyp Ref Expression
1 mirval.p ⊢ P = Base G
2 mirval.d ⊢ - ˙ = dist ⁡ G
3 mirval.i ⊢ I = Itv ⁡ G
4 mirval.l ⊢ L = Line 𝒢 ⁡ G
5 mirval.s ⊢ S = pInv 𝒢 ⁡ G
6 mirval.g ⊢ φ → G ∈ 𝒢 Tarski
7 mirln2.m ⊢ M = S ⁡ A
8 mirln2.d ⊢ φ → D ∈ ran ⁡ L
9 mirln2.a ⊢ φ → A ∈ P
10 mirln2.1 ⊢ φ → B ∈ D
11 mirln2.2 ⊢ φ → M ⁡ B ∈ D
12 1 4 3 6 8 10 tglnpt ⊢ φ → B ∈ P
13 1 2 3 4 5 6 9 7 12 mirinv ⊢ φ → M ⁡ B = B ↔ A = B
14 13 biimpa ⊢ φ ∧ M ⁡ B = B → A = B
15 10 adantr ⊢ φ ∧ M ⁡ B = B → B ∈ D
16 14 15 eqeltrd ⊢ φ ∧ M ⁡ B = B → A ∈ D
17 6 adantr ⊢ φ ∧ M ⁡ B ≠ B → G ∈ 𝒢 Tarski
18 1 4 3 6 8 11 tglnpt ⊢ φ → M ⁡ B ∈ P
19 18 adantr ⊢ φ ∧ M ⁡ B ≠ B → M ⁡ B ∈ P
20 12 adantr ⊢ φ ∧ M ⁡ B ≠ B → B ∈ P
21 9 adantr ⊢ φ ∧ M ⁡ B ≠ B → A ∈ P
22 simpr ⊢ φ ∧ M ⁡ B ≠ B → M ⁡ B ≠ B
23 1 2 3 4 5 17 21 7 20 mirbtwn ⊢ φ ∧ M ⁡ B ≠ B → A ∈ M ⁡ B I B
24 1 3 4 17 19 20 21 22 23 btwnlng1 ⊢ φ ∧ M ⁡ B ≠ B → A ∈ M ⁡ B L B
25 8 adantr ⊢ φ ∧ M ⁡ B ≠ B → D ∈ ran ⁡ L
26 11 adantr ⊢ φ ∧ M ⁡ B ≠ B → M ⁡ B ∈ D
27 10 adantr ⊢ φ ∧ M ⁡ B ≠ B → B ∈ D
28 1 3 4 17 19 20 22 22 25 26 27 tglinethru ⊢ φ ∧ M ⁡ B ≠ B → D = M ⁡ B L B
29 24 28 eleqtrrd ⊢ φ ∧ M ⁡ B ≠ B → A ∈ D
30 16 29 pm2.61dane ⊢ φ → A ∈ D