Metamath Proof Explorer


Theorem oppmir

Description: The mirror point with regard to a point X on a line A lies on the other side of A . (Contributed by Thierry Arnoux, 5-Jul-2026)

Ref Expression
Hypotheses oppmir.p ⊢ P = Base G
oppmir.i ⊢ I = Itv ⁡ G
oppmir.s ⊢ S = pInv 𝒢 ⁡ G
oppmir.m ⊢ M = S ⁡ X
oppmir.o ⊢ O = a b | a ∈ P ∖ A ∧ b ∈ P ∖ A ∧ ∃ t ∈ A t ∈ a I b
oppmir.g ⊢ φ → G ∈ 𝒢 Tarski
oppmir.a ⊢ φ → A ∈ ran ⁡ L
oppmir.x ⊢ φ → X ∈ A
oppmir.y ⊢ φ → Y ∈ P ∖ A
oppmir.1 ⊢ L = Line 𝒢 ⁡ G
Assertion oppmir ⊢ φ → Y O M ⁡ Y

Proof

Step Hyp Ref Expression
1 oppmir.p ⊢ P = Base G
2 oppmir.i ⊢ I = Itv ⁡ G
3 oppmir.s ⊢ S = pInv 𝒢 ⁡ G
4 oppmir.m ⊢ M = S ⁡ X
5 oppmir.o ⊢ O = a b | a ∈ P ∖ A ∧ b ∈ P ∖ A ∧ ∃ t ∈ A t ∈ a I b
6 oppmir.g ⊢ φ → G ∈ 𝒢 Tarski
7 oppmir.a ⊢ φ → A ∈ ran ⁡ L
8 oppmir.x ⊢ φ → X ∈ A
9 oppmir.y ⊢ φ → Y ∈ P ∖ A
10 oppmir.1 ⊢ L = Line 𝒢 ⁡ G
11 eqid ⊢ dist ⁡ G = dist ⁡ G
12 9 eldifad ⊢ φ → Y ∈ P
13 1 10 2 6 7 8 tglnpt ⊢ φ → X ∈ P
14 1 11 2 10 3 6 13 4 12 mircl ⊢ φ → M ⁡ Y ∈ P
15 9 eldifbd ⊢ φ → ¬ Y ∈ A
16 1 11 2 10 3 6 13 4 12 mirmir ⊢ φ → M ⁡ M ⁡ Y = Y
17 16 adantr ⊢ φ ∧ M ⁡ Y ∈ A → M ⁡ M ⁡ Y = Y
18 6 adantr ⊢ φ ∧ M ⁡ Y ∈ A → G ∈ 𝒢 Tarski
19 7 adantr ⊢ φ ∧ M ⁡ Y ∈ A → A ∈ ran ⁡ L
20 8 adantr ⊢ φ ∧ M ⁡ Y ∈ A → X ∈ A
21 simpr ⊢ φ ∧ M ⁡ Y ∈ A → M ⁡ Y ∈ A
22 1 11 2 10 3 18 4 19 20 21 mirln ⊢ φ ∧ M ⁡ Y ∈ A → M ⁡ M ⁡ Y ∈ A
23 17 22 eqeltrrd ⊢ φ ∧ M ⁡ Y ∈ A → Y ∈ A
24 15 23 mtand ⊢ φ → ¬ M ⁡ Y ∈ A
25 1 11 2 10 3 6 13 4 12 mirbtwn ⊢ φ → X ∈ M ⁡ Y I Y
26 1 11 2 6 14 13 12 25 tgbtwncom ⊢ φ → X ∈ Y I M ⁡ Y
27 1 11 2 5 12 14 8 15 24 26 islnoppd ⊢ φ → Y O M ⁡ Y