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Theorem List for Metamath Proof Explorer - 26601-26700   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
15.4.1  Geometry in the complex plane
 
Theoremcchhllem 26601* Lemma for chlbas and chlvsca . (Contributed by Thierry Arnoux, 15-Apr-2019.)
𝐶 = (((subringAlg ‘ℂfld)‘ℝ) sSet ⟨(·𝑖‘ndx), (𝑥 ∈ ℂ, 𝑦 ∈ ℂ ↦ (𝑥 · (∗‘𝑦)))⟩)    &   𝐸 = Slot 𝑁    &   𝑁 ∈ ℕ    &   (𝑁 < 5 ∨ 8 < 𝑁)       (𝐸‘ℂfld) = (𝐸𝐶)
 
15.4.2  Geometry in Euclidean spaces
 
15.4.2.1  Definition of the Euclidean space
 
Syntaxcee 26602 Declare the syntax for the Euclidean space generator.
class 𝔼
 
Syntaxcbtwn 26603 Declare the syntax for the Euclidean betweenness predicate.
class Btwn
 
Syntaxccgr 26604 Declare the syntax for the Euclidean congruence predicate.
class Cgr
 
Definitiondf-ee 26605 Define the Euclidean space generator. For details, see elee 26608. (Contributed by Scott Fenton, 3-Jun-2013.)
𝔼 = (𝑛 ∈ ℕ ↦ (ℝ ↑m (1...𝑛)))
 
Definitiondf-btwn 26606* Define the Euclidean betweenness predicate. For details, see brbtwn 26613. (Contributed by Scott Fenton, 3-Jun-2013.)
Btwn = {⟨⟨𝑥, 𝑧⟩, 𝑦⟩ ∣ ∃𝑛 ∈ ℕ ((𝑥 ∈ (𝔼‘𝑛) ∧ 𝑧 ∈ (𝔼‘𝑛) ∧ 𝑦 ∈ (𝔼‘𝑛)) ∧ ∃𝑡 ∈ (0[,]1)∀𝑖 ∈ (1...𝑛)(𝑦𝑖) = (((1 − 𝑡) · (𝑥𝑖)) + (𝑡 · (𝑧𝑖))))}
 
Definitiondf-cgr 26607* Define the Euclidean congruence predicate. For details, see brcgr 26614. (Contributed by Scott Fenton, 3-Jun-2013.)
Cgr = {⟨𝑥, 𝑦⟩ ∣ ∃𝑛 ∈ ℕ ((𝑥 ∈ ((𝔼‘𝑛) × (𝔼‘𝑛)) ∧ 𝑦 ∈ ((𝔼‘𝑛) × (𝔼‘𝑛))) ∧ Σ𝑖 ∈ (1...𝑛)((((1st𝑥)‘𝑖) − ((2nd𝑥)‘𝑖))↑2) = Σ𝑖 ∈ (1...𝑛)((((1st𝑦)‘𝑖) − ((2nd𝑦)‘𝑖))↑2))}
 
Theoremelee 26608 Membership in a Euclidean space. We define Euclidean space here using Cartesian coordinates over 𝑁 space. We later abstract away from this using Tarski's geometry axioms, so this exact definition is unimportant. (Contributed by Scott Fenton, 3-Jun-2013.)
(𝑁 ∈ ℕ → (𝐴 ∈ (𝔼‘𝑁) ↔ 𝐴:(1...𝑁)⟶ℝ))
 
Theoremmptelee 26609* A condition for a mapping to be an element of a Euclidean space. (Contributed by Scott Fenton, 7-Jun-2013.)
(𝑁 ∈ ℕ → ((𝑘 ∈ (1...𝑁) ↦ (𝐴𝐹𝐵)) ∈ (𝔼‘𝑁) ↔ ∀𝑘 ∈ (1...𝑁)(𝐴𝐹𝐵) ∈ ℝ))
 
Theoremeleenn 26610 If 𝐴 is in (𝔼‘𝑁), then 𝑁 is a natural. (Contributed by Scott Fenton, 1-Jul-2013.)
(𝐴 ∈ (𝔼‘𝑁) → 𝑁 ∈ ℕ)
 
Theoremeleei 26611 The forward direction of elee 26608. (Contributed by Scott Fenton, 1-Jul-2013.)
(𝐴 ∈ (𝔼‘𝑁) → 𝐴:(1...𝑁)⟶ℝ)
 
Theoremeedimeq 26612 A point belongs to at most one Euclidean space. (Contributed by Scott Fenton, 1-Jul-2013.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐴 ∈ (𝔼‘𝑀)) → 𝑁 = 𝑀)
 
Theorembrbtwn 26613* The binary relation form of the betweenness predicate. The statement 𝐴 Btwn ⟨𝐵, 𝐶 should be informally read as "𝐴 lies on a line segment between 𝐵 and 𝐶. This exact definition is abstracted away by Tarski's geometry axioms later on. (Contributed by Scott Fenton, 3-Jun-2013.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) → (𝐴 Btwn ⟨𝐵, 𝐶⟩ ↔ ∃𝑡 ∈ (0[,]1)∀𝑖 ∈ (1...𝑁)(𝐴𝑖) = (((1 − 𝑡) · (𝐵𝑖)) + (𝑡 · (𝐶𝑖)))))
 
Theorembrcgr 26614* The binary relation form of the congruence predicate. The statement 𝐴, 𝐵⟩Cgr⟨𝐶, 𝐷 should be read informally as "the 𝑁 dimensional point 𝐴 is as far from 𝐵 as 𝐶 is from 𝐷, or "the line segment 𝐴𝐵 is congruent to the line segment 𝐶𝐷. This particular definition is encapsulated by Tarski's axioms later on. (Contributed by Scott Fenton, 3-Jun-2013.)
(((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝐷 ∈ (𝔼‘𝑁))) → (⟨𝐴, 𝐵⟩Cgr⟨𝐶, 𝐷⟩ ↔ Σ𝑖 ∈ (1...𝑁)(((𝐴𝑖) − (𝐵𝑖))↑2) = Σ𝑖 ∈ (1...𝑁)(((𝐶𝑖) − (𝐷𝑖))↑2)))
 
Theoremfveere 26615 The function value of a point is a real. (Contributed by Scott Fenton, 10-Jun-2013.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐼 ∈ (1...𝑁)) → (𝐴𝐼) ∈ ℝ)
 
Theoremfveecn 26616 The function value of a point is a complex. (Contributed by Scott Fenton, 10-Jun-2013.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐼 ∈ (1...𝑁)) → (𝐴𝐼) ∈ ℂ)
 
Theoremeqeefv 26617* Two points are equal iff they agree in all dimensions. (Contributed by Scott Fenton, 10-Jun-2013.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → (𝐴 = 𝐵 ↔ ∀𝑖 ∈ (1...𝑁)(𝐴𝑖) = (𝐵𝑖)))
 
Theoremeqeelen 26618* Two points are equal iff the square of the distance between them is zero. (Contributed by Scott Fenton, 10-Jun-2013.) (Revised by Mario Carneiro, 22-May-2014.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → (𝐴 = 𝐵 ↔ Σ𝑖 ∈ (1...𝑁)(((𝐴𝑖) − (𝐵𝑖))↑2) = 0))
 
Theorembrbtwn2 26619* Alternate characterization of betweenness, with no existential quantifiers. (Contributed by Scott Fenton, 24-Jun-2013.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) → (𝐴 Btwn ⟨𝐵, 𝐶⟩ ↔ (∀𝑖 ∈ (1...𝑁)(((𝐵𝑖) − (𝐴𝑖)) · ((𝐶𝑖) − (𝐴𝑖))) ≤ 0 ∧ ∀𝑖 ∈ (1...𝑁)∀𝑗 ∈ (1...𝑁)(((𝐵𝑖) − (𝐴𝑖)) · ((𝐶𝑗) − (𝐴𝑗))) = (((𝐵𝑗) − (𝐴𝑗)) · ((𝐶𝑖) − (𝐴𝑖))))))
 
Theoremcolinearalglem1 26620 Lemma for colinearalg 26624. Expand out a multiplication. (Contributed by Scott Fenton, 24-Jun-2013.)
(((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐷 ∈ ℂ ∧ 𝐸 ∈ ℂ ∧ 𝐹 ∈ ℂ)) → (((𝐵𝐴) · (𝐹𝐷)) = ((𝐸𝐷) · (𝐶𝐴)) ↔ ((𝐵 · 𝐹) − ((𝐴 · 𝐹) + (𝐵 · 𝐷))) = ((𝐶 · 𝐸) − ((𝐴 · 𝐸) + (𝐶 · 𝐷)))))
 
Theoremcolinearalglem2 26621* Lemma for colinearalg 26624. Translate between two forms of the colinearity condition. (Contributed by Scott Fenton, 24-Jun-2013.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) → (∀𝑖 ∈ (1...𝑁)∀𝑗 ∈ (1...𝑁)(((𝐵𝑖) − (𝐴𝑖)) · ((𝐶𝑗) − (𝐴𝑗))) = (((𝐵𝑗) − (𝐴𝑗)) · ((𝐶𝑖) − (𝐴𝑖))) ↔ ∀𝑖 ∈ (1...𝑁)∀𝑗 ∈ (1...𝑁)(((𝐶𝑖) − (𝐵𝑖)) · ((𝐴𝑗) − (𝐵𝑗))) = (((𝐶𝑗) − (𝐵𝑗)) · ((𝐴𝑖) − (𝐵𝑖)))))
 
Theoremcolinearalglem3 26622* Lemma for colinearalg 26624. Translate between two forms of the colinearity condition. (Contributed by Scott Fenton, 24-Jun-2013.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) → (∀𝑖 ∈ (1...𝑁)∀𝑗 ∈ (1...𝑁)(((𝐵𝑖) − (𝐴𝑖)) · ((𝐶𝑗) − (𝐴𝑗))) = (((𝐵𝑗) − (𝐴𝑗)) · ((𝐶𝑖) − (𝐴𝑖))) ↔ ∀𝑖 ∈ (1...𝑁)∀𝑗 ∈ (1...𝑁)(((𝐴𝑖) − (𝐶𝑖)) · ((𝐵𝑗) − (𝐶𝑗))) = (((𝐴𝑗) − (𝐶𝑗)) · ((𝐵𝑖) − (𝐶𝑖)))))
 
Theoremcolinearalglem4 26623* Lemma for colinearalg 26624. Prove a disjunction that will be needed in the final proof. (Contributed by Scott Fenton, 27-Jun-2013.)
(((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) ∧ 𝐾 ∈ ℝ) → (∀𝑖 ∈ (1...𝑁)((((𝐾 · ((𝐶𝑖) − (𝐴𝑖))) + (𝐴𝑖)) − (𝐴𝑖)) · ((𝐶𝑖) − (𝐴𝑖))) ≤ 0 ∨ ∀𝑖 ∈ (1...𝑁)(((𝐶𝑖) − ((𝐾 · ((𝐶𝑖) − (𝐴𝑖))) + (𝐴𝑖))) · ((𝐴𝑖) − ((𝐾 · ((𝐶𝑖) − (𝐴𝑖))) + (𝐴𝑖)))) ≤ 0 ∨ ∀𝑖 ∈ (1...𝑁)(((𝐴𝑖) − (𝐶𝑖)) · (((𝐾 · ((𝐶𝑖) − (𝐴𝑖))) + (𝐴𝑖)) − (𝐶𝑖))) ≤ 0))
 
Theoremcolinearalg 26624* An algebraic characterization of colinearity. Note the similarity to brbtwn2 26619. (Contributed by Scott Fenton, 24-Jun-2013.)
((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) → ((𝐴 Btwn ⟨𝐵, 𝐶⟩ ∨ 𝐵 Btwn ⟨𝐶, 𝐴⟩ ∨ 𝐶 Btwn ⟨𝐴, 𝐵⟩) ↔ ∀𝑖 ∈ (1...𝑁)∀𝑗 ∈ (1...𝑁)(((𝐵𝑖) − (𝐴𝑖)) · ((𝐶𝑗) − (𝐴𝑗))) = (((𝐵𝑗) − (𝐴𝑗)) · ((𝐶𝑖) − (𝐴𝑖)))))
 
Theoremeleesub 26625* Membership of a subtraction mapping in a Euclidean space. (Contributed by Scott Fenton, 17-Jul-2013.)
𝐶 = (𝑖 ∈ (1...𝑁) ↦ ((𝐴𝑖) − (𝐵𝑖)))       ((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → 𝐶 ∈ (𝔼‘𝑁))
 
Theoremeleesubd 26626* Membership of a subtraction mapping in a Euclidean space. Deduction form of eleesub 26625. (Contributed by Scott Fenton, 17-Jul-2013.)
(𝜑𝐶 = (𝑖 ∈ (1...𝑁) ↦ ((𝐴𝑖) − (𝐵𝑖))))       ((𝜑𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → 𝐶 ∈ (𝔼‘𝑁))
 
15.4.2.2  Tarski's axioms for geometry for the Euclidean space
 
Theoremaxdimuniq 26627 The unique dimension axiom. If a point is in 𝑁 dimensional space and in 𝑀 dimensional space, then 𝑁 = 𝑀. This axiom is not traditionally presented with Tarski's axioms, but we require it here as we are considering spaces in arbitrary dimensions. (Contributed by Scott Fenton, 24-Sep-2013.)
(((𝑁 ∈ ℕ ∧ 𝐴 ∈ (𝔼‘𝑁)) ∧ (𝑀 ∈ ℕ ∧ 𝐴 ∈ (𝔼‘𝑀))) → 𝑁 = 𝑀)
 
Theoremaxcgrrflx 26628 𝐴 is as far from 𝐵 as 𝐵 is from 𝐴. Axiom A1 of [Schwabhauser] p. 10. (Contributed by Scott Fenton, 3-Jun-2013.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → ⟨𝐴, 𝐵⟩Cgr⟨𝐵, 𝐴⟩)
 
Theoremaxcgrtr 26629 Congruence is transitive. Axiom A2 of [Schwabhauser] p. 10. (Contributed by Scott Fenton, 3-Jun-2013.)
((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) ∧ (𝐷 ∈ (𝔼‘𝑁) ∧ 𝐸 ∈ (𝔼‘𝑁) ∧ 𝐹 ∈ (𝔼‘𝑁))) → ((⟨𝐴, 𝐵⟩Cgr⟨𝐶, 𝐷⟩ ∧ ⟨𝐴, 𝐵⟩Cgr⟨𝐸, 𝐹⟩) → ⟨𝐶, 𝐷⟩Cgr⟨𝐸, 𝐹⟩))
 
Theoremaxcgrid 26630 If there is no distance between 𝐴 and 𝐵, then 𝐴 = 𝐵. Axiom A3 of [Schwabhauser] p. 10. (Contributed by Scott Fenton, 3-Jun-2013.)
((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁))) → (⟨𝐴, 𝐵⟩Cgr⟨𝐶, 𝐶⟩ → 𝐴 = 𝐵))
 
Theoremaxsegconlem1 26631* Lemma for axsegcon 26641. Handle the degenerate case. (Contributed by Scott Fenton, 7-Jun-2013.)
((𝐴 = 𝐵 ∧ ((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝐷 ∈ (𝔼‘𝑁)))) → ∃𝑥 ∈ (𝔼‘𝑁)∃𝑡 ∈ (0[,]1)(∀𝑖 ∈ (1...𝑁)(𝐵𝑖) = (((1 − 𝑡) · (𝐴𝑖)) + (𝑡 · (𝑥𝑖))) ∧ Σ𝑖 ∈ (1...𝑁)(((𝐵𝑖) − (𝑥𝑖))↑2) = Σ𝑖 ∈ (1...𝑁)(((𝐶𝑖) − (𝐷𝑖))↑2)))
 
Theoremaxsegconlem2 26632* Lemma for axsegcon 26641. Show that the square of the distance between two points is a real number. (Contributed by Scott Fenton, 17-Sep-2013.)
𝑆 = Σ𝑝 ∈ (1...𝑁)(((𝐴𝑝) − (𝐵𝑝))↑2)       ((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → 𝑆 ∈ ℝ)
 
Theoremaxsegconlem3 26633* Lemma for axsegcon 26641. Show that the square of the distance between two points is nonnegative. (Contributed by Scott Fenton, 17-Sep-2013.)
𝑆 = Σ𝑝 ∈ (1...𝑁)(((𝐴𝑝) − (𝐵𝑝))↑2)       ((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → 0 ≤ 𝑆)
 
Theoremaxsegconlem4 26634* Lemma for axsegcon 26641. Show that the distance between two points is a real number. (Contributed by Scott Fenton, 17-Sep-2013.)
𝑆 = Σ𝑝 ∈ (1...𝑁)(((𝐴𝑝) − (𝐵𝑝))↑2)       ((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → (√‘𝑆) ∈ ℝ)
 
Theoremaxsegconlem5 26635* Lemma for axsegcon 26641. Show that the distance between two points is nonnegative. (Contributed by Scott Fenton, 17-Sep-2013.)
𝑆 = Σ𝑝 ∈ (1...𝑁)(((𝐴𝑝) − (𝐵𝑝))↑2)       ((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → 0 ≤ (√‘𝑆))
 
Theoremaxsegconlem6 26636* Lemma for axsegcon 26641. Show that the distance between two distinct points is positive. (Contributed by Scott Fenton, 17-Sep-2013.)
𝑆 = Σ𝑝 ∈ (1...𝑁)(((𝐴𝑝) − (𝐵𝑝))↑2)       ((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐴𝐵) → 0 < (√‘𝑆))
 
Theoremaxsegconlem7 26637* Lemma for axsegcon 26641. Show that a particular ratio of distances is in the closed unit interval. (Contributed by Scott Fenton, 18-Sep-2013.)
𝑆 = Σ𝑝 ∈ (1...𝑁)(((𝐴𝑝) − (𝐵𝑝))↑2)    &   𝑇 = Σ𝑝 ∈ (1...𝑁)(((𝐶𝑝) − (𝐷𝑝))↑2)       (((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐴𝐵) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝐷 ∈ (𝔼‘𝑁))) → ((√‘𝑆) / ((√‘𝑆) + (√‘𝑇))) ∈ (0[,]1))
 
Theoremaxsegconlem8 26638* Lemma for axsegcon 26641. Show that a particular mapping generates a point. (Contributed by Scott Fenton, 18-Sep-2013.)
𝑆 = Σ𝑝 ∈ (1...𝑁)(((𝐴𝑝) − (𝐵𝑝))↑2)    &   𝑇 = Σ𝑝 ∈ (1...𝑁)(((𝐶𝑝) − (𝐷𝑝))↑2)    &   𝐹 = (𝑘 ∈ (1...𝑁) ↦ (((((√‘𝑆) + (√‘𝑇)) · (𝐵𝑘)) − ((√‘𝑇) · (𝐴𝑘))) / (√‘𝑆)))       (((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐴𝐵) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝐷 ∈ (𝔼‘𝑁))) → 𝐹 ∈ (𝔼‘𝑁))
 
Theoremaxsegconlem9 26639* Lemma for axsegcon 26641. Show that 𝐵𝐹 is congruent to 𝐶𝐷. (Contributed by Scott Fenton, 19-Sep-2013.)
𝑆 = Σ𝑝 ∈ (1...𝑁)(((𝐴𝑝) − (𝐵𝑝))↑2)    &   𝑇 = Σ𝑝 ∈ (1...𝑁)(((𝐶𝑝) − (𝐷𝑝))↑2)    &   𝐹 = (𝑘 ∈ (1...𝑁) ↦ (((((√‘𝑆) + (√‘𝑇)) · (𝐵𝑘)) − ((√‘𝑇) · (𝐴𝑘))) / (√‘𝑆)))       (((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐴𝐵) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝐷 ∈ (𝔼‘𝑁))) → Σ𝑖 ∈ (1...𝑁)(((𝐵𝑖) − (𝐹𝑖))↑2) = Σ𝑖 ∈ (1...𝑁)(((𝐶𝑖) − (𝐷𝑖))↑2))
 
Theoremaxsegconlem10 26640* Lemma for axsegcon 26641. Show that the scaling constant from axsegconlem7 26637 produces the betweenness condition for 𝐴, 𝐵 and 𝐹. (Contributed by Scott Fenton, 21-Sep-2013.)
𝑆 = Σ𝑝 ∈ (1...𝑁)(((𝐴𝑝) − (𝐵𝑝))↑2)    &   𝑇 = Σ𝑝 ∈ (1...𝑁)(((𝐶𝑝) − (𝐷𝑝))↑2)    &   𝐹 = (𝑘 ∈ (1...𝑁) ↦ (((((√‘𝑆) + (√‘𝑇)) · (𝐵𝑘)) − ((√‘𝑇) · (𝐴𝑘))) / (√‘𝑆)))       (((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐴𝐵) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝐷 ∈ (𝔼‘𝑁))) → ∀𝑖 ∈ (1...𝑁)(𝐵𝑖) = (((1 − ((√‘𝑆) / ((√‘𝑆) + (√‘𝑇)))) · (𝐴𝑖)) + (((√‘𝑆) / ((√‘𝑆) + (√‘𝑇))) · (𝐹𝑖))))
 
Theoremaxsegcon 26641* Any segment 𝐴𝐵 can be extended to a point 𝑥 such that 𝐵𝑥 is congruent to 𝐶𝐷. Axiom A4 of [Schwabhauser] p. 11. (Contributed by Scott Fenton, 4-Jun-2013.)
((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝐷 ∈ (𝔼‘𝑁))) → ∃𝑥 ∈ (𝔼‘𝑁)(𝐵 Btwn ⟨𝐴, 𝑥⟩ ∧ ⟨𝐵, 𝑥⟩Cgr⟨𝐶, 𝐷⟩))
 
Theoremax5seglem1 26642* Lemma for ax5seg 26652. Rexpress a one congruence sum given betweenness. (Contributed by Scott Fenton, 11-Jun-2013.)
((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) ∧ (𝑇 ∈ (0[,]1) ∧ ∀𝑖 ∈ (1...𝑁)(𝐵𝑖) = (((1 − 𝑇) · (𝐴𝑖)) + (𝑇 · (𝐶𝑖))))) → Σ𝑗 ∈ (1...𝑁)(((𝐴𝑗) − (𝐵𝑗))↑2) = ((𝑇↑2) · Σ𝑗 ∈ (1...𝑁)(((𝐴𝑗) − (𝐶𝑗))↑2)))
 
Theoremax5seglem2 26643* Lemma for ax5seg 26652. Rexpress another congruence sum given betweenness. (Contributed by Scott Fenton, 11-Jun-2013.)
((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) ∧ (𝑇 ∈ (0[,]1) ∧ ∀𝑖 ∈ (1...𝑁)(𝐵𝑖) = (((1 − 𝑇) · (𝐴𝑖)) + (𝑇 · (𝐶𝑖))))) → Σ𝑗 ∈ (1...𝑁)(((𝐵𝑗) − (𝐶𝑗))↑2) = (((1 − 𝑇)↑2) · Σ𝑗 ∈ (1...𝑁)(((𝐴𝑗) − (𝐶𝑗))↑2)))
 
Theoremax5seglem3a 26644 Lemma for ax5seg 26652. (Contributed by Scott Fenton, 7-May-2015.)
(((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) ∧ (𝐷 ∈ (𝔼‘𝑁) ∧ 𝐸 ∈ (𝔼‘𝑁) ∧ 𝐹 ∈ (𝔼‘𝑁))) ∧ 𝑗 ∈ (1...𝑁)) → (((𝐴𝑗) − (𝐶𝑗)) ∈ ℝ ∧ ((𝐷𝑗) − (𝐹𝑗)) ∈ ℝ))
 
Theoremax5seglem3 26645* Lemma for ax5seg 26652. Combine congruences for points on a line. (Contributed by Scott Fenton, 11-Jun-2013.)
(((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) ∧ (𝐷 ∈ (𝔼‘𝑁) ∧ 𝐸 ∈ (𝔼‘𝑁) ∧ 𝐹 ∈ (𝔼‘𝑁))) ∧ ((𝑇 ∈ (0[,]1) ∧ 𝑆 ∈ (0[,]1)) ∧ (∀𝑖 ∈ (1...𝑁)(𝐵𝑖) = (((1 − 𝑇) · (𝐴𝑖)) + (𝑇 · (𝐶𝑖))) ∧ ∀𝑖 ∈ (1...𝑁)(𝐸𝑖) = (((1 − 𝑆) · (𝐷𝑖)) + (𝑆 · (𝐹𝑖))))) ∧ (⟨𝐴, 𝐵⟩Cgr⟨𝐷, 𝐸⟩ ∧ ⟨𝐵, 𝐶⟩Cgr⟨𝐸, 𝐹⟩)) → Σ𝑗 ∈ (1...𝑁)(((𝐴𝑗) − (𝐶𝑗))↑2) = Σ𝑗 ∈ (1...𝑁)(((𝐷𝑗) − (𝐹𝑗))↑2))
 
Theoremax5seglem4 26646* Lemma for ax5seg 26652. Given two distinct points, the scaling constant in a betweenness statement is nonzero. (Contributed by Scott Fenton, 11-Jun-2013.)
(((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁))) ∧ ∀𝑖 ∈ (1...𝑁)(𝐵𝑖) = (((1 − 𝑇) · (𝐴𝑖)) + (𝑇 · (𝐶𝑖))) ∧ 𝐴𝐵) → 𝑇 ≠ 0)
 
Theoremax5seglem5 26647* Lemma for ax5seg 26652. If 𝐵 is between 𝐴 and 𝐶, and 𝐴 is distinct from 𝐵, then 𝐴 is distinct from 𝐶. (Contributed by Scott Fenton, 11-Jun-2013.)
(((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁))) ∧ (𝐴𝐵𝑇 ∈ (0[,]1) ∧ ∀𝑖 ∈ (1...𝑁)(𝐵𝑖) = (((1 − 𝑇) · (𝐴𝑖)) + (𝑇 · (𝐶𝑖))))) → Σ𝑗 ∈ (1...𝑁)(((𝐴𝑗) − (𝐶𝑗))↑2) ≠ 0)
 
Theoremax5seglem6 26648* Lemma for ax5seg 26652. Given two line segments that are divided into pieces, if the pieces are congruent, then the scaling constant is the same. (Contributed by Scott Fenton, 12-Jun-2013.)
(((𝑁 ∈ ℕ ∧ ((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) ∧ (𝐷 ∈ (𝔼‘𝑁) ∧ 𝐸 ∈ (𝔼‘𝑁) ∧ 𝐹 ∈ (𝔼‘𝑁)))) ∧ (𝐴𝐵 ∧ (𝑇 ∈ (0[,]1) ∧ 𝑆 ∈ (0[,]1)) ∧ (∀𝑖 ∈ (1...𝑁)(𝐵𝑖) = (((1 − 𝑇) · (𝐴𝑖)) + (𝑇 · (𝐶𝑖))) ∧ ∀𝑖 ∈ (1...𝑁)(𝐸𝑖) = (((1 − 𝑆) · (𝐷𝑖)) + (𝑆 · (𝐹𝑖))))) ∧ (⟨𝐴, 𝐵⟩Cgr⟨𝐷, 𝐸⟩ ∧ ⟨𝐵, 𝐶⟩Cgr⟨𝐸, 𝐹⟩)) → 𝑇 = 𝑆)
 
Theoremax5seglem7 26649 Lemma for ax5seg 26652. An algebraic calculation needed further down the line. (Contributed by Scott Fenton, 12-Jun-2013.)
𝐴 ∈ ℂ    &   𝑇 ∈ ℂ    &   𝐶 ∈ ℂ    &   𝐷 ∈ ℂ       (𝑇 · ((𝐶𝐷)↑2)) = ((((((1 − 𝑇) · 𝐴) + (𝑇 · 𝐶)) − 𝐷)↑2) + ((1 − 𝑇) · ((𝑇 · ((𝐴𝐶)↑2)) − ((𝐴𝐷)↑2))))
 
Theoremax5seglem8 26650 Lemma for ax5seg 26652. Use the weak deduction theorem to eliminate the hypotheses from ax5seglem7 26649. (Contributed by Scott Fenton, 11-Jun-2013.)
(((𝐴 ∈ ℂ ∧ 𝑇 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝑇 · ((𝐶𝐷)↑2)) = ((((((1 − 𝑇) · 𝐴) + (𝑇 · 𝐶)) − 𝐷)↑2) + ((1 − 𝑇) · ((𝑇 · ((𝐴𝐶)↑2)) − ((𝐴𝐷)↑2)))))
 
Theoremax5seglem9 26651* Lemma for ax5seg 26652. Take the calculation in ax5seglem8 26650 and turn it into a series of measurements. (Contributed by Scott Fenton, 12-Jun-2013.) (Revised by Mario Carneiro, 22-May-2014.)
(((𝑁 ∈ ℕ ∧ ((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝐷 ∈ (𝔼‘𝑁)))) ∧ (𝑇 ∈ (0[,]1) ∧ ∀𝑖 ∈ (1...𝑁)(𝐵𝑖) = (((1 − 𝑇) · (𝐴𝑖)) + (𝑇 · (𝐶𝑖))))) → (𝑇 · Σ𝑗 ∈ (1...𝑁)(((𝐶𝑗) − (𝐷𝑗))↑2)) = (Σ𝑗 ∈ (1...𝑁)(((𝐵𝑗) − (𝐷𝑗))↑2) + ((1 − 𝑇) · ((𝑇 · Σ𝑗 ∈ (1...𝑁)(((𝐴𝑗) − (𝐶𝑗))↑2)) − Σ𝑗 ∈ (1...𝑁)(((𝐴𝑗) − (𝐷𝑗))↑2)))))
 
Theoremax5seg 26652 The five segment axiom. Take two triangles 𝐴𝐷𝐶 and 𝐸𝐻𝐺, a point 𝐵 on 𝐴𝐶, and a point 𝐹 on 𝐸𝐺. If all corresponding line segments except for 𝐶𝐷 and 𝐺𝐻 are congruent, then so are 𝐶𝐷 and 𝐺𝐻. Axiom A5 of [Schwabhauser] p. 11. (Contributed by Scott Fenton, 12-Jun-2013.)
(((𝑁 ∈ ℕ ∧ 𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝐷 ∈ (𝔼‘𝑁) ∧ 𝐸 ∈ (𝔼‘𝑁)) ∧ (𝐹 ∈ (𝔼‘𝑁) ∧ 𝐺 ∈ (𝔼‘𝑁) ∧ 𝐻 ∈ (𝔼‘𝑁))) → (((𝐴𝐵𝐵 Btwn ⟨𝐴, 𝐶⟩ ∧ 𝐹 Btwn ⟨𝐸, 𝐺⟩) ∧ (⟨𝐴, 𝐵⟩Cgr⟨𝐸, 𝐹⟩ ∧ ⟨𝐵, 𝐶⟩Cgr⟨𝐹, 𝐺⟩) ∧ (⟨𝐴, 𝐷⟩Cgr⟨𝐸, 𝐻⟩ ∧ ⟨𝐵, 𝐷⟩Cgr⟨𝐹, 𝐻⟩)) → ⟨𝐶, 𝐷⟩Cgr⟨𝐺, 𝐻⟩))
 
Theoremaxbtwnid 26653 Points are indivisible. That is, if 𝐴 lies between 𝐵 and 𝐵, then 𝐴 = 𝐵. Axiom A6 of [Schwabhauser] p. 11. (Contributed by Scott Fenton, 3-Jun-2013.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) → (𝐴 Btwn ⟨𝐵, 𝐵⟩ → 𝐴 = 𝐵))
 
Theoremaxpaschlem 26654* Lemma for axpasch 26655. Set up coefficents used in the proof. (Contributed by Scott Fenton, 5-Jun-2013.)
((𝑇 ∈ (0[,]1) ∧ 𝑆 ∈ (0[,]1)) → ∃𝑟 ∈ (0[,]1)∃𝑝 ∈ (0[,]1)(𝑝 = ((1 − 𝑟) · (1 − 𝑇)) ∧ 𝑟 = ((1 − 𝑝) · (1 − 𝑆)) ∧ ((1 − 𝑟) · 𝑇) = ((1 − 𝑝) · 𝑆)))
 
Theoremaxpasch 26655* The inner Pasch axiom. Take a triangle 𝐴𝐶𝐸, a point 𝐷 on 𝐴𝐶, and a point 𝐵 extending 𝐶𝐸. Then 𝐴𝐸 and 𝐷𝐵 intersect at some point 𝑥. Axiom A7 of [Schwabhauser] p. 12. (Contributed by Scott Fenton, 3-Jun-2013.)
((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) ∧ (𝐷 ∈ (𝔼‘𝑁) ∧ 𝐸 ∈ (𝔼‘𝑁))) → ((𝐷 Btwn ⟨𝐴, 𝐶⟩ ∧ 𝐸 Btwn ⟨𝐵, 𝐶⟩) → ∃𝑥 ∈ (𝔼‘𝑁)(𝑥 Btwn ⟨𝐷, 𝐵⟩ ∧ 𝑥 Btwn ⟨𝐸, 𝐴⟩)))
 
Theoremaxlowdimlem1 26656 Lemma for axlowdim 26675. Establish a particular constant function as a function. (Contributed by Scott Fenton, 29-Jun-2013.)
((3...𝑁) × {0}):(3...𝑁)⟶ℝ
 
Theoremaxlowdimlem2 26657 Lemma for axlowdim 26675. Show that two sets are disjoint. (Contributed by Scott Fenton, 29-Jun-2013.)
((1...2) ∩ (3...𝑁)) = ∅
 
Theoremaxlowdimlem3 26658 Lemma for axlowdim 26675. Set up a union property for an interval of integers. (Contributed by Scott Fenton, 29-Jun-2013.)
(𝑁 ∈ (ℤ‘2) → (1...𝑁) = ((1...2) ∪ (3...𝑁)))
 
Theoremaxlowdimlem4 26659 Lemma for axlowdim 26675. Set up a particular constant function. (Contributed by Scott Fenton, 17-Apr-2013.)
𝐴 ∈ ℝ    &   𝐵 ∈ ℝ       {⟨1, 𝐴⟩, ⟨2, 𝐵⟩}:(1...2)⟶ℝ
 
Theoremaxlowdimlem5 26660 Lemma for axlowdim 26675. Show that a particular union is a point in Euclidean space. (Contributed by Scott Fenton, 29-Jun-2013.)
𝐴 ∈ ℝ    &   𝐵 ∈ ℝ       (𝑁 ∈ (ℤ‘2) → ({⟨1, 𝐴⟩, ⟨2, 𝐵⟩} ∪ ((3...𝑁) × {0})) ∈ (𝔼‘𝑁))
 
Theoremaxlowdimlem6 26661 Lemma for axlowdim 26675. Show that three points are non-colinear. (Contributed by Scott Fenton, 29-Jun-2013.)
𝐴 = ({⟨1, 0⟩, ⟨2, 0⟩} ∪ ((3...𝑁) × {0}))    &   𝐵 = ({⟨1, 1⟩, ⟨2, 0⟩} ∪ ((3...𝑁) × {0}))    &   𝐶 = ({⟨1, 0⟩, ⟨2, 1⟩} ∪ ((3...𝑁) × {0}))       (𝑁 ∈ (ℤ‘2) → ¬ (𝐴 Btwn ⟨𝐵, 𝐶⟩ ∨ 𝐵 Btwn ⟨𝐶, 𝐴⟩ ∨ 𝐶 Btwn ⟨𝐴, 𝐵⟩))
 
Theoremaxlowdimlem7 26662 Lemma for axlowdim 26675. Set up a point in Euclidean space. (Contributed by Scott Fenton, 29-Jun-2013.)
𝑃 = ({⟨3, -1⟩} ∪ (((1...𝑁) ∖ {3}) × {0}))       (𝑁 ∈ (ℤ‘3) → 𝑃 ∈ (𝔼‘𝑁))
 
Theoremaxlowdimlem8 26663 Lemma for axlowdim 26675. Calculate the value of 𝑃 at three. (Contributed by Scott Fenton, 21-Apr-2013.)
𝑃 = ({⟨3, -1⟩} ∪ (((1...𝑁) ∖ {3}) × {0}))       (𝑃‘3) = -1
 
Theoremaxlowdimlem9 26664 Lemma for axlowdim 26675. Calculate the value of 𝑃 away from three. (Contributed by Scott Fenton, 21-Apr-2013.)
𝑃 = ({⟨3, -1⟩} ∪ (((1...𝑁) ∖ {3}) × {0}))       ((𝐾 ∈ (1...𝑁) ∧ 𝐾 ≠ 3) → (𝑃𝐾) = 0)
 
Theoremaxlowdimlem10 26665 Lemma for axlowdim 26675. Set up a family of points in Euclidean space. (Contributed by Scott Fenton, 21-Apr-2013.)
𝑄 = ({⟨(𝐼 + 1), 1⟩} ∪ (((1...𝑁) ∖ {(𝐼 + 1)}) × {0}))       ((𝑁 ∈ ℕ ∧ 𝐼 ∈ (1...(𝑁 − 1))) → 𝑄 ∈ (𝔼‘𝑁))
 
Theoremaxlowdimlem11 26666 Lemma for axlowdim 26675. Calculate the value of 𝑄 at its distinguished point. (Contributed by Scott Fenton, 21-Apr-2013.)
𝑄 = ({⟨(𝐼 + 1), 1⟩} ∪ (((1...𝑁) ∖ {(𝐼 + 1)}) × {0}))       (𝑄‘(𝐼 + 1)) = 1
 
Theoremaxlowdimlem12 26667 Lemma for axlowdim 26675. Calculate the value of 𝑄 away from its distinguished point. (Contributed by Scott Fenton, 21-Apr-2013.)
𝑄 = ({⟨(𝐼 + 1), 1⟩} ∪ (((1...𝑁) ∖ {(𝐼 + 1)}) × {0}))       ((𝐾 ∈ (1...𝑁) ∧ 𝐾 ≠ (𝐼 + 1)) → (𝑄𝐾) = 0)
 
Theoremaxlowdimlem13 26668 Lemma for axlowdim 26675. Establish that 𝑃 and 𝑄 are different points. (Contributed by Scott Fenton, 21-Apr-2013.)
𝑃 = ({⟨3, -1⟩} ∪ (((1...𝑁) ∖ {3}) × {0}))    &   𝑄 = ({⟨(𝐼 + 1), 1⟩} ∪ (((1...𝑁) ∖ {(𝐼 + 1)}) × {0}))       ((𝑁 ∈ ℕ ∧ 𝐼 ∈ (1...(𝑁 − 1))) → 𝑃𝑄)
 
Theoremaxlowdimlem14 26669 Lemma for axlowdim 26675. Take two possible 𝑄 from axlowdimlem10 26665. They are the same iff their distinguished values are the same. (Contributed by Scott Fenton, 21-Apr-2013.)
𝑄 = ({⟨(𝐼 + 1), 1⟩} ∪ (((1...𝑁) ∖ {(𝐼 + 1)}) × {0}))    &   𝑅 = ({⟨(𝐽 + 1), 1⟩} ∪ (((1...𝑁) ∖ {(𝐽 + 1)}) × {0}))       ((𝑁 ∈ ℕ ∧ 𝐼 ∈ (1...(𝑁 − 1)) ∧ 𝐽 ∈ (1...(𝑁 − 1))) → (𝑄 = 𝑅𝐼 = 𝐽))
 
Theoremaxlowdimlem15 26670* Lemma for axlowdim 26675. Set up a one-to-one function of points. (Contributed by Scott Fenton, 21-Apr-2013.)
𝐹 = (𝑖 ∈ (1...(𝑁 − 1)) ↦ if(𝑖 = 1, ({⟨3, -1⟩} ∪ (((1...𝑁) ∖ {3}) × {0})), ({⟨(𝑖 + 1), 1⟩} ∪ (((1...𝑁) ∖ {(𝑖 + 1)}) × {0}))))       (𝑁 ∈ (ℤ‘3) → 𝐹:(1...(𝑁 − 1))–1-1→(𝔼‘𝑁))
 
Theoremaxlowdimlem16 26671* Lemma for axlowdim 26675. Set up a summation that will help establish distance. (Contributed by Scott Fenton, 21-Apr-2013.)
𝑃 = ({⟨3, -1⟩} ∪ (((1...𝑁) ∖ {3}) × {0}))    &   𝑄 = ({⟨(𝐼 + 1), 1⟩} ∪ (((1...𝑁) ∖ {(𝐼 + 1)}) × {0}))       ((𝑁 ∈ (ℤ‘3) ∧ 𝐼 ∈ (2...(𝑁 − 1))) → Σ𝑖 ∈ (3...𝑁)((𝑃𝑖)↑2) = Σ𝑖 ∈ (3...𝑁)((𝑄𝑖)↑2))
 
Theoremaxlowdimlem17 26672 Lemma for axlowdim 26675. Establish a congruence result. (Contributed by Scott Fenton, 22-Apr-2013.) (Proof shortened by Mario Carneiro, 22-May-2014.)
𝑃 = ({⟨3, -1⟩} ∪ (((1...𝑁) ∖ {3}) × {0}))    &   𝑄 = ({⟨(𝐼 + 1), 1⟩} ∪ (((1...𝑁) ∖ {(𝐼 + 1)}) × {0}))    &   𝐴 = ({⟨1, 𝑋⟩, ⟨2, 𝑌⟩} ∪ ((3...𝑁) × {0}))    &   𝑋 ∈ ℝ    &   𝑌 ∈ ℝ       ((𝑁 ∈ (ℤ‘3) ∧ 𝐼 ∈ (2...(𝑁 − 1))) → ⟨𝑃, 𝐴⟩Cgr⟨𝑄, 𝐴⟩)
 
Theoremaxlowdim1 26673* The lower dimension axiom for one dimension. In any dimension, there are at least two distinct points. Theorem 3.13 of [Schwabhauser] p. 32, where it is derived from axlowdim2 26674. (Contributed by Scott Fenton, 22-Apr-2013.)
(𝑁 ∈ ℕ → ∃𝑥 ∈ (𝔼‘𝑁)∃𝑦 ∈ (𝔼‘𝑁)𝑥𝑦)
 
Theoremaxlowdim2 26674* The lower two-dimensional axiom. In any space where the dimension is greater than one, there are three non-colinear points. Axiom A8 of [Schwabhauser] p. 12. (Contributed by Scott Fenton, 15-Apr-2013.)
(𝑁 ∈ (ℤ‘2) → ∃𝑥 ∈ (𝔼‘𝑁)∃𝑦 ∈ (𝔼‘𝑁)∃𝑧 ∈ (𝔼‘𝑁) ¬ (𝑥 Btwn ⟨𝑦, 𝑧⟩ ∨ 𝑦 Btwn ⟨𝑧, 𝑥⟩ ∨ 𝑧 Btwn ⟨𝑥, 𝑦⟩))
 
Theoremaxlowdim 26675* The general lower dimension axiom. Take a dimension 𝑁 greater than or equal to three. Then, there are three non-colinear points in 𝑁 dimensional space that are equidistant from 𝑁 − 1 distinct points. Derived from remarks in Tarski's System of Geometry, Alfred Tarski and Steven Givant, Bulletin of Symbolic Logic, Volume 5, Number 2 (1999), 175-214. (Contributed by Scott Fenton, 22-Apr-2013.)
(𝑁 ∈ (ℤ‘3) → ∃𝑝𝑥 ∈ (𝔼‘𝑁)∃𝑦 ∈ (𝔼‘𝑁)∃𝑧 ∈ (𝔼‘𝑁)(𝑝:(1...(𝑁 − 1))–1-1→(𝔼‘𝑁) ∧ ∀𝑖 ∈ (2...(𝑁 − 1))(⟨(𝑝‘1), 𝑥⟩Cgr⟨(𝑝𝑖), 𝑥⟩ ∧ ⟨(𝑝‘1), 𝑦⟩Cgr⟨(𝑝𝑖), 𝑦⟩ ∧ ⟨(𝑝‘1), 𝑧⟩Cgr⟨(𝑝𝑖), 𝑧⟩) ∧ ¬ (𝑥 Btwn ⟨𝑦, 𝑧⟩ ∨ 𝑦 Btwn ⟨𝑧, 𝑥⟩ ∨ 𝑧 Btwn ⟨𝑥, 𝑦⟩)))
 
Theoremaxeuclidlem 26676* Lemma for axeuclid 26677. Handle the algebraic aspects of the theorem. (Contributed by Scott Fenton, 9-Sep-2013.)
((((𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁)) ∧ (𝐶 ∈ (𝔼‘𝑁) ∧ 𝑇 ∈ (𝔼‘𝑁))) ∧ (𝑃 ∈ (0[,]1) ∧ 𝑄 ∈ (0[,]1) ∧ 𝑃 ≠ 0) ∧ ∀𝑖 ∈ (1...𝑁)(((1 − 𝑃) · (𝐴𝑖)) + (𝑃 · (𝑇𝑖))) = (((1 − 𝑄) · (𝐵𝑖)) + (𝑄 · (𝐶𝑖)))) → ∃𝑥 ∈ (𝔼‘𝑁)∃𝑦 ∈ (𝔼‘𝑁)∃𝑟 ∈ (0[,]1)∃𝑠 ∈ (0[,]1)∃𝑢 ∈ (0[,]1)∀𝑖 ∈ (1...𝑁)((𝐵𝑖) = (((1 − 𝑟) · (𝐴𝑖)) + (𝑟 · (𝑥𝑖))) ∧ (𝐶𝑖) = (((1 − 𝑠) · (𝐴𝑖)) + (𝑠 · (𝑦𝑖))) ∧ (𝑇𝑖) = (((1 − 𝑢) · (𝑥𝑖)) + (𝑢 · (𝑦𝑖)))))
 
Theoremaxeuclid 26677* Euclid's axiom. Take an angle 𝐵𝐴𝐶 and a point 𝐷 between 𝐵 and 𝐶. Now, if you extend the segment 𝐴𝐷 to a point 𝑇, then 𝑇 lies between two points 𝑥 and 𝑦 that lie on the angle. Axiom A10 of [Schwabhauser] p. 13. (Contributed by Scott Fenton, 9-Sep-2013.)
((𝑁 ∈ ℕ ∧ (𝐴 ∈ (𝔼‘𝑁) ∧ 𝐵 ∈ (𝔼‘𝑁) ∧ 𝐶 ∈ (𝔼‘𝑁)) ∧ (𝐷 ∈ (𝔼‘𝑁) ∧ 𝑇 ∈ (𝔼‘𝑁))) → ((𝐷 Btwn ⟨𝐴, 𝑇⟩ ∧ 𝐷 Btwn ⟨𝐵, 𝐶⟩ ∧ 𝐴𝐷) → ∃𝑥 ∈ (𝔼‘𝑁)∃𝑦 ∈ (𝔼‘𝑁)(𝐵 Btwn ⟨𝐴, 𝑥⟩ ∧ 𝐶 Btwn ⟨𝐴, 𝑦⟩ ∧ 𝑇 Btwn ⟨𝑥, 𝑦⟩)))
 
Theoremaxcontlem1 26678* Lemma for axcont 26690. Change bound variables for later use. (Contributed by Scott Fenton, 20-Jun-2013.)
𝐹 = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐷 ∧ (𝑡 ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑥𝑖) = (((1 − 𝑡) · (𝑍𝑖)) + (𝑡 · (𝑈𝑖)))))}       𝐹 = {⟨𝑦, 𝑠⟩ ∣ (𝑦𝐷 ∧ (𝑠 ∈ (0[,)+∞) ∧ ∀𝑗 ∈ (1...𝑁)(𝑦𝑗) = (((1 − 𝑠) · (𝑍𝑗)) + (𝑠 · (𝑈𝑗)))))}
 
Theoremaxcontlem2 26679* Lemma for axcont 26690. The idea here is to set up a mapping 𝐹 that will allow us to transfer dedekind 10792 to two sets of points. Here, we set up 𝐹 and show its domain and range. (Contributed by Scott Fenton, 17-Jun-2013.)
𝐷 = {𝑝 ∈ (𝔼‘𝑁) ∣ (𝑈 Btwn ⟨𝑍, 𝑝⟩ ∨ 𝑝 Btwn ⟨𝑍, 𝑈⟩)}    &   𝐹 = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐷 ∧ (𝑡 ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑥𝑖) = (((1 − 𝑡) · (𝑍𝑖)) + (𝑡 · (𝑈𝑖)))))}       (((𝑁 ∈ ℕ ∧ 𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈 ∈ (𝔼‘𝑁)) ∧ 𝑍𝑈) → 𝐹:𝐷1-1-onto→(0[,)+∞))
 
Theoremaxcontlem3 26680* Lemma for axcont 26690. Given the separation assumption, 𝐵 is a subset of 𝐷. (Contributed by Scott Fenton, 18-Jun-2013.)
𝐷 = {𝑝 ∈ (𝔼‘𝑁) ∣ (𝑈 Btwn ⟨𝑍, 𝑝⟩ ∨ 𝑝 Btwn ⟨𝑍, 𝑈⟩)}       (((𝑁 ∈ ℕ ∧ (𝐴 ⊆ (𝔼‘𝑁) ∧ 𝐵 ⊆ (𝔼‘𝑁) ∧ ∀𝑥𝐴𝑦𝐵 𝑥 Btwn ⟨𝑍, 𝑦⟩)) ∧ (𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈𝐴𝑍𝑈)) → 𝐵𝐷)
 
Theoremaxcontlem4 26681* Lemma for axcont 26690. Given the separation assumption, 𝐴 is a subset of 𝐷. (Contributed by Scott Fenton, 18-Jun-2013.)
𝐷 = {𝑝 ∈ (𝔼‘𝑁) ∣ (𝑈 Btwn ⟨𝑍, 𝑝⟩ ∨ 𝑝 Btwn ⟨𝑍, 𝑈⟩)}       (((𝑁 ∈ ℕ ∧ (𝐴 ⊆ (𝔼‘𝑁) ∧ 𝐵 ⊆ (𝔼‘𝑁) ∧ ∀𝑥𝐴𝑦𝐵 𝑥 Btwn ⟨𝑍, 𝑦⟩)) ∧ ((𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈𝐴𝐵 ≠ ∅) ∧ 𝑍𝑈)) → 𝐴𝐷)
 
Theoremaxcontlem5 26682* Lemma for axcont 26690. Compute the value of 𝐹. (Contributed by Scott Fenton, 18-Jun-2013.)
𝐷 = {𝑝 ∈ (𝔼‘𝑁) ∣ (𝑈 Btwn ⟨𝑍, 𝑝⟩ ∨ 𝑝 Btwn ⟨𝑍, 𝑈⟩)}    &   𝐹 = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐷 ∧ (𝑡 ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑥𝑖) = (((1 − 𝑡) · (𝑍𝑖)) + (𝑡 · (𝑈𝑖)))))}       ((((𝑁 ∈ ℕ ∧ 𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈 ∈ (𝔼‘𝑁)) ∧ 𝑍𝑈) ∧ 𝑃𝐷) → ((𝐹𝑃) = 𝑇 ↔ (𝑇 ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑃𝑖) = (((1 − 𝑇) · (𝑍𝑖)) + (𝑇 · (𝑈𝑖))))))
 
Theoremaxcontlem6 26683* Lemma for axcont 26690. State the defining properties of the value of 𝐹. (Contributed by Scott Fenton, 19-Jun-2013.)
𝐷 = {𝑝 ∈ (𝔼‘𝑁) ∣ (𝑈 Btwn ⟨𝑍, 𝑝⟩ ∨ 𝑝 Btwn ⟨𝑍, 𝑈⟩)}    &   𝐹 = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐷 ∧ (𝑡 ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑥𝑖) = (((1 − 𝑡) · (𝑍𝑖)) + (𝑡 · (𝑈𝑖)))))}       ((((𝑁 ∈ ℕ ∧ 𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈 ∈ (𝔼‘𝑁)) ∧ 𝑍𝑈) ∧ 𝑃𝐷) → ((𝐹𝑃) ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑃𝑖) = (((1 − (𝐹𝑃)) · (𝑍𝑖)) + ((𝐹𝑃) · (𝑈𝑖)))))
 
Theoremaxcontlem7 26684* Lemma for axcont 26690. Given two points in 𝐷, one preceeds the other iff its scaling constant is less than the other point's. (Contributed by Scott Fenton, 18-Jun-2013.)
𝐷 = {𝑝 ∈ (𝔼‘𝑁) ∣ (𝑈 Btwn ⟨𝑍, 𝑝⟩ ∨ 𝑝 Btwn ⟨𝑍, 𝑈⟩)}    &   𝐹 = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐷 ∧ (𝑡 ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑥𝑖) = (((1 − 𝑡) · (𝑍𝑖)) + (𝑡 · (𝑈𝑖)))))}       ((((𝑁 ∈ ℕ ∧ 𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈 ∈ (𝔼‘𝑁)) ∧ 𝑍𝑈) ∧ (𝑃𝐷𝑄𝐷)) → (𝑃 Btwn ⟨𝑍, 𝑄⟩ ↔ (𝐹𝑃) ≤ (𝐹𝑄)))
 
Theoremaxcontlem8 26685* Lemma for axcont 26690. A point in 𝐷 is between two others if its function value falls in the middle. (Contributed by Scott Fenton, 18-Jun-2013.)
𝐷 = {𝑝 ∈ (𝔼‘𝑁) ∣ (𝑈 Btwn ⟨𝑍, 𝑝⟩ ∨ 𝑝 Btwn ⟨𝑍, 𝑈⟩)}    &   𝐹 = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐷 ∧ (𝑡 ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑥𝑖) = (((1 − 𝑡) · (𝑍𝑖)) + (𝑡 · (𝑈𝑖)))))}       ((((𝑁 ∈ ℕ ∧ 𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈 ∈ (𝔼‘𝑁)) ∧ 𝑍𝑈) ∧ (𝑃𝐷𝑄𝐷𝑅𝐷)) → (((𝐹𝑃) ≤ (𝐹𝑄) ∧ (𝐹𝑄) ≤ (𝐹𝑅)) → 𝑄 Btwn ⟨𝑃, 𝑅⟩))
 
Theoremaxcontlem9 26686* Lemma for axcont 26690. Given the separation assumption, all values of 𝐹 over 𝐴 are less than or equal to all values of 𝐹 over 𝐵. (Contributed by Scott Fenton, 20-Jun-2013.)
𝐷 = {𝑝 ∈ (𝔼‘𝑁) ∣ (𝑈 Btwn ⟨𝑍, 𝑝⟩ ∨ 𝑝 Btwn ⟨𝑍, 𝑈⟩)}    &   𝐹 = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐷 ∧ (𝑡 ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑥𝑖) = (((1 − 𝑡) · (𝑍𝑖)) + (𝑡 · (𝑈𝑖)))))}       (((𝑁 ∈ ℕ ∧ (𝐴 ⊆ (𝔼‘𝑁) ∧ 𝐵 ⊆ (𝔼‘𝑁) ∧ ∀𝑥𝐴𝑦𝐵 𝑥 Btwn ⟨𝑍, 𝑦⟩)) ∧ ((𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈𝐴𝐵 ≠ ∅) ∧ 𝑍𝑈)) → ∀𝑛 ∈ (𝐹𝐴)∀𝑚 ∈ (𝐹𝐵)𝑛𝑚)
 
Theoremaxcontlem10 26687* Lemma for axcont 26690. Given a handful of assumptions, derive the conclusion of the final theorem. (Contributed by Scott Fenton, 20-Jun-2013.)
𝐷 = {𝑝 ∈ (𝔼‘𝑁) ∣ (𝑈 Btwn ⟨𝑍, 𝑝⟩ ∨ 𝑝 Btwn ⟨𝑍, 𝑈⟩)}    &   𝐹 = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐷 ∧ (𝑡 ∈ (0[,)+∞) ∧ ∀𝑖 ∈ (1...𝑁)(𝑥𝑖) = (((1 − 𝑡) · (𝑍𝑖)) + (𝑡 · (𝑈𝑖)))))}       (((𝑁 ∈ ℕ ∧ (𝐴 ⊆ (𝔼‘𝑁) ∧ 𝐵 ⊆ (𝔼‘𝑁) ∧ ∀𝑥𝐴𝑦𝐵 𝑥 Btwn ⟨𝑍, 𝑦⟩)) ∧ ((𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈𝐴𝐵 ≠ ∅) ∧ 𝑍𝑈)) → ∃𝑏 ∈ (𝔼‘𝑁)∀𝑥𝐴𝑦𝐵 𝑏 Btwn ⟨𝑥, 𝑦⟩)
 
Theoremaxcontlem11 26688* Lemma for axcont 26690. Eliminate the hypotheses from axcontlem10 26687. (Contributed by Scott Fenton, 20-Jun-2013.)
(((𝑁 ∈ ℕ ∧ (𝐴 ⊆ (𝔼‘𝑁) ∧ 𝐵 ⊆ (𝔼‘𝑁) ∧ ∀𝑥𝐴𝑦𝐵 𝑥 Btwn ⟨𝑍, 𝑦⟩)) ∧ ((𝑍 ∈ (𝔼‘𝑁) ∧ 𝑈𝐴𝐵 ≠ ∅) ∧ 𝑍𝑈)) → ∃𝑏 ∈ (𝔼‘𝑁)∀𝑥𝐴𝑦𝐵 𝑏 Btwn ⟨𝑥, 𝑦⟩)
 
Theoremaxcontlem12 26689* Lemma for axcont 26690. Eliminate the trivial cases from the previous lemmas. (Contributed by Scott Fenton, 20-Jun-2013.)
(((𝑁 ∈ ℕ ∧ (𝐴 ⊆ (𝔼‘𝑁) ∧ 𝐵 ⊆ (𝔼‘𝑁) ∧ ∀𝑥𝐴𝑦𝐵 𝑥 Btwn ⟨𝑍, 𝑦⟩)) ∧ 𝑍 ∈ (𝔼‘𝑁)) → ∃𝑏 ∈ (𝔼‘𝑁)∀𝑥𝐴𝑦𝐵 𝑏 Btwn ⟨𝑥, 𝑦⟩)
 
Theoremaxcont 26690* The axiom of continuity. Take two sets of points 𝐴 and 𝐵. If all the points in 𝐴 come before the points of 𝐵 on a line, then there is a point separating the two. Axiom A11 of [Schwabhauser] p. 13. (Contributed by Scott Fenton, 20-Jun-2013.)
((𝑁 ∈ ℕ ∧ (𝐴 ⊆ (𝔼‘𝑁) ∧ 𝐵 ⊆ (𝔼‘𝑁) ∧ ∃𝑎 ∈ (𝔼‘𝑁)∀𝑥𝐴𝑦𝐵 𝑥 Btwn ⟨𝑎, 𝑦⟩)) → ∃𝑏 ∈ (𝔼‘𝑁)∀𝑥𝐴𝑦𝐵 𝑏 Btwn ⟨𝑥, 𝑦⟩)
 
15.4.2.3  EE^n fulfills Tarski's Axioms
 
Syntaxceeng 26691 Extends class notation with the Tarski geometry structure for 𝔼↑𝑁.
class EEG
 
Definitiondf-eeng 26692* Define the geometry structure for 𝔼↑𝑁. (Contributed by Thierry Arnoux, 24-Aug-2017.)
EEG = (𝑛 ∈ ℕ ↦ ({⟨(Base‘ndx), (𝔼‘𝑛)⟩, ⟨(dist‘ndx), (𝑥 ∈ (𝔼‘𝑛), 𝑦 ∈ (𝔼‘𝑛) ↦ Σ𝑖 ∈ (1...𝑛)(((𝑥𝑖) − (𝑦𝑖))↑2))⟩} ∪ {⟨(Itv‘ndx), (𝑥 ∈ (𝔼‘𝑛), 𝑦 ∈ (𝔼‘𝑛) ↦ {𝑧 ∈ (𝔼‘𝑛) ∣ 𝑧 Btwn ⟨𝑥, 𝑦⟩})⟩, ⟨(LineG‘ndx), (𝑥 ∈ (𝔼‘𝑛), 𝑦 ∈ ((𝔼‘𝑛) ∖ {𝑥}) ↦ {𝑧 ∈ (𝔼‘𝑛) ∣ (𝑧 Btwn ⟨𝑥, 𝑦⟩ ∨ 𝑥 Btwn ⟨𝑧, 𝑦⟩ ∨ 𝑦 Btwn ⟨𝑥, 𝑧⟩)})⟩}))
 
Theoremeengv 26693* The value of the Euclidean geometry for dimension 𝑁. (Contributed by Thierry Arnoux, 15-Mar-2019.)
(𝑁 ∈ ℕ → (EEG‘𝑁) = ({⟨(Base‘ndx), (𝔼‘𝑁)⟩, ⟨(dist‘ndx), (𝑥 ∈ (𝔼‘𝑁), 𝑦 ∈ (𝔼‘𝑁) ↦ Σ𝑖 ∈ (1...𝑁)(((𝑥𝑖) − (𝑦𝑖))↑2))⟩} ∪ {⟨(Itv‘ndx), (𝑥 ∈ (𝔼‘𝑁), 𝑦 ∈ (𝔼‘𝑁) ↦ {𝑧 ∈ (𝔼‘𝑁) ∣ 𝑧 Btwn ⟨𝑥, 𝑦⟩})⟩, ⟨(LineG‘ndx), (𝑥 ∈ (𝔼‘𝑁), 𝑦 ∈ ((𝔼‘𝑁) ∖ {𝑥}) ↦ {𝑧 ∈ (𝔼‘𝑁) ∣ (𝑧 Btwn ⟨𝑥, 𝑦⟩ ∨ 𝑥 Btwn ⟨𝑧, 𝑦⟩ ∨ 𝑦 Btwn ⟨𝑥, 𝑧⟩)})⟩}))
 
Theoremeengstr 26694 The Euclidean geometry as a structure. (Contributed by Thierry Arnoux, 15-Mar-2019.)
(𝑁 ∈ ℕ → (EEG‘𝑁) Struct ⟨1, 17⟩)
 
Theoremeengbas 26695 The Base of the Euclidean geometry. (Contributed by Thierry Arnoux, 15-Mar-2019.)
(𝑁 ∈ ℕ → (𝔼‘𝑁) = (Base‘(EEG‘𝑁)))
 
Theoremebtwntg 26696 The betweenness relation used in the Tarski structure for the Euclidean geometry is the same as Btwn. (Contributed by Thierry Arnoux, 15-Mar-2019.)
(𝜑𝑁 ∈ ℕ)    &   𝑃 = (Base‘(EEG‘𝑁))    &   𝐼 = (Itv‘(EEG‘𝑁))    &   (𝜑𝑋𝑃)    &   (𝜑𝑌𝑃)    &   (𝜑𝑍𝑃)       (𝜑 → (𝑍 Btwn ⟨𝑋, 𝑌⟩ ↔ 𝑍 ∈ (𝑋𝐼𝑌)))
 
Theoremecgrtg 26697 The congruence relation used in the Tarski structure for the Euclidean geometry is the same as Cgr. (Contributed by Thierry Arnoux, 15-Mar-2019.)
(𝜑𝑁 ∈ ℕ)    &   𝑃 = (Base‘(EEG‘𝑁))    &    = (dist‘(EEG‘𝑁))    &   (𝜑𝐴𝑃)    &   (𝜑𝐵𝑃)    &   (𝜑𝐶𝑃)    &   (𝜑𝐷𝑃)       (𝜑 → (⟨𝐴, 𝐵⟩Cgr⟨𝐶, 𝐷⟩ ↔ (𝐴 𝐵) = (𝐶 𝐷)))
 
Theoremelntg 26698* The line definition in the Tarski structure for the Euclidean geometry. (Contributed by Thierry Arnoux, 7-Apr-2019.)
𝑃 = (Base‘(EEG‘𝑁))    &   𝐼 = (Itv‘(EEG‘𝑁))       (𝑁 ∈ ℕ → (LineG‘(EEG‘𝑁)) = (𝑥𝑃, 𝑦 ∈ (𝑃 ∖ {𝑥}) ↦ {𝑧𝑃 ∣ (𝑧 ∈ (𝑥𝐼𝑦) ∨ 𝑥 ∈ (𝑧𝐼𝑦) ∨ 𝑦 ∈ (𝑥𝐼𝑧))}))
 
Theoremelntg2 26699* The line definition in the Tarski structure for the Euclidean geometry. In contrast to elntg 26698, the betweenness can be strengthened by excluding 1 resp. 0 from the related intervals (because of 𝑥𝑦). (Contributed by AV, 14-Feb-2023.)
𝑃 = (Base‘(EEG‘𝑁))    &   𝐼 = (1...𝑁)       (𝑁 ∈ ℕ → (LineG‘(EEG‘𝑁)) = (𝑥𝑃, 𝑦 ∈ (𝑃 ∖ {𝑥}) ↦ {𝑝𝑃 ∣ (∃𝑘 ∈ (0[,]1)∀𝑖𝐼 (𝑝𝑖) = (((1 − 𝑘) · (𝑥𝑖)) + (𝑘 · (𝑦𝑖))) ∨ ∃𝑙 ∈ (0[,)1)∀𝑖𝐼 (𝑥𝑖) = (((1 − 𝑙) · (𝑝𝑖)) + (𝑙 · (𝑦𝑖))) ∨ ∃𝑚 ∈ (0(,]1)∀𝑖𝐼 (𝑦𝑖) = (((1 − 𝑚) · (𝑥𝑖)) + (𝑚 · (𝑝𝑖))))}))
 
Theoremeengtrkg 26700 The geometry structure for 𝔼↑𝑁 is a Tarski geometry. (Contributed by Thierry Arnoux, 15-Mar-2019.)
(𝑁 ∈ ℕ → (EEG‘𝑁) ∈ TarskiG)
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