By Shang-Ching Chou; Xiao-Shan Gao; Jingzhong Zhang

ISBN-10: 9810215843

ISBN-13: 9789810215842

Pt. I. the idea of computer facts. 1. Geometry Preliminaries. 2. the realm procedure. three. laptop facts in airplane Geometry. four. computer evidence in strong Geometry. five. Vectors and computer Proofs -- Pt. II. subject matters From Geometry: a suite of four hundred robotically Proved Theorems. 6. themes From Geometry

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**Extra info for Machine Proofs In Geometry: Automated Production of Readable Proofs for Geometry Theorems**

**Example text**

Moreover, l (a, b) := z ∈ S\{b} | a ∈ [z, b] ∪ [a, b] ∪ z ∈ S\{a} | b ∈ [a, z] is called a (Menger) line of (S, d). In the euclidean case (X, eucl), the interval [a, b] consists of all x ∈ X with (a − x) + (x − b) = a − b = a − x + x − b . 7) Hence, by Lemma 2, the elements a − x and x − b are linearly dependent. e. x= λ 1 b−a a+ b=a+ . 7) holds true, but not for λ ∈] − 1, 0[ or λ < −1. Hence [a, b] = {a + µ (b − a) | 0 ≤ µ ≤ 1}, and l (a, b) = {a + µ (b − a) | µ ∈ R}. In the case (X, eucl) the Menger lines are thus exactly the previous lines.

C + x x = > . ∈ B (c, ) implies (c − c ) x 1 = x 2 2 − 2 − (c − c )2 for all elements x = 0 of X. If c − c were = 0, the left-hand side of this equation would be 0 for 0 = x ⊥ (c − c ) and = 0 for x = c − c which is impossible, since the right-hand side of the equation does not depend on x. ) Hence c − c = 0, and thus 0= Proposition 9. Let B (c, ), 2 − 2 − (c − c )2 = 2 − 2 . > 0, be a ball of (X, hyp). Then B (c, ) = {x ∈ X | x − a + x − b = 2α} √ with a := ce− , b := ce and α := sinh · 1 + c2 , where et denotes the exponential function exp (t) for t ∈ R.

Proof. If a, b are linearly dependent, then there exists a real λ = 0 with b = λa since a, b are both unequal to 0. Put x0 a2 := αa. e. β = bx0 = λa · x0 = λα, and thus H (a, α) = H (b, β). e. e. e. b − ab a2 ab a a2 2 = b2 − (ab)2 = b (q − x0 ) = 0, a2 a = 0 would hold true. If a = 0 is in X and a2 = 1, then the hyperplanes of (X, hyp) can also be deﬁned by αTt β (a⊥ ) with α, β ∈ O (X) and t ∈ R : take ω ∈ O (X) with a = ω (e) and observe αTt β [ω (e)]⊥ = αTt β ω (e⊥ ) = αTt βω (e⊥ ). Obviously, ω H (a, α) = H ω (a), α for ω ∈ O (X), where H (a, α) is a euclidean hyperplane.

### Machine Proofs In Geometry: Automated Production of Readable Proofs for Geometry Theorems by Shang-Ching Chou; Xiao-Shan Gao; Jingzhong Zhang

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