{{interfaces | imports = Interface:Orthogonality 2 | exports = none yet }}

This is part of a series of modules which prove a variety of geometrical theorems starting with Tarski's axioms for geometry. We follow the formalization of Julien Narboux<ref>The formal proofs are at http://www.lix.polytechnique.fr/Labo/Julien.Narboux/tarski.html Formalization of Tarski's geometry in the Coq proof assistant and are described in Julien Narboux (2007), "http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.158.8614 Mechanical Theorem Proving in Tarski’s Geometry", F. Botana and T. Recio (Eds.): ADG 2006, LNAI 4869, pp. 139–156</ref> which itself closely follows a treatise by Schwabhäuser, Szmielew, and Tarski.<ref>W. Schwabhäuser, W Szmielew, and A. Tarski (1983), ''Metamathematische Methoden in der Geometrie'', ISBN 0387129588</ref>

This page is one of several involving perpendicular lines. We start with some theorems about the (is perpendicular) predicate.

We import the theorems of propositional logic and predicate logic, and the geometry results so far and define some variables:

Perpendicular lines

Saying that lines are perpendicular at a point is the main step towards saying those lines are perpendicular. The connection is pretty simple: A B ⟂ C D is defined as ∃ x A B C D ⟂at x.

Theorem form of the definition

As usual, we need a theorem corresponding to the definition.

No line is perpendicular to itself

Here we show ¬ A B ⟂ A B. The interesting part of the proof was already done as PerpendicularIrreflexivityAt.<ref>l8_14_1 in Narboux</ref>

Being perpendicular at any point implies being perpendicular

That is, A B C D ⟂at X → A B ⟂ C D.<ref>l8_14_2_1a in Narboux</ref>

Uniqueness of intersection point

If two lines are perpendicular at a point, they intersect only at that point. In symbols, A B C D ⟂at X ∧ collinear Y A B ∧ collinear Y C D → X = Y.<ref>l8_14_2_1b in Narboux</ref>

To prove this we first show is-right-angle Y X Y which follows directly from the ∀ p ∀ q (collinear p A B ∧ collinear q C D → is-right-angle p X q) part of PerpendicularAt.

From is-right-angle Y X Y, it is a short step to X = Y.

Perpendicular lines which intersect are perpendicular at that point

If two lines are perpendicular and they intersect at a point, they are perpendicular at that point. In symbols, A B ⟂ C D ∧ collinear X A B ∧ collinear X C D → A B C D ⟂at X.<ref>l8_14_2_1b_bis in Narboux</ref>

The proof is a straightforward application of PerpendicularAtUniqueIntersection{{{. We start with a lemma performing a substitution. * #(→ (∧ (∧ (⟂at A B C D Y) (collinear X A B)) (collinear X C D)) (⟂at A B C D X))# ([/general/Orthogonality_3.gh/PerpendicularIntersectionPerpendicularAt-1 | PerpendicularIntersectionPerpendicularAt-1]) * #(→ (∧ (∧ (⟂ A B C D) (collinear X A B)) (collinear X C D)) (⟂at A B C D X))# ([/general/Orthogonality_3.gh/PerpendicularIntersectionPerpendicularAt | PerpendicularIntersectionPerpendicularAt]) === Perpendicular lines which intersect at one point are perpendicular at that point === Here is another theorem which relates being perpendicular to being perpendicular at a point: {{{A B ⟂ C D ∧ ∀ y (collinear y A B ∧ collinear y C D → X = y) → A B C D ⟂at X.<ref>l8_14_2_2 in Narboux</ref>

The proof only relies on PerpendicularAt, the definition of , and logic. Basically, we expand A B ⟂ C D to A B C D ⟂at X0, use the other hypothesis to conclude X = X0, and then substitute to turn A B C D ⟂at X0 into A B C D ⟂at X.

References

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{{DEFAULTSORT:{{PAGENAME}}}} Category:Euclidean geometries (general) and generalizations Category:Foundations of classical theories (including reverse mathematics)

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