Axiom:Five-Segment Axiom

From ProofWiki
Jump to navigation Jump to search

Axiom

Let $\equiv$ be the relation of equidistance.

Let $\mathsf B$ be the relation of betweenness.

Let $=$ be the relation of equality.


This axiom asserts that:

$\forall a, b, c, d, a', b', c', d':$
$\left({\neg \left({a = b}\right) \land \left({\mathsf B a b c \land \mathsf B a' b' c'}\right) \land \left({a b \equiv a' b' \land b c \equiv b' c' \land a d \equiv a' d' \land b d \equiv b' d'}\right)}\right)$
$\implies c d \equiv c' d'$

where $a, b, c, d, a', b', c', d'$ are points.


Intuition

Note that the following section does not cover degenerate cases.

Tarski's Five Segment Axiom.png

Let $a, b, c$ and $a', b', c'$ be collinear.

Let $a c d$ and $a' c' d'$ be triangles.

Draw a line connecting $d$ to $b$ and $d'$ to $b'$.

Suppose that every corresponding pair of line segments so constructed have been confirmed to be congruent except for segments $c d$ and $c' d'$.

Then $c d$ and $c' d'$ are also congruent.


Also see


Sources