I am interested in the interplay between the Playfair and Proclus Axioms in linear spaces.


By a *[linear space](https://mathworld.wolfram.com/LinearSpace.html)* I understand a pair $(X,\mathcal L)$ consisting of a set $X$ and a family $\mathcal L$ of subsets of $X$ such that the following axioms are satisfied:


(L1) for any distinct points $x,y\in X$ there exists a unique set $L\in\mathcal L$ containing the points $x,y$;


(L2) every set $L\in\mathcal L$ contains at least two distinct points.


Elements of the family $\mathcal L$ are called *lines*. By the axiom (L1), for every distinct points $p,q\in X$ there exists a unique line containing those points. This unique line will be denoted by $L(a,b)$. Also we put $L(a,a)=\{a\}$.


Given a linear space $(X,\mathcal L)$, consider the following (well-known) axioms:


**[Playfair](https://en.wikipedia.org/wiki/Playfair%27s_axiom):** For every line $L\in\mathcal L$, points $a,b\in L$, $c\in X\setminus L$ and $x\in L(a,c)\setminus L$ there exists a unique point $y\in L(c,b)$ such that $L(x,y)\cap L=\emptyset$;


**[Proclus](https://mathworld.wolfram.com/ProclusAxiom.html):** For every lines $L\in \mathcal L$, points $a,b\in L$, $c\in X\setminus L$, $x\in L(a,c)\setminus\{c\}$, $y\in L(c,b)\setminus\{c\}$ and $z\in L(x,y)$, if $L(a,b)\cap L(x,y)=\emptyset$, then $L(c,z)\cap L(a,b)\ne\emptyset$.



> 
> **Question.** Does **Playfair** imply **Proclus** in linear spaces?
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> 
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Does Playfair imply Proclus?
I think the following construction gives a counterexample. It stems from the observation that the Playfair axiom is quite weak in the case where all lines only have three points (it produces some pairs of parallel lines, but doesn't force any new intersections between lines).


The simplest geometry in which lines have three points is a vector space ${\bf F}\_3^n$ over the field of three elements, where the lines $L$ take the form $\{a,b,c\}$ where $a,b,c \in {\bf F}\_3^n$ are distinct with $a+b+c=0$. This geometry turns out to obey both the Playfair and Proclus axioms, so we don't have a counterexample yet. But one can modify this geometry in a flexible fashion by a skew-product construction. Let $f: {\bf F}\_3^n \to {\bf F}\_3^m$ be an arbitrary even function with $f(0)=0$. We can then define a geometry on ${\bf F}\_3^n \times {\bf F}\_3^m$ by defining the lines to be the set of triples $\{ (x,x'), (y,y'), (z,z')\}$ of three distinct points $(x,x'), (y,y'), (z,z')$ in ${\bf F}\_3^n \times {\bf F}\_3^m$ with
$$ x+y+z=0$$
and
$$ x'+y'+z' = f(y-x)$$
(note that $f(y-x)=f(x-y)=f(z-y)=f(y-z)=f(x-z)=f(z-x)$ since $f$ is even and $x,y,z$ are in arithmetic progression). One easily checks that this is a linear system. It also obeys the Playfair axiom. To see this, one just has to check that if $(a,a'), (b,b'), (c,c')$ are not collinear, $(x,x')$ is the third point on $L((a,a'), (c,c'))$, and $(y,y')$ is the third point on $L((b,b'), (c,c'))$, then $L((a,a'),(b,b'))$ and $L((x,x'),(y,y'))$ do not intersect. Suppose for contradiction that they intersected in $(z,z')$. Then we have the equations
$$ a + b + z = a + c + x = b + c + y = x + y + z = 0$$
$$ a' + b' + z' = f(b-a)$$
$$ a' + c' + x' = f(c-a)$$
$$ b' + c' + y' = f(c-b)$$
$$ x' + y' + z' = f(y-x).$$
The first equation (noting that one can divide by $2$ in ${\bf F}\_3^n$) implies that $z=c$, $b=x$, $a=y$, so that the right-hand sides of the next four equations all agree; this then implies that $z'=c', b'=x', a'=y'$, and then $(a,a'), (b,b'), (c,c')$ are collinear, a contradiction.


On the other hand, the Proclus axiom does not need to be obeyed. Take three non-collinear points $(a,a'), (b,b'), (c,c')$, let $(x,x')$ be the third point in $L((a,a'), (c,c'))$, let $(y,y')$ be the third point in $L((b,b'), (c,c'))$, let $(z,z')$ be the third point in $L((a,a'), (b,b'))$, and let $(w,w')$ be the third point in $L((x,x'), (y,y'))$. Playfair gives that $(w,w'), (z,z'), (c,c')$ are distinct; Proclus would force these three points to be collinear. This gives the equations
$$ a+c+x=b+c+y=a+b+z=x+y+w=w+z+c=0$$
$$ a'+c'+x' = f(c-a)$$
$$ b'+c'+y' = f(c-b)$$
$$ a'+b'+z' = f(b-a)$$
$$ x'+y'+w' = f(y-z)$$
$$ w'+z'+c' = f(z-w).$$
The last four equations imply
$$ f(c-a)+f(c-b) - f(b-a) - f(y-z) + f(z-w) = 3c' = 0.$$
On the other hand, $y-z = a-c$ and $z-w = a+b+c$ so we simplify to
$$ f(c-b) - f(b-a) + f(a+b+c) = 0.$$
But one can easily construct an $f$ for which this equation fails for some non-collinear $a,b,c$, so the Proclus axiom can fail.


