Proving this axiom system is consistent. I want to make sure I understand the correct notation and expressions for proving that an axiom system is consistent.  I have the axioms 


*

*Every line is a collection of points.

*There are at least two points.

*For any two points there is exactly one line containing them.

*For any line there is a point not on the line.

*For any line and point not on the line, there is a line parallel to the first line which contains the point.


To prove this consistent I want to exhibit a model.  I this case, I think I want two points $P,Q$ that lay on one line, and an infinity of other (distinct) points $A_{1}, A_{2}, ...$ each of which lies on lines that are pairwise parallel.  Each of these lines will, themselves, be a collection of points.  While I could have each line be an infinity of points, a minimal model would have all lines contain just one point except for the first line I mentioned, which has two.  
Do I need to say more than this?  Is there some notation or form of expression that is more appropriate for the assignment?
 A: $
\newcommand{L}{\mathbb L}
\newcommand{P}{\mathbb P}
$

Every line is a collection of points.

$$\forall l \in \L ~~ a \in l \rightarrow a \in \P$$

There are at least two points.

$$\exists a,b \in \P ~~  a \ne b$$

For any two points there is exactly one line containing them.

$$\forall a,b \in \P ~~\exists_1 l \in \L ~~a \in l \land b \in l $$

For any line there is a point not on the line.

$$\forall l \in \L~~ \exists p \in \P ~~p \not \in l$$

For any line and point not on the line, there is a line parallel to the first line which contains the point.

$$\forall l \in \L , p \in \P ~~ p \not \in \L \rightarrow \exists m \in \L ~~ p \in m \land Z(m,l)$$
These axioms require 2 sets, $\P$ and $\L$.  They require 1 function $Z(,)$ (for parallel) which has to be defined over $\L$.
To show consistency in terms of model theory, you show a structure (a collection of objects,sets,functions) for which the axioms are true.  This structure doesn't have to have anything to do with geometry as you intuitively think of it.
For example , a structure which satisfies the axioms is:


*

*$\P$ is the set of natural numbers $\{0,~ 1,~ 2,~ \dots\}$

*$\L$ is the set of pairs of natural numbers $\{(0,1),~ (0, 2),~ (1, 2),~ (0, 3),~ \dots \}$

*$Z( (a,b) , (x,y) )$ is true if and only if $a + 1 = x$ and $b + 1 = y$



each of which lies on lines that are pairwise parallel.

This seems to defy the axiom that all pairs of points can make a line.  You have partitioned the points among lines, but the axioms that any 2 points make a line imply that there must be a "chain" among the line containment predicate : in other words, for any 2 points $p_1, p_2$, there are a collections of points $q$ and line $L$ such that $p_1 L_0 q_0 \land q_0 L_1 q_1 \land q_1 L_2 q_2 \land q_2 L_3 q_3 \dots q_n L_m p_2$.  Where $pLq$ means points $p$ and $q$ are on $L$.

Each of these lines will, themselves, be a collection of points.

Why?  What axiom is this satisfying?

a minimal model would have all lines contain just one point except for the first line I mentioned, which has two.

I believe a minimal model would be


*

*$\P = \{a, b, c\}$

*$\L = \{(a,b) , (b,c), (a,c)\}$

*$Z( x , y ) = 
\begin{cases}
 x = (a,b) \land y = (b,c) \\ \text{ or } \\ 
 x = (b,c) \land y = (a,c) \\ \text{ or } \\
 x = (a,c) \land y = (a,b) 
\end{cases}$


Since "For any line there is a point not on the line." demands that there is at least 3 points.
