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\import{nat.tex}
\import{order/order.tex}
\import{relation.tex}
\section{The real numbers}
\begin{signature}
$\reals$ is a set.
\end{signature}
\begin{signature}
$x + y$ is a set.
\end{signature}
\begin{signature}
$x \times y$ is a set.
\end{signature}
\begin{axiom}\label{one_in_reals}
$1 \in \reals$.
\end{axiom}
\begin{axiom}\label{reals_axiom_order}
$\lt[\reals]$ is an order on $\reals$.
%$\reals$ is an ordered set.
\end{axiom}
\begin{axiom}\label{reals_axiom_strictorder}
$\lt[\reals]$ is a strict order.
\end{axiom}
\begin{axiom}\label{reals_axiom_dense}
For all $x,y \in \reals$ if $(x,y)\in \lt[\reals]$ then
there exist $z \in \reals$ such that $(x,z) \in \lt[\reals]$ and $(z,y) \in \lt[\reals]$.
%For all $X,Y \subseteq \reals$ if for all $x,y$ $x\in X$ and $y \in Y$ such that $x \lt[\reals] y$
%then there exist a $z \in \reals$ such that if $x \neq z$ and $y \neq z$ $x \lt[\reals] z$ and $z \lt[\reals] y$.
\end{axiom}
\begin{axiom}\label{reals_axiom_order_def}
$(x,y) \in \lt[\reals]$ iff there exist $z \in \reals$ such that $(\zero, z) \in \lt[\reals]$ and $x + z = y$.
\end{axiom}
\begin{lemma}\label{reals_one_bigger_than_zero}
$(\zero,1) \in \lt[\reals]$.
\end{lemma}
\begin{axiom}\label{reals_axiom_assoc}
For all $x,y,z \in \reals$ $(x + y) + z = x + (y + z)$ and $(x \times y) \times z = x \times (y \times z)$.
\end{axiom}
\begin{axiom}\label{reals_axiom_kommu}
For all $x,y \in \reals$ $x + y = y + x$ and $x \times y = y \times x$.
\end{axiom}
\begin{axiom}\label{reals_axiom_zero_in_reals}
$\zero \in \reals$.
\end{axiom}
%\begin{axiom}\label{reals_axiom_one_in_reals}
% $\one \in \reals$.
%\end{axiom}
\begin{axiom}\label{reals_axiom_zero}
%There exist $\zero \in \reals$ such that
For all $x \in \reals$ $x + \zero = x$.
\end{axiom}
\begin{axiom}\label{reals_axiom_one}
%There exist $1 \in \reals$ such that
For all $x \in \reals$ $1 \neq \zero$ and $x \times 1 = x$.
\end{axiom}
\begin{axiom}\label{reals_axiom_add_invers}
For all $x \in \reals$ there exist $y \in \reals$ such that $x + y = \zero$.
\end{axiom}
%TODO: Implementing Notion for negativ number such as -x.
%\begin{abbreviation}\label{reals_notion_minus}
% $y = -x$ iff $x + y = \zero$.
%\end{abbreviation} %This abbrevation result in a killed process.
\begin{axiom}\label{reals_axiom_mul_invers}
For all $x \in \reals$ there exist $y \in \reals$ such that $x \neq \zero$ and $x \times y = 1$.
\end{axiom}
\begin{axiom}\label{reals_axiom_disstro1}
For all $x,y,z \in \reals$ $x \times (y + z) = (x \times y) + (x \times z)$.
\end{axiom}
\begin{proposition}\label{reals_disstro2}
For all $x,y,z \in \reals$ $(y + z) \times x = (y \times x) + (z \times x)$.
\end{proposition}
\begin{proposition}\label{reals_reducion_on_addition}
For all $x,y,z \in \reals$ if $x + y = x + z$ then $y = z$.
\end{proposition}
\begin{signature}
$\invers$ is a set.
\begin{signature}
%TODO:
%x \rless y in einer signatur hinzufügen und dann axiom x+z = y und dann \rlt in def per iff
%\inv{} für inverse benutzen. Per Signatur einfüheren und dann axiomatisch absicher
%\cdot für multiklikation verwenden.
%< für die relation benutzen.
%\begin{signature}
% $y^{\rightarrow}$ is a function.
%\end{signature}
%\begin{axiom}\label{notion_multi_invers}
% If $y \in \reals$ then $\invers{y} \in \reals$ and $y \times y^{\rightarrow} = 1$.
%\end{axiom}
%\begin{abbreviation}\label{notion_fraction}
% $\frac{x}{y} = x \times y^{\rightarrow}$.
%\end{abbreviation}
\begin{lemma}\label{order_reals_lemma1}
For all $x,y,z \in \reals$ such that $(\zero,x) \in \lt[\reals]$
if $(y,z) \in \lt[\reals]$
then $((y \times x), (z \times x)) \in \lt[\reals]$.
\end{lemma}
\begin{lemma}\label{order_reals_lemma2}
For all $x,y,z \in \reals$ such that $(\zero,x) \in \lt[\reals]$
if $(y,z) \in \lt[\reals]$
then $((x \times y), (x \times z)) \in \lt[\reals]$.
\end{lemma}
\begin{lemma}\label{order_reals_lemma3}
For all $x,y,z \in \reals$ such that $(x,\zero) \in \lt[\reals]$
if $(y,z) \in \lt[\reals]$
then $((x \times z), (x \times y)) \in \lt[\reals]$.
\end{lemma}
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