\import{nat.tex}
\import{order/order.tex}
\import{relation.tex}

\section{The real numbers}

%TODO: Implementing Notion for negativ number such as -x.

%TODO:
%\inv{} für inverse benutzen. Per Signatur einfüheren und dann axiomatisch absicher
%\cdot für multiklikation verwenden. 
%< für die relation benutzen.


\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 $\reals$.
\end{axiom}

\begin{axiom}\label{reals_axiom_strictorder}
    $\lt[\reals]$ is a strict order.
\end{axiom}

\begin{abbreviation}\label{less_on_reals}
    $x < y$ iff $(x,y) \in \lt[\reals]$.
\end{abbreviation}

\begin{abbreviation}\label{greater_on_reals}
    $x > y$ iff $y < x$.
\end{abbreviation}

\begin{abbreviation}\label{lesseq_on_reals}
    $x \leq y$ iff it is wrong that $x > y$.
\end{abbreviation}

\begin{abbreviation}\label{greatereq_on_reals}
    $x \geq y$ iff it is wrong that $x < y$.
\end{abbreviation}

\begin{axiom}\label{reals_axiom_dense}
    For all $x,y \in \reals$ if $x < y$ then 
    there exist $z \in \reals$ such that $x < z$ and $z < y$.
\end{axiom}

\begin{axiom}\label{reals_axiom_order_def}
    $x < y$ iff there exist $z \in \reals$ such that $\zero < z$ and $x + z = y$.
\end{axiom}

\begin{lemma}\label{reals_one_bigger_than_zero}
    $\zero < 1$.
\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_zero}
    For all $x \in \reals$ $x + \zero = x$. 
\end{axiom}

\begin{axiom}\label{reals_axiom_one}
    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}


\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{lemma}\label{order_reals_lemma1}
    For all $x,y,z \in \reals$ such that $\zero < x$ 
    if $y < z$ 
    then $(y \times x) < (z \times x)$.
\end{lemma}

\begin{lemma}\label{order_reals_lemma2}
    For all $x,y,z \in \reals$ such that $\zero < x$ 
    if $y < z$ 
    then $(x \times y) < (x \times z)$.
\end{lemma}


\begin{lemma}\label{order_reals_lemma3}
    For all $x,y,z \in \reals$ such that $x < \zero$ 
    if $y < z$ 
    then $(x \times z) < (x \times y)$.
\end{lemma}

\begin{lemma}\label{a}
    For all $x,y \in \reals$ if $x > y$ then $x \geq y$.
\end{lemma}

\begin{lemma}\label{aa}
    For all $x,y \in \reals$ if $x < y$ then $x \leq y$.
\end{lemma}

\begin{lemma}\label{aaa}
    For all $x,y \in \reals$ if $x \leq y \leq x$ then $x=y$.
\end{lemma}