\documentclass{article}
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\begin{document}

\title{High school geometry theorems}
\author{Hilbert's axiomatic system.\\
Formalizovano od strane: Sana Stojanovic Djurdjevic \\
Dokaz generisan uz pomo\'{c}: ArgoGeoChecker.}
\date{30.03.2018.}
\maketitle

\begin{theorem}[th_4_01.]
Pod pretpostavkom da va\v{z}i $\alpha \neq \beta$ i $A \in \alpha$ i $A \in \beta$ i $p \in \alpha$ i $p \in \beta$ pokazati da postoji ta\v{c}ka $B$  i ta\v{c}ka $C$ tako da va\v{z}i $B \neq C$ i $B \in p$ i $C \in p$.
\end{theorem}

{\em Dokaz:}
\setcounter{proofstepnum}{0} 


\proofstep{0} {Postoje ta\v{c}ka $B$  i ta\v{c}ka $C$  tako da va\v{z}i $B \neq C$ i $B \in p$ i $C \in p$ (aksioma $I3a$).} 

\proofstep{0} {Zaklju\v{c}ak teoreme sledi iz \v{c}injenica $B \neq C$ i $B \in p$ i $C \in p$.} 
QED
\hrule 
\begin{theorem}[th_4_02.]
Pod pretpostavkom da va\v{z}i $\alpha \neq \beta$ i $A \in \alpha$ i $A \in \beta$ i $p \in \alpha$ i $p \in \beta$ i $B \neq C$ i $B \in p$ i $C \in p$ pokazati da va\v{z}i $A \in p$.
\end{theorem}

{\em Dokaz:}
\setcounter{proofstepnum}{0} 


\proofstep{0} {Na osnovu \v{c}injenica $p \in \alpha$ i $B \in p$ va\v{z}i $B \in \alpha$ (aksioma $D11$).} 
\proofstep{0} {Na osnovu \v{c}injenica $p \in \alpha$ i $C \in p$ va\v{z}i $C \in \alpha$ (aksioma $D11$).} 
\proofstep{0} {Na osnovu \v{c}injenica $p \in \beta$ i $B \in p$ va\v{z}i $B \in \beta$ (aksioma $D11$).} 
\proofstep{0} {Na osnovu \v{c}injenica $p \in \beta$ i $C \in p$ va\v{z}i $C \in \beta$ (aksioma $D11$).} 
\proofstep{0} {Va\v{z}i $A \in p$ ili $A \not \in p$.}  

\proofstep{3}  {Pretpostavimo da va\v{z}i: $A \in p$.} 

\proofstep{6} {Zaklju\v{c}ak teoreme sledi iz \v{c}injenice $A \in p$.}  

\proofstep{3}  {Pretpostavimo da va\v{z}i: $A \not \in p$.} 
\proofstep{6} {Na osnovu \v{c}injenica $B \neq C$ i $B \in p$ i $C \in p$ i $A \not \in p$ va\v{z}i $\neg col(B, C, A)$ (aksioma $D1a$).} 
\proofstep{6} {Na osnovu \v{c}injenica $\neg col(B, C, A)$ i $B \in \alpha$ i $C \in \alpha$ i $A \in \alpha$ i $B \in \beta$ i $C \in \beta$ i $A \in \beta$ va\v{z}i $\alpha = \beta$ (aksioma $I5$).} 
\proofstep{6} {Na osnovu \v{c}injenica $\alpha \neq \beta$ i $\alpha = \beta$ dobijamo kontradikciju.} 
\proofstep{0} {Teorema je dokazana u svim slu\v{c}ajevima.}
QED
\hrule 

\end{document}
