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sav08:dpll_algorithm_for_sat [2013/04/17 17:35]
vkuncak 		sav08:dpll_algorithm_for_sat [2015/04/21 17:30] (current)
Line 90: Line 90:
  
 **Decision step** (picking variable value): from proofs for **Decision step** (picking variable value): from proofs for
-\[+\begin{equation*}
     S' \cup \{p\} \vdash {\it false}     S' \cup \{p\} \vdash {\it false}
-\] +\end{equation*} 
-\[+\begin{equation*}
     S' \cup \{\lnot p\} \vdash {\it false}     S' \cup \{\lnot p\} \vdash {\it false}
-\]+\end{equation*}
 we would like to construct the proof for we would like to construct the proof for
-\[+\begin{equation*}
     S' \vdash {\it false}     S' \vdash {\it false}
-\]+\end{equation*}
 From From
-\[+\begin{equation*}
     S' \cup \{p\} \vdash {\it false}     S' \cup \{p\} \vdash {\it false}
-\]+\end{equation*}
 derive proof tree for derive proof tree for
-\[+\begin{equation*}
     S' \vdash \lnot p     S' \vdash \lnot p
-\]+\end{equation*}
 and from and from
-\[+\begin{equation*}
     S' \cup \{\lnot p\} \vdash {\it false}     S' \cup \{\lnot p\} \vdash {\it false}
-\]+\end{equation*}
 derive proof tree for derive proof tree for
-\[+\begin{equation*}
     S' \vdash p     S' \vdash p
-\]+\end{equation*}
 Then combine these two trees with resolution on $\{p\}$ and $\{\lnot p\}$ to get ${\it false}$. Then combine these two trees with resolution on $\{p\}$ and $\{\lnot p\}$ to get ${\it false}$.