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DPLL algorithm:
\scalebox{0.9}{\setlength\tabcolsep{3pt}\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|}\hlineStep & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hline \hlineDecision Level & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 2 \\ \hlineAssignment & - & $a$ & $a,b$ & $a,b,c$ & $a,b,c,d$ & $a,b,c,d,e$ & $\lnot a$ & $\lnot a, b$ & $\lnot a,b,c$ & $\lnot a,b,c,d$ & $\lnot a,b,c,d,e$ \\ \hlineCl. 1: $\lnot a, b$ & 1 & $b$ & \cmark & \cmark & \cmark & \cmark & \cmark & \cmark & \cmark & \cmark & \cmark \\ \hlineCl. 2: $\lnot b, c$ & 2 & 2 & $c$ & \cmark & \cmark & \cmark & 2 & $c$ & \cmark & \cmark & \cmark \\ \hlineCl. 3: $\lnot c, d$ & 3 & 3 & 3 & $d$ & \cmark & \cmark & 3 & 3 & $d$ & \cmark & \cmark \\ \hlineCl. 4: $\lnot d, e$ & 4 & 4 & 4 & 4 & $e$ & \cmark & 4 & 4 & 4 & $e$ & \cmark \\ \hlineCl. 5: $\lnot e, \lnot a$ & 5 &$\lnot e$& $\lnot e$ & $\lnot e$ & $\lnot e$ & $\{\}$ \xmark & \cmark & \cmark & \cmark & \cmark & \cmark \\ \hline \hlineBCP & - & $b$ & $c$ & $d$ & $e$ & - & - & $c$ & $d$ & $e$ & \cmark \\ \hlineDecision &$a$& - & - & - & - & $\lnot a$ & $b$ & - & - & - & SAT \\ \hline\end{tabular}}
Model:
$a$ = F, $b$ = T, $c$ = T, $d$ = T, $e$ = T
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