\begin{dplltabular}{6}
  \dpllStep{1|2|3|4|5|6}
  \dpllDecL{0|1|2|2|2|2}

  \dpllAssi{-|
            $\lnot a$|
            $\lnot a, \lnot b$|
            $\lnot a, \lnot b, e$|
            \makecell{$\lnot a, \lnot b, e$,\\ $\lnot c$}|
            \makecell{$\lnot a, \lnot b, e$,\\ $\lnot c, \lnot d $}}


  \dpllClause{1}{$a, \lnot c, \lnot e$}
  {$a, \lnot c, \lnot e$|$\lnot c,\lnot e$|$\lnot c,\lnot e$|$\lnot c$|\done|\done}

  \dpllClause{2}{$\lnot a, \lnot e$}
  {$\lnot a, \lnot e$|\done|\done|\done|\done|\done}

  \dpllClause{3}{$b,e$}
  {$b,e$|$b,e$|$e$|\done|\done|\done}

  \dpllClause{4}{$\lnot b,d,e$}
  {$\lnot b,d,e$|$\lnot b,d,e$|\done|\done|\done|\done}

  \dpllClause{5}{$\lnot b,\lnot d$}
  {$\lnot b,\lnot d$|$\lnot b,\lnot d$|\done|\done|\done|\done}

  \dpllClause{6}{$c,\lnot d$}
  {$c,\lnot d$|$c,\lnot d$|$c,\lnot d$|$c,\lnot d$|$\lnot d$|\done}

  \dpllClause{7}{$c,d$}
  {$c,d$|$c,d$|$c,d$|$c,d$|$d$|\conflict}

  \dpllBCP {-|-|$e$|$\lnot c$|$\lnot d$|-}
  \dpllPL  {-|-|-|-|-|-}
  \dpllDeci{$\lnot a$|$\lnot b$|-|-|-|-}
\end{dplltabular}

\begin{conflictgraph}
  \node[base node] (notA)                       {$\lnot a$};
  \node[base node] (notB) [below of=notA]       {$\lnot b$};
  \node[base node] (E)    [right of=notB]       {$e$};
  \node[base node] (notC) [above right of=E]    {$\lnot c$};
  \node[base node] (D)    [above right of=notC] {$d$};
  \node[base node] (notD) [below right of=notC] {$\lnot d$};
  \node[base node] (bot)  [above right of=notD] {$\bot$};
  \path[]
      (notB) edge [] node {$3$} (E)
      (notA) edge [] node {$1$} (notC)
      (E)    edge [] node {$1$} (notC)
      (notC) edge [] node {$6$} (notD)
      (notC) edge [] node {$7$} (D)
      (notD) edge [] node {}    (bot)
      (D)    edge [] node {}    (bot);
\end{conflictgraph}

\begin{prooftree}
\AxiomC{$6. \; c \lor \lnot d$}
\AxiomC{$7. \; c \lor d$}
	\BinaryInfC{$c$}
	\AxiomC{$1. \; a\lor\lnot c \lor\lnot e$}
		\BinaryInfC{$a\lor\lnot e$}
	  \AxiomC{$3. \; b \lor e$}
	  	\BinaryInfC{$a \lor b$}
\end{prooftree}

\begin{dplltabular}{6}
  \dpllStep{(2)|7|8|9|10}
  \dpllDecL{1  |1|1|1|1}

  \dpllAssi{$\lnot a$|
            $\lnot a, b$|
            $\lnot a, b, \lnot d$|
            \makecell{$\lnot a, b, \lnot d$,\\ $c$}|
            \makecell{$\lnot a, b, \lnot d$,\\ $c, \lnot e $}}


  \dpllClause{1}{$a, \lnot c, \lnot e$}
  {$\lnot c, \lnot e$|$\lnot c, \lnot e$|$\lnot c, \lnot e$|$\lnot e$|\done}

  \dpllClause{2}{$\lnot a, \lnot e$}
  {\done|\done|\done|\done|\done}

  \dpllClause{3}{$b,e$}
  {$b,e$|\done|\done|\done|\done}

  \dpllClause{4}{$\lnot b,d,e$}
  {$\lnot b,d,e$|$d,e$|$e$|$e$|\conflict}

  \dpllClause{5}{$\lnot b,\lnot d$}
  {$\lnot b,\lnot d$|$\lnot d$|\done|\done|\done}

  \dpllClause{6}{$c,\lnot d$}
  {$c,\lnot d$|$c,\lnot d$|\done|\done|\done}

  \dpllClause{7}{$c,d$}
  {$c,d$|$c,d$|$c$|\done|\done}

  \dpllClause{8}{$a,b$}
  {$b$|\done|\done|\done|\done}

  \dpllBCP {$b$|$\lnot d$|$c$|$\lnot e$|-}
  \dpllPL  {-|-|-|-|-}
  \dpllDeci{-|-|-|-|-}
\end{dplltabular}


\begin{conflictgraph}
  \node[base node] (notA)                 {$\lnot a$};
  \node[base node] (B)    [right of=notA] {$b$};
  \node[base node] (notD) [right of=B]    {$\lnot d$};
  \node[base node] (C)    [right of=notD] {$c$};
  \node[base node] (notE) [below of=C]    {$\lnot e$};
  \node[base node] (E)    [above of=C]    {$e$};
  \node[base node] (bot)  [right of=C]    {$\bot$};
  \path[]
      (notA) edge [] node {8} (B)
      (B)    edge [] node {5} (notD)
      (notA) edge [bend right] node {1} (notE)
      (C)    edge [] node {1} (notE)
      (notD) edge [] node {7} (C)
      (notD) edge [] node {4} (E)
      (B)    edge [bend left] node {4} (E)
      (notE) edge [] node {}  (bot)
      (E)    edge [] node {}  (bot);
\end{conflictgraph}