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\def\classtitle{Propositional Proof Complexity}
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\begin{document}
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\begin{center}
\Large\bf\doctitle\\[1em]
\Large\bf\classid\\[1em]
{\large\bf Lecturer: \classinstructor}\\[.5em]
{\large\bf Scribe Notes by: \scribeone}
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\vspace*{.4in}

% HERE IS WHERE YOUR SCRIBE NOTES SHOULD START
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\section{Monotone Craig Interpolation}

Monotone Craig Interpolation provides another way to obtain
exponential lower bounds on Resolution proofs.

\subsection{Propositional Case}

\begin{theorem}
Let $\phi=\phi(\vec{p},\vec{q})$ be a formula in which the $\vec{p}$
variables occur only positively.  Also, suppose that
$\models\phi\rightarrow\psi$ where $\psi(\vec{p},\vec{r})$.  Then
there exists an interpolant $C(\vec{p})$ such that the $\vec{p}$
variables occur only positively in $C$.
\end{theorem}

Formulae are built with $\wedge$, $\vee$, $\neg$.  

\begin{definition}
An occurrence is positive if and only if it is under the scope of an
even number of $\neg$ signs.  
\end{definition}

Note that application of De Morgan's Laws or the distributive laws
does not affect whether a particular occurrence is positive.

\begin{lemma}
Let the variables in $\vec{p}$ be $p_1,p_2,\ldots,p_k$ and let the
variables $\vec{p}\,'$ be $p_1',p_2',\ldots,p_k'$.  If
$p_i\rightarrow p_i'$ is true $\forall i$ for some truth assignment,
then $C(\vec{p})$ is true $\Rightarrow C(\vec{p}\,')$ is true,
assuming that the variables of $\vec{p}$ occur only positively in $C$.
(This last property is called \emph{monotonicity}.)
\end{lemma}

\begin{proof}
By induction on size of $C$.  
\end{proof}

\begin{proof}
(of Theorem 1) $\phi(\vec{p},\vec{q})\rightarrow
\psi(\vec{p},\vec{r})$ is the same as $\exists\vec{q}$
$\phi(\vec{p},\vec{q})\rightarrow\forall \vec{r} \;
\psi(\vec{p},\vec{r})$.  
Then let 
\begin{equation}
C(\vec{p})\doteq
\Bigvee_{\substack{\text{all T/F settings}\\ 
\text{of the variables in } \vec{q}}}
\phi(\vec{p},\vec{q}). 
\end{equation}
The $p$'s occur only positively so we can see that this interpolant has
the proper form.

Alternately, it can be shown that a fitting interpolant is 
\begin{equation}
C(\vec{p})\doteq\Bigwedge_{\substack{\text{all T/F settings} \\ 
\text{of the variables in } \vec{r}}}\psi(\vec{p},\vec{r}).
\end{equation}

\end{proof}

\subsection{Resolution Case}

\begin{theorem}
Let $\Gamma=\Gamma(\vec{p},\vec{q})$ be a set of clauses, and
let $\Delta=\Delta(\vec{p},\vec{r})$ be a set of clauses.  

Assume that the $\vec{p}$ variables occur only positively in $\Delta$
(that is, there are no $\neg p_i$'s in clauses in $\Delta$), or assume
that the $\vec{p}$ variables occur only negatively in $\Gamma$ (that
is, there are no $p_i$'s in clauses in $\Gamma$). Also assume
$\Gamma\cup\Delta$ is unsatisfiable (i.e. has a refutation).  Then
there is an interpolant $C(\vec{p})$ such that $\forall$ truth
assignments $\tau$
\begin{list}{}{}
  \item if $\bar{\tau}(C(\vec{p}))=T$, then $\exists$ clause
	$C\in\Gamma$ such that $\tau(C)=$False,
  \item if $\bar{\tau}(C(\vec{p}))=F$, then $\exists$ clause
	$C\in\Delta$ such that $\tau(C)=$False,
  \end{list}
and such that the $\vec{p}$ variables occur only positively in $C$. 
\end{theorem}

\begin{proof}

Let 
\begin{equation}
C(\vec{p})\doteq\Bigwedge_{\substack{\text{all T/F settings} \\ 
\text{of the variables in } \vec{q}}}\Bigvee_{C\in\Gamma} 
\neg C(\vec{p},\vec{q}).  
\end{equation}
That $C(\vec{p})$ will be a fitting interpolant is immediate from the
fact that if $\Delta\cup\Gamma$ is unsatisfiable
\begin{equation}
\Bigwedge_{C\in\Delta}C\Rightarrow\Bigvee_{C\in\Gamma}\neg C.
\end{equation}
Alternately we may let 
\begin{equation}
C(\vec{p})\doteq\Bigvee_{\substack{\text{all T/F settings} \\ 
\text{of the variables in } \vec{r}}}
\Bigwedge_{C\in\Delta} C(\vec{p},\vec{r}). 
\end{equation} 
\end{proof}

\begin{theorem}
\label{thm:smthg}
Let $R$ be a refutation of $\Gamma\cup\Delta$ of size $s$.  Then
$C(\vec{p})$ can be written with monotone circuit size $O(s)$.  If $R$
is tree-like, $C(\vec{p})$ has monotone formula size $O(s)$.
\end{theorem}

\begin{theorem}
(Restatement of part of Theorem \ref{thm:smthg}) Let $\Gamma$ consist
of clauses containing $\vec{q}$'s and negative occurrences of
$\vec{p}$'s.  Let $\Delta$ consist of clauses containing $\vec{r}$'s
and $\vec{p}$'s.  (Note that we make no assumption on whether these
$\vec{p}$ occur positively or negatively in the clauses of $\Delta$.)
If $R$ is a refutation of $\Gamma\cup\Delta$, then there exists an
interpolant $\phi$, such that the size of $\phi$ is $O(\text{number of
steps in }R)$.
\end{theorem}

\begin{definition}
For $C$ a clause in $R$, we let $\phi_C$ be defined by
\begin{enumerate}
\item $\phi_C=T$ if $C\in\Gamma$
\item $\phi_C=F$ if $C\in\Delta$
\item $\phi_C=\phi_{C_1\cup\{p_i\}} \vee  
(p_i\wedge\phi_{C_2\cup\{\bar{p}_i\}})$ if $C$ is such that 
\begin{equation*}
\frac{C_1\cup\{p_i\}\quad C_2\cup\{\bar{p}_i\}}%
{C=C_1\cup C_2}
\end{equation*}
\item $\phi_C=\phi_{C_1\cup\{q_i\}} \wedge  
\phi_{C_2\cup\{\bar{q}_i\}}$ if $C$ is such that 
\begin{equation*}
\frac{C_1\cup\{q_i\}\quad C_2\cup\{\bar{q}_i\}}%
{C=C_1\cup C_2}
\end{equation*}
\item $\phi_C=\phi_{C_1\cup\{r_i\}} \vee
\phi_{C_2\cup\{\bar{r}_i\}}$ if $C$ is such that 
\begin{equation*}
\frac{C_1\cup\{r_i\}\quad C_2\cup\{\bar{r}_i\}}%
{C=C_1\cup C_2}
\end{equation*}
\end{enumerate}
\end{definition}

\begin{definition}
For $C$ a clause, let $C^{\Gamma}=C\cap\{q_i,\bar{q}_i,\bar{p}_i:i\geq
0\}$ and let $C^{\Delta}=C\cap\{p_i,\bar{p}_i,r_i,\bar{r}_i:i\geq0\}$.
\end{definition}

\begin{claim}
For all $\tau$, $\forall C\in R$, 
\begin{list}{}{}
\item A. if $\tau \nmodels C^{\Gamma}$ and $\tau(\phi_C)=T$, then
$\exists D \in\Gamma$ such that $\tau\nmodels D$.  
\item B. if $\tau \nmodels C^{\Delta}$ and $\tau(\phi_C)=F$, then
$\exists D \in\Delta$ such that $\tau\nmodels D$.
\end{list}
\end{claim}

Proof of this claim will imply that $\phi_{\emptyset}$ works as an
interpolant, since
$\emptyset^{\Gamma}=\emptyset=\emptyset^{\Delta}$, which is not
satisfied by any $\tau$.

\begin{proof} (of Claim) 

Proof is by induction on inferences in $R$.  
\begin{enumerate}
\item $C\in\Gamma$.  Then $\phi_C=T$.  $C\in\Gamma\Rightarrow
C^{\Gamma}=C\cap\{q_i,\bar{q}_i,\bar{p}_i:i\geq0\}=C$.  Then if
$\tau\nmodels C^{\Gamma}$, $\tau\nmodels C$, and so trivially 
$\exists C\in\Gamma$ such that $\tau\nmodels C$.  
\item $C\in\Delta$.  Then $\phi_C=F$.  $C\in\Delta\Rightarrow
C^{\Delta}=C$.  Hence if $\tau\nmodels C^{\Delta}$, trivially $\exists
C\in\Delta$ such that $\tau \nmodels C$.  
\item $C=C_1\cup C_2$ with 
$\frac{C_1\cup\{p_i\}\quad C_2\cup\{\bar{p}_i\}}{C=C_1\cup C_2}$.
Then $\phi_C=\phi_{C_1\cup\{p_i\}} \vee  
(p_i\wedge\phi_{C_2\cup\{\bar{p}_i\}})$.
  \begin{enumerate}
    \item Suppose $\tau\nmodels C^{\Gamma}$ and $\tau(\phi_C)=T$.
    Note that $(C_1\cup\{p_i\})^{\Gamma}= C_1^{\Gamma}$. Then
    $\tau\nmodels(C_1\cup\{p_i\})^{\Gamma}$, since $C_1^{\Gamma} \subseteq
    C^{\Gamma}$ and $\tau\nmodels C^{\Gamma}$ imply that $\tau\nmodels
    C_1^{\Gamma}$.   
      \begin{enumerate}
        \item If $\tau(\phi_{C_1\cup\{p_i\}})= T$, then since we have
        $\tau\nmodels(C_1\cup\{p_i\})^{\Gamma}$, by the
        induction hypothesis we are done.   
        \item Otherwise, $\tau(p_i\wedge\phi_{C_2\cup\{\bar{p}_i\}})=T$. 
        Thus $\tau(\phi_{C_2\cup\{\bar{p}_i\}})=T$.  
        Also $\tau(p_i)=T$, so $\tau\nmodels \{\bar{p}_i\}$. Note
        $\{\bar{p}_i\}=\{\bar{p}_i\}^{\Gamma}$, so $\tau\nmodels
	\{\bar{p}_i\}^{\Gamma}$.  Also $C_2^{\Gamma}\subseteq
        C^{\Gamma}$ and $\tau\nmodels C^{\Gamma}$ imply $\tau\nmodels
        C_2^{\Gamma}$.  Thus $\tau\nmodels
        C_2^{\Gamma}\cup\{\bar{p}_i\}^{\Gamma}$ and hence as
	$C_2^{\Gamma}\cup\{\bar{p}_i\}^{\Gamma}= 
        (C_2\cup\{\bar{p}_i\})^{\Gamma}$, we get 
        $\tau\nmodels(C_2\cup\{\bar{p}_i\})^{\Gamma}$.  Since we have
        $\tau(\phi_{C_2\cup\{\bar{p}_i\}})=T$ and
        $\tau\nmodels(C_2\cup\{\bar{p}_i\})^{\Gamma}$, the
	induction hypothesis applies. 
      \end{enumerate}
    \item Suppose $\tau\nmodels C^{\Delta}$ and $\tau(\phi_C)=F$
      \begin{enumerate}
        \item If $\tau(p_i)=T$, then
        $\tau\nmodels(C_2\cup\{\bar{p}_i\})^{\Delta}$.  Also
        $\tau(\phi_{C_2\cup\{\bar{p}_i\}})=F$.  Then the induction
        hypothesis applies.  
        \item If $\tau(p_i)=F$, then
        $\tau\nmodels(C_1\cup\{p_i\})^{\Delta}$ and
        $\tau(\phi_{C_1\cup\{p_i\}})=F$.  Then the induction
        hypothesis applies.  
      \end{enumerate}
  \end{enumerate}
\item $C=C_1\cup C_2$ with 
$\frac{C_1\cup\{q_i\}\quad C_2\cup\{\bar{q}_i\}}{C=C_1\cup C_2}$. Then
$\phi_C=\phi_{C_1\cup\{q_i\}}\wedge\phi_{C_2\cup\{\bar{q}_i\}}$. 
  \begin{enumerate}
    \item Suppose $\tau\nmodels C^{\Gamma}$ and
    $\tau(\phi_C)=T$. $\tau(\phi_C)=T$ implies that 
    $\tau(\phi_{C_1\cup\{q_i\}})=\tau(\phi_{C_2\cup\{\bar{q}_i\}})=T$. 
    Either
    $\tau\nmodels(C_1\cup\{q_i\})^{\Gamma}$ (if $\tau(q_i)=F$) or 
    $\tau\nmodels(C_2\cup\{\bar{q}_i\})^{\Gamma}$ (if $\tau(q_i)=T$).  
    In either case, the induction hypothesis applies.
    \item Suppose $\tau\nmodels C^{\Delta}$ and $\tau(\phi_{C})=F$. 
    Note that $(C_1\cup\{q_i\})^{\Delta}=C_1^{\Delta}$ and 
    $(C_2\cup\{\bar{q}_i\})^{\Delta}=C_2^{\Delta}$.  So 
    $\tau\nmodels(C_1\cup\{q_i\})^{\Delta}$ and 
    $\tau\nmodels(C_2\cup\{\bar{q}_i\})^{\Delta}$.  Also, since
    $\tau(\phi_C)=F$, either $\tau(\phi_{C_1\cup\{q_i\}})=F$ or 
    $\tau(\phi_{C_2\cup\{\bar{q}_i\}})=F$ so in either case the 
    induction hypothesis applies.  
  \end{enumerate} 
\item $C=C_1\cup C_2$ with 
$\frac{C_1\cup\{r_i\}\quad C_2\cup\{\bar{r}_i\}}{C=C_1\cup C_2}$.
Then $\phi_C=\phi_{C_1\cup\{r_i\}} \vee\phi_{C_2\cup\{\bar{r}_i\}}$. 
  \begin{enumerate}
    \item Suppose $\tau\nmodels C^{\Gamma}$ and $\tau(\phi_C)=T$.
    Note that $(C_1\cup\{r_i\})^{\Gamma}=C_1^{\Gamma}$, and 
    $(C_2\cup\{\bar{r}_i\})^{\Gamma}=C_2^{\Gamma}$.  Hence 
    $\tau\nmodels(C_1\cup\{r_i\})^{\Gamma}$ and 
    $\tau\nmodels(C_2\cup\{\bar{r}_i\})^{\Gamma}$. Also, 
    $\tau(\phi_{C_1\cup\{r_i\}})=T$ or 
    $\tau(\phi_{C_2\cup\{\bar{r}_i\}})=T$.  Then the 
    induction hypothesis applies to whichever one equals $T$.  
    \item Suppose $\tau\nmodels C^{\Delta}$ and $\tau(\phi_{C})=F$.
    $\tau(\phi_C)=F$ implies 
    $\tau(\phi_{C_1\cup\{r_i\}})=\tau(\phi_{C_2\cup\{\bar{r}_i\}})=F$. 
    Either 
    $\tau\nmodels(C_1\cup\{r_i\})^{\Delta}$ (if $\tau(r_i)=F$) or 
    $\tau\nmodels(C_2\cup\{\bar{r}_i\})^{\Delta}$ (if $\tau(r_i)=T$). 
    In either case, the induction hypothesis applies.  
  \end{enumerate}
\end{enumerate}

Notice that $\phi$ is monotone in the $p_i$'s.  

\end{proof}

\subsection{Exponential Lower Bounds on Resolution Proofs for Clique
and Coloring}

\begin{definition}
A $k$-coloring of a graph is an assignment of $k$ colors to vertices
such that no adjacent vertices have the same color.  
\end{definition}

\begin{definition}
A $k$-clique in a graph is a subset consisting of $k$ nodes such that
for any pair of nodes in the subset there is an edge in the graph
which joins them.
\end{definition}

\begin{theorem}
A graph cannot have both a $k+1$ clique and a $k$-coloring.
\end{theorem}

\end{document}


