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\noindent {\bf CS 4995 \hfill Problem Set 1\hfill Fall, 2022}\\
\begin{center}
{\bf Due: Tuesday, October 11, 11:59pm}
\end{center}


%\noindent {\bf CS 4995 \hfill Problem Set 1 \hfill Fall, 2022}\\


\medskip

\begin{enumerate}




\item (12 points) In class, we discussed the DLL procedure for finding
Resolution refutations. This procedure was inherently bottom-up
in that we started at the root, and worked our way to the
leaves (initial clauses). Another procedure for finding
Resolution refutations is top-down and is
called the Davis-Putnam (DP) procedure. The procedure is as follows.
First, order the variables $x_1,...,x_n$. Let $C_0$ be the original
set of clauses.
Apply all possible resolution steps to the initial clauses where we
resolve only on the variable $x_1$. Let $C_1$ be all clauses
obtained so far (including the original clauses) and not containing
the literals $x_1$ or $\neg x_1$. Now apply all possible
resolution steps to clauses in $C_1$ where now we resolve
only on the variable $x_2$, and let $C_2$ be the resulting set
of all clauses obtained from $C_1$ (including clauses in $C_1$)
and not containing the literals $x_2$ or $\neg x_2$.
Continue in this fashion until we have either derived the
empty clause, or until we have used up all of our variables
Note that the clauses in $C_i$ will involve only the variables
$x_{i+1},...,x_n$.

\bigskip


\begin{enumerate}
\item Use the above DP procedure to obtain a
refutation of the following formula.
$$(x_1 \lor x_2 \lor x_3) \land (\neg x_1 \lor x_2 \lor x_3)\land
(\neg x_2 \lor x_3) \land (\neg x_3 \lor x_2)\land(\neg x_2 \lor \neg x_3)$$
\item Show that the DP procedure, when applied to 2CNF formulas
always terminates in polynomial time.
\item Prove completeness of the DP procedure.
That is, show that for every unsatisfiable CNF formula $f$,
there exists a DP refutation of $f$.
\end{enumerate}

\bigskip

\item (10 points)
Recall that a tree-like Resolution refutation of an unsatisfiable CNF formula $f$
is a Resolution refutation where each derived clause is used at most once.
Prove that tree-like Resolution refutations are closed under
restrictions. Let $f(x_1,\ldots,x_n)$ be an unsatisfiable CNF formula.
Let $S$ be a subset of the underlying variables, and let
$\rho$ be a partial truth assignment that sets each
variable in $S$ to either 0 or 1.
Prove that if $T$ is tree-like Resolution refutation of $F$ of size $s$, then
we can apply the restriction $\rho$ to $T$ to obtain
a tree-like Resolution refutation of $F|_{\rho}$ of size at most $s$.

\bigskip


%Exercise 2 of Lecture Notes on Propositional Calculus (Craig interpolation lemma)
\item (5 points)
Let $A$ and $B$ be propositional formulas, and let $S$ be the the set of 
propositional variables/atoms that occur in both $A$ and $B$.
Prove that if $A \rightarrow B$ is valid, then there is a formula
$C$ involving on the variables from $S$ such that 
$A \rightarrow C$ and $C \rightarrow B$.

\bigskip



\item (10 points) Exercise 14, page 17 of notes on propositional calculus.


\bigskip

\item (5 points) Exercise 5, page 25 of notes.
(Show that $\forall x(gfx=x)$ is not a logical consequence of
$\forall x(fgx=x)$.)

\bigskip

\item (5 points)
Let $\Phi =\{A_1,A_2,...\}$ be an infinite set of sentences.  Suppose that
for all $n$, $A_{n+1}$ is not a logical consequence of $\{A_1,...,A_n\}$.
Now let $B$ be any sentence such that $\Phi \models B$.  Prove that
there exists $n$ such that $A_n$ is not a logical consequence of $B$.

\bigskip

\item  (5 points) Exercise 10, page 25 of notes on Predicate Calculus

\bigskip

\item ({\bf Extra Credit}) The
pigeonhole principle, $PHP^{n+1}_n$ asserts
that $n+1$ pigeons cannot be mapped in a one-to-one way to $n$ holes.
The negation of the propositional principle, $\neg PHP^{n+1}_n$
is a CNF formula with underlying variables
$P_{i,j}$ for $i \leq n+1$ and $j \leq n$. $P_{i,j}$ is intended
to represent whether or not pigeon $i$ is mapped to hole $j$.
The clauses of $\neg PHP^{n+1}_n$ are of two types:
First, for every $i \leq n+1$ there are pigeon clauses
${\cal P}_i$: $(P_{i,1} \lor P_{i,2} \lor \cdots \lor P_{i,n})$
stating that each pigeon goes to at least one hole.
Secondly, for every $i1, i2 \leq n+1$, $i1 \neq i2$ and $j \leq n$,
there are hole clauses ${\cal H}_{i1,i2,j}$: $(\neg P_{i1,j} \lor \neg P_{i2,j})$
stating that each hole has at most one pigeon mapped to it.

\begin{enumerate}
\item Prove that for $n$ sufficiently large, any DPLL refutation of
$\neg PHP^{n+1}_n$ requires size $2^{O(n)}$.

\medskip

{\bf Hint:}
Recall the proof of completeness for Resolution discussed in class.
In the proof, we showed that a tree-like Resolution refutation
of a CNF formula $f(x_1,\ldots,x_n)$ can
be viewed as a decision tree that queries
the variables $x_1,\ldots,x_n$ of $f$, and such that
each leaf node $l$ of the
decision tree is labelled with a clause from $f$ that is falsified by the partial assignment to that leaf node.


\item Prove the stronger lower bound of $2^{\Omega(n \log n)}$ on the size
of any tree-like Resolution refutation of $PHP^{n+1}_n$.
\end{enumerate}

\end{enumerate}
\end{document}