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\section{Proof Complexity Lower Bounds}

Last class we proved communication complexity lower bounds
for lifted search problems, from lower bounds for UDISJ.
We saw how this
implies monotone depth lower bounds.

Today we will show that from communication complexity
lower bounds for lifted search problems, we can also obtain
proof complexity lower bounds.


Recall that , $Tseitin$ on a graph with $n$ vertices
has $n$ constraints, and we will select a gadget $g$ with $c=2$.
Thus, our monotone function will have $4n$ inputs
and its monotone depth will be at least the communication
complexity of $S(Tseitin(G))og^n$.

Last lecture, we proved that
for appropriate graphs $G$, the communication complexity 
the lifted Tseitin CNF search problem, $S(Tseitin(G))og^n$, is
$\Omega(n/\log n)$.

\subsection{Resolution and Cutting Planes}

We will prove lower bounds for Resolution proofs as
well as Cutting Planes Proofs. Both are refutation systems for
showing that a CNF formula is unsatisfiable.

Resolution has one rule, the resolution rule:
from the clause $(A \lor x)$ and $(B \lor \neg x)$ we
can derive $(A \lor B)$ where $A$, $B$ are disjunctions of literals.
Given a CNF formula $f=C_1 \land C_2 \land \ldots \land C_m$ a resolution
refutation of $f$ is a sequence of clauses where each clause in the
sequence is either a clause of $f$ or follows from two previous clauses
by the resolution rule, and where the final clause is the empty clause.
Resolution is known to be sound and complete.
The {\it size} of a resolution proof is the total number of clauses 
occuring in the proof; the {\it depth} is the maximum height of the underlying
proof dag. A resolution refutation is {\it tree-like} if the underlying
proof dag is a tree (input clauses from $f$ can be repeated any number of times).

Like Resolution, Cutting Planes is a refutation system.
However lines in a CP proof are linear inequalities $\sum_i a_i x_i \geq b$,
where $a_i$ are rational numbers.
There are three CP rules:
\begin{itemize}
\item[(1)] (Addition) Two previously derived inequalities can be added together;
\item[(2)] (Multiplication) Multiply a previously derived inequality by an integer;
\item[(3)] (Division by 2 with rounding) From the inequality $\sum_i a_i x_i \geq b$, we can derive
$\sum_i a_i/2 x_i \geq \lceil b/2 \rceil$, as long as each $a_i$ is divisible by 2. 
\end{itemize}

Let $f$ be a CNF formula. Then it can easily be converted into an equivalent set of inequalities.
First, for each variable $x_i$ we have the inequalities $x_i \geq 0$ and $x_i \leq 1$.
Then we convert each clause into an inequality in the obvious way.
For example, $(x_1 \lor \neg x_2)$ becomes $x_1 + (1-x_2) \geq 1$.
It is easy to see that $f$ is satisfiable (over boolean assignments) if and only if the corresponding set of
inequalities is satisfiable (over boolean assignments).
A CP refutation of $f$ is a sequence of inequalities where each inequality
is either one from $f$, or derived from previous inequalities by one of the CP rules,
and where the final inequality is $0 \geq 1$.
The size is the total number of inequalities in the proof.
(Note: actual size would measure the total length of all coefficients in the proof;
however it turns out that the coefficients without loss of generality have
size at most exponential in $n$, and thus their bit length is $O(n)$ and thus
we can just use the number of inequalities as a fairly good measure of the size.)
Again we can define the depth of a CP proof, and the notion of a tree-like CP proof.

\subsection{Reducing Depth and Tree-Size Lower bounds to Communication Complexity}

\begin{theorem}
For both Resolution and CPs, if $f$ has a refutation of depth $h$,
then the CNF search problem for $f$ has a (randomized)
communication complexity protocol of cost $O(\log n)h$.
\end{theorem}

The proof idea is as follows.
We will design an efficient communication complexity protocol
for solving the CNF search problem for $f$ from a small-depth Resolution/CPs
refutation of $f$.
The players will work their way from the root of the proof dag to a leaf of the dag,
evaluating lines along the way. The key property is that for any partition of
the variables underlying $f$, every line in a Resolution or CP  refutation
can be evaluated by the players with an efficient protocol, and thus the overall
complexity is the complexity of evaluating a line times the
depth.
For example, for Resolution, given a clause $C$, some of the variables
underlying $C$ belong to Alice and some to Bob. Alice evaluates the
literals in $C$ that she owns, and sends "1" if one of these literals
evaluates to true, and otherwise she sends "0"; Bob does the same.
If at least one of the players announced "1" then the clause is true
under their assignment, and otherwise it is false.
They follow a path from the root to a leaf with the property that
all clauses along the path evaluate to false.
Since the rules are sound, and the root clause is identically false,
it follows that for each clause that they encounter (which inductively
is set to false by their assignment), one of the two children clauses
will be set to false by their assignment.
When they arrive at a leaf they have found a clause of $f$ that
is falsified.
The argument for CPs is similar but is a bit more complicated because
of the high weights. (In the case where the weights have length at most
$O(\log n)$, the players can just compute the weighted sum of their variables
and send the sum to the other player.)

Note that this lower  bound method works much more generally
for any sound proof system whose lines can be evaluated efficiently by
a small 2-player communication protocol.
The proof also easily generalizes to the case of sound
proof systems (such as Sherali-Adams, Lasserre) where the lines
in the proof are efficiently evaluated by a small $k$-player communication
protocol.

\subsection{Lower Bounds on depth and tree-like size for Resolution and CPs}

The idea is as follows.
We start with an unsatisfiable family of CNF formulas, $F_n(z_1,\ldots,z_n)$, whose underlying
search problem has high (randomized) communication complexity.
We lift $F_n$ to obtain a new CNF formula $F' = F_n o g^n$, where
$g$ is our 2-player versatile function defined in the last lecture.
(Recall that $g$ takes $x,y$, $|x|=|y|=2$ and outputs 1 if and only if
the mod 2 sum of the numbers is either 2 or 3, modulo 4.)
The lifted CNF, $F'$ will be on variables $x,y$, $|x|=|y|=2n$,
and will be defined in the obvious way.
If $F_n$ is a $k$-CNF, then $F'$ will be a $2k$-CNF formula.
By the way that we define $F'$, it will turn out that the CNF search problem for $F'$ will
be the lifted CNF search problem associated with $F_n,g$.


Let $F$ be the CNF search problem, with clauses $C_1, \ldots, C_m$ over
variables $z_1,\ldots, z_n$.
Using $g$, the lifted CNF, $F'$, will be over variables
$.x_1,\ldots,x_{2n}$, $y_1,\ldots,y_{2n}$.
We will denote the $i^{th}$ variable in the $j^{th}$ block of $x$
by $x^j_i$. So for example, $x_3$, the first variable in the second
block, will be denoted by $x^2_1$.
By $x^j$ we mean the $j^{th}$ block of $x$.
(And similarly for $y$.)
By way of example, suppose that $C_1 = (z_1 \lor \neg z_2 \lor z_4)$.
Then $C_1$ will convert to a bunch of new clauses.
For each assignment $\alpha$ to $x^1,x^2,x^4$ and 
assignment $\beta$ to $y^1,y^2,y^4$ such that
$g({\overline \alpha^1},{\overline \beta^1})=0$,
and $g({\overline \alpha^2},{\overline \beta^2})=1$ and
$g({\overline \alpha^4},{\overline \beta^4})=0$, we
have the clause
$$(x^{1,\alpha^1} \lor y^{1,\beta^1} \lor x^{2,\alpha^2} \lor y^{2,\beta^2} \lor
  x^{4,\alpha^4} \lor y^{4,\beta^4}).$$

For example, let $\alpha^1=\beta^1 = 11$,
$\alpha^2=\beta^2=00$,
$\alpha^4=\beta^4 = 11$.
Then $g(00,00)=0$, and $g(11,11)=1$ and $g(00,00) =0$.
(Because $00$ is binary for 0, and $0+0 (mod 4) =0$
and similarly, $11$ is binary for 3, and $3+3 (mod 4) = 2$.)
we add the following clause:
$$(x^1_1 \lor x^2_2 \lor y^1_1 \lor y^1_2 \lor \neg x^2_1 \lor \neg x^2_2
 \lor \neg y^2_1 \lor \neg y^2_2 \lor x^4_1 \lor x^4_2 \lor y^4_1 \lor y^4_2).$$

It is easy to check that the CNF search problem associated with $F'$ is
none other than the lifted search problem associated with $F,g$.

To prove our lower bound, we will start with $F_n$ that has very
high (randomized) communication complexity, namely the Tseitin formulas.

Thus, it follows from our lower bound for the lifted Tseitin search
problem together with the above theorem,
that any Resolution or CP refutation of the lifted Tseitin formula
requires depth $\Omega(n/\log n)$.

It can also be proven using a slightly more complicated argument
that any tree-like Resolution or CP refutation for the lifted Tseitin formula
requires exponential size.






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