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\begin{center}
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\Large\bf\classid\\[1em]
{\large\bf Lecturer: \classinstructor}\\[.5em]
{\large\bf Scribe Notes by: \scribeone}
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\vspace*{.4in}

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\section{Lower bound for bounded-depth Frege proofs of $PHP_n^{n+1}$}

In this lecture we will continue the proof of the following theorem.

\begin{theorem}
Any bounded-depth Frege proof of $PHP_n^{n+1}$ requires exponential size.
\end{theorem}

We have seen in the previous lectures the definitions of matching
restrictions, matching disjunctions, matching decision trees.

\subsection{Overview}

We will prove the theorem by contradiction. Assuming there is a short
proof $\mathcal{P}$ of $PHP_n^{n+1}$ in which all formulas have depth
at most $d$, we will apply mathcing restrictions in order to turn the
formulas into matching decision trees. The assignment of matching
decision trees to formulas is a $k$-evaluation. We consider the
formulas in $\mathcal{P}$ in order of increasing depth (recall that
depth is defined as the maximum number of alternations of
quantifiers).

If $S = \neg{A}$ is a formula in our proof, assume we have assigned a
matching decision tree $T(A)$ to formula $A$. We assign it a decision
tree $T(S)$ by turning all leaf labels in $T(A)$ from 0 to 1 and
from 1 to 0.

If $S = A_1 \vee \cdots \vee A_k$ is a disjunction in $\mathcal{P}$,
we construct $T(S)$ by taking the OR path of 1 leaves in all $T(A_i)$,
applying a nice restriction to that DNF formula, and building a
canonical matching decision tree for that formula. A switching Lemma
will guarantee that nice restrictions exist.

The contradiction will come in the following manner.

\begin{enumerate}

\item Axioms of the Frege system will be turned into 1-trees (ie,
trees which have only leaves labelled by 1).

\item The rules of the Frege system preserve 1-trees.

\item However, any formula in $PHP_n^{n+1}$ is transformed into a
0-tree.

\end{enumerate}



\subsection{Analogy}

The assignment of trees to formulas creates an analogy with the proof
that bounded-depth circuit computing Parity requires super-polynomial
size. However, the analogy is broken in the sense that the trees in
there compute the exact function, while the matching decision trees in
the $k$-evaluations used for the proof of Theorem 1 do not, not even
on restrictions compatible with that tree. That is, assuming $\rho =
\rho_1 \dots \rho_k$ is a ``good'' restriction, compare $f$ only
on assignments which extend $\rho$ to $T$.

The tree is equivalent to the formula for only one level. However,
when $S = A_1 \vee \cdots A_k$, rewriting the matching decision trees
as matching disjunctions will not preserve the equivalence. Consider
$\sigma$ a partial matching. Even if there exists one path in all
trees $T(A_i)$ consistent with $\sigma$, the trees might have nothing
in common. Each one is querying only \emph{some} pigeons and we are
trying to build something about \emph{all} pigeons. Eg, in a tree
which starts by quering $P_{1,1}$ and $\sigma$ sends pigeon 2 to hole
5, there might be many paths consistent wth $\sigma$.

% erase this
\noindent TONI: I didn't quite get the argument above.



\subsection{$PHP_n^{n+1}$ consists of 0-trees}

$PHP$ is the disjunction of the following formulas:

\begin{enumerate}

\item $\neg{( \neg{P_{i,k}} \vee \neg{P_{j,k}} )}, \forall i \not = j
\leq n+1, \forall k \leq n$

\item $\neg{( P_{i,1} \vee \cdots \vee P_{i,n} )}, \forall i \leq n+1$

\end{enumerate}

Restrictions reduce PHP to fewer pigeons and holes. After the second block
of $\vee$, the tree is no longer equivalent to the formula.


Consider formulas of the first type. In order to show that
$T(\neg{( \neg{P_{i,k}} \vee \neg{P_{j,k}} )})$ is a 0-tree, it's enough
to show that $T(\neg{P_{i,k}} \vee \neg{P_{j,k}})$ is a 1-tree. By
definition

$$ T(\neg{P_{i,k}} \vee \neg{P_{j,k}}) = T( Disj( T^c( P_{i,k} ) )
\vee Disj( T^c( P_{j,k} ) ) ) $$


$T^c( P_{i,k} )$ is a tree of size 2 which has 1's for all assignments
where $i$ and $k$ are mapped to something, and only one 0
corresponding to mapping pigeon $i$ to hole $k$ [picture?].

The DNF $Disj( T^c( P_{i,k} ) ) \vee Disj( T^c( P_{j,k} ) )$ will
contain all terms where $i$, $j$ and $k$ are mapped to something,
because, eg, mapping $k$ to $i$ is always a leaf labelled with 1 in
$T^c(P_{j,k})$. [picture?]


For formulas of the second kind, it is enough to show that $T(P_{i,1}
\vee \cdots \vee P{i,n})$ is a 1-tree. By definition,

$$T(P_{i,1} \vee \cdots \vee P{i,n}) = T( \vee_{j=1}^n Disj( T(
P_{i,j} ) ) ) $$

But each $Disj( T( P_{i,j} ) )$ contains only one term, namely
$P_{i,j}$. Then the DNF is $P_{i,1} \vee \cdots \vee P{i,n}$ and its
associated tree starts by querying pigeon $i$ and will have all leaves
labelled with 1 at one level below root, as the formula is true no
matter where this pigeon is mapped.


\subsection{All formulas in a bounded-depth Frege proofs get assigned
1-trees}

This is Lemma 5.1 in the paper. The Frege system we are considering
has axiom $A \vee \neg{A}$, and rules

$$ \frac{A}{A \vee B}, \frac{A \vee A}{A}, \frac{A \vee (B \vee C)}{(A
\vee B) \vee C}, \frac{A \vee B, \neg{A} \vee C}{B \vee C} $$

The proof is in the paper, using as parameter the maximum number of
subformulas in each rule.

\begin{theorem}[Lemma 5.1]
Let $f$ be the maximum number of subformulas appearing in a rule (this
is a constant, 7?). Let $\mathcal{P}$ be a proof of $PHP_n^{n+1}$, $T$
a $k$-evaluation for all subformulas in $\mathcal{P}$ and $k < n/f$,
then any formula occuring as a line in $\mathcal{P}$ gets converted to
a 1-tree.
\end{theorem}

The proof is by induction on the number of lines, if we start with
axioms and keep applying sound rules (as the ones above), all formulas
convert to 1-trees.

After applying a restriction the number of variables we are left is
$n' = n^{\epsilon}$. Since we might be applying $d$ restrictions (the
bound on the depth of formulas), we want $k << n^{\epsilon^d}$.

\noindent TONI: Here you argued that the proof works for two of the
rules, the axiom and $\frac{A}{A \vee B}$ but I didn't understand the
argument for either.

\noindent TONI: Next you quickly considered how the parameters look
like. What I have is very vague.





The entire argument also works for onto-$PHP$ or func-$PHP$ because
they also convert to 0-trees.



\end{document}