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\section{Substitution Frege systems}
In a previous lecture we encountered certain proof systems
for which it is {\em conjectured} that they
are stronger than Frege systems (in the sense that Frege systems do not
p-simulate them). These were the extended Frege systems.
Here we will define other such proof systems: the substitution
Frege systems. We will see that these systems are as strong
as extended Frege systems; they p-simulate extended Frege systems
and vice versa. As in the case of extended Frege systems,
it is an open question whether Frege systems p-simulate substitution
Frege systems or not.
\begin{definition} A {\em substitution Frege system}, $s\mathcal{F}$,
is a Frege system
augmented with a substitution rule $A/ A\sigma $. Thus a proof in
a substitution Frege system $s\mathcal{F}$ is a sequence
$\varphi _1, \ldots ,\varphi _n$, where every $\varphi _i$
is an axiom of $\mathcal F$, or inferred from earlier $\varphi _j$ by
a rule of $\mathcal F$, or $\varphi _i = \varphi _j\sigma$, for some
$j*m$,
$\varphi _i$ is not an instance of the extension rule.
Clearly, we have
\[\varphi _1, \ldots ,\varphi _m\vdash _{\mathcal{F}} \varphi _n.\]
Observe that this Frege proof has size at most $k$.
By the Deduction Theorem (Lecture $\# 2$) we have
\begin{equation}
\label{obladi}
\vdash _{\mathcal{F}}\left( {\bigwedge \!\!\!\! \bigwedge}_{i=1}^m
(z_i \leftrightarrow \chi _i)\right) \rightarrow \varphi _n,
\end{equation}
where the proof has size polynomial in $k$. W.l.o.g.\ we associate
the brackets in the big conjunction to the left. From (\ref{obladi})
we get
\begin{equation}
\label{oblada}
\vdash _{\mathcal{F}}\left( {\bigwedge \!\!\!\! \bigwedge}_{i=1}^{m-1}
(z_i \leftrightarrow \chi _i)
\wedge (z_m \leftrightarrow \chi _m)\right) \rightarrow \varphi _n.
\end{equation}
Note that the variable $z_m$ only occurs at one place in the whole
formula. Now we hit the formula
in (\ref{oblada}) with a substitution $\sigma (z_m) =\chi _m$ and obtain
\[\vdash _{\mathcal{F}}\left( {\bigwedge \!\!\!\! \bigwedge}_{i=1}^{m-1}
(z_i \leftrightarrow \chi _i)
\wedge (\chi _m\leftrightarrow \chi _m)\right) \rightarrow \varphi _n.\]
From this we get
\[\vdash _{\mathcal{F}}\left( {\bigwedge \!\!\!\! \bigwedge}_{i=1}^{m-1}
(z_i \leftrightarrow \chi _i)\right) \rightarrow \varphi _n.\]
Then we repeat this procedure for $m-1$, then for $m-2$, etc., so that
we finally end up with
\[\vdash _{\mathcal{F}}\varphi _n. \]
About the size of the proof: for (\ref{obladi}) we have
already observed that the proof is of size polynomial in $k$.
From this it is easy to
conclude that the other proofs have size polynomial in $k$ as well.
\end{proof}
It is interesting to ask whether
one can restrict the substitutions in the substitution rule
and still get systems that are as strong as
$s\mathcal{F}$ systems. Renaming Frege systems are an example of this.
A substitution is a {\em renaming substitution} if its range
is contained in the set of propositional variables.
A {\em renaming Frege system}, $r\mathcal{F}$, is a substitution
Frege system for which in the substitution rule only renaming substitutions
are allowed. A renaming substitutions $\sigma$
replaces the variables in $A$ by (possibly) other variables (hence the name
renaming).
At first sight renaming Frege systems may seem weaker than substitution Frege
systems, but it turns out that both systems p-simulate each other (a proof
can be found in \cite{Buss:propproofs}). Hence
extended Frege systems and renaming Frege systems p-simulate each
other as well.
One could try to restrict the substitutions that are allowed in the
substitution rule even further than in renaming Frege systems. For example,
one could require that the substitutions map variables only to variables
that occur in the last line of the proof, or one could require
that the renaming substitutions are injective. For these systems
it is no longer known whether they are as strong as extended
Frege systems or not.
A propositional proof systems that is considered stronger than
extended Frege systems, is
quantified propositional logic $QBF$. $QBF$ is an extension of
propositional logic by propositional quantifiers
\[\forall x \mbox{ (intended meaning $A(x/ \top)\wedge A(x/ \bot )$)}\]
\[\exists x \mbox{ (intended meaning $A(x/ \top)\vee A(x/ \bot )$)}\]
It is known that $QBF$ p-simulates extended Frege systems
\cite{Krajicek:book}. It is
conjectured that the converse does not hold, but this is open.
\section{The best known lower bounds on proof lengths}
In this section we will discuss the best known lower bounds on
Frege and extended Frege proofs. For Frege proofs we have
a linear lower bound on the number of lines
and a quadratic lower bound on the size. Hence for extended Frege proofs
we have a linear lower bound on the size of proofs. Since the best known
upper bounds on Frege proofs are exponential, there is still a large
gap between the lower bounds and the upper bounds. To
prove the lower bounds we need some terminology.
\begin{definition} Given a Frege proof $P$, a formula $A$ is called
{\em active} in $P$ if it occurs in $P$ as a subformula in an
inference that explicitly uses the principal connective of $A$.
We tacitly assume here that there is no ambiguity
as to what rule is applied in a certain inference; one can always label
the rules to avoid ambiguity of this kind.
\end{definition}
\begin{example}
In the rule Modus Ponens
\[\frac{\textstyle A\ \ \ \ A\rightarrow B}{B}\]
the formula $A\rightarrow B$ is the only formula that is made active by
this inference. The axiom $A\rightarrow (B\rightarrow A)$ makes
the formulas $A\rightarrow (B\rightarrow A)$ and $(B\rightarrow A)$ active.
\end{example}
The intuition here is that in a proof the inactive formulas can be
changed while the proof remains valid. This is the content of the
next claim, of which we omit the proof.
\begin{claim}
\label{inactive-subformulas}
If $A$ is not active in a proof $P$, then if $A$
is everywhere replaced by $B$, the result is still a valid proof.
\end{claim}
\begin{claim}
Let $c$ be the maximum number of connectives shown in any inference
rule or axiom of a Frege system $\mathcal F$ (for $\mathcal{F}_0$, $c=6$),
and let $A_1, \ldots ,A_k$ be
all the distinct active formulas in an $\mathcal F$-proof $P$, then
$|P|\geq \frac{1}{c}(\sum _{i=1}^k |A_i|)$.
\end{claim}
\begin{proof} Observe that any inference activates at most $c$ formulas, and
that if a subformula of a formula is active, then so is the formula of
which it is a subformula.
Therefore, given a symbol in the proof $P$, it lies in at most $c$ activated
occurrences of $A_1, \ldots , A_k$. Moreover, every $A_i$ is activated
somewhere. This implies
that $c\cdot |P| \geq \sum _{i=1}^k |A_i|$. Hence
$|P|\geq \frac{1}{c}(\sum _{i=1}^k |A_i|)$.
\end{proof}
Let $\varphi _n$ be the formula
\[\bot \vee (\bot \vee ( \ldots (\bot \vee \top)\ldots )\]
in which $n$ $\bot$'s occur. $\top$ denotes ``true'': $\top = x\vee \neg x$.
$\bot$ denotes ``falsum'': $\bot =x\wedge \neg x$ (sometimes these
symbols are added to the language of propositional logic).
\begin{claim} In any proof of $\varphi _n$ all the $n$ subformulas
of $\varphi _n$ that are distinct from $\bot$ and $\top$ are active.
\end{claim}
\begin{proof}
If not, by Claim~\ref{inactive-subformulas} we could replace such an
inactive subformula by $\bot$ and obtain a valid formula as well.
This cannot be, as such a formula would not be a tautology.
\end{proof}
By the previous two claims, any Frege proof of $\varphi _n$ has
at least $\sum _{i=1}^n i$ symbols. Thus any Frege proof of $\varphi _n$
has $\Omega (n^2)$ symbols.
\begin{claim} Let $c$ be the maximum number of connectives shown
in any inference rule or axiom of a Frege system $\mathcal F$.
If an $\mathcal{F}$-proof $P$ has $k$ distinct active subformulas,
then $P$ has at least $\frac{k}{c}$ lines.
\end{claim}
Thus any Frege proof of $\varphi _n$ has $\Omega (n)$ lines. Hence
any extended Frege proof of $\varphi _n$ has size $\Omega (n)$.
\section{Resolution}
Resolution is an algorithm to prove formulas that are of
a certain syntactic form. It arose in the 50's when people
were looking for efficient theorem provers. The algorithm is simple; it
has only one rule, the so-called Resolution Rule. The
drawback is that certain formulas have long Resolution proofs
compared to their Frege proofs. The Pigeonhole Principle is an example of
this. In one of the next lectures we will see that it has no polynomial size
Resolution proof, while it has polynomial size Frege proofs \cite{Buss:PHP}.
As said, Resolution can only be applied to formulas of a special
kind. Namely, the formulas in Conjunctive Normal Form (CNF).
A formula is said to be in
CNF if it is the conjunction of disjunctions of variables and negated
variables, i.e.
if it is of the form
\[{\bigwedge \!\!\!\! \bigwedge}_i \left( {\bigvee \!\!\!\! \bigvee}_j A_{ij}
\right) \]
where the $A_{ij}$'s are of the form $x$ or $\neg x$, for some variable $x$.
Note that every formula can be written in CNF. For example,
\[\begin{array}{lll}
(x\rightarrow y) &\leftrightarrow & (\neg x \vee y) \\
(\neg (x\wedge y) \wedge z) &\leftrightarrow &((\neg x \vee \neg y)
\wedge z) \\
(x \vee (y \wedge z)) & \leftrightarrow & ((x \vee y)\wedge (x\vee z))
\end{array}\]
where the formulas at the right side are in CNF.
Observe that in going to CNF the size of a formula can increase
exponentially.
\begin{definition} We define what a Resolution Refutation is.
\noindent Syntax: variables $x_1,x_2, \ldots $
\noindent Literals: $x_i$, $\bar{x_i}$ (the intended meaning of $\bar{x_i}$
is $\neg x_i$). If $x$ is a literal, then
$\bar{x}$ is defined so that $\bar{\bar{x}}=x$: $\bar{x}=y$, when
$x=\bar{y}$, and $\bar{x}=\bar{x}$ otherwise.
\noindent A {\em Clause} is a set of literals. The intended meaning of a
clause is the disjunction of its members. We call a set of clauses
$\Gamma$ satisfiable if there exists a truth-assignment that satisfies all
clauses in $\Gamma$.
\begin{example}
\begin{itemize}
\item $\{\bar{x}, y\}$ means $\neg x \vee y$.
\item $\{x,\bar{x}\}$ is always valid.
\item $\{\} =\emptyset$ is the unsatisfiable clause.
\end{itemize}
\end{example}
The {\em Resolution Rule}: ($C$ and $D$ denote clauses)
\[\frac{\textstyle C\cup \{x\} \ \ \ \ D \cup \{\bar{x}\} }{C\cup D}\]
The clause $C\cup D$ is called the {\em resolvent} of the rule.
A {\em Resolution Refutation} of a set of clauses $\Gamma$ consists of a
sequence of clauses $C_1,\ldots ,C_n$, where $C_n=\emptyset$, and for
each $i\leq n$ either
\begin{enumerate}
\item $C_i \in \Gamma$
\item $C_i$ is inferred by the Resolution Rule from $C_j$ and $C_h$,
for some $j,h < i$.
\end{enumerate}
\end{definition}
The idea behind a Resolution Refutation is the following.
Assuming that $\Gamma$ is
satisfiable we infer other clauses that are also
satisfiable till we end up with the empty clause. Since the empty clause is
not satisfiable, the assumption that $\Gamma$ is satisfiable is refuted.
Thus $\Gamma$ is not satisfiable. The following claims make this precise.
W.l.o.g.\ we can disallow having
both $x$ and $\bar{x}$ in a clause in a proof.
\begin{claim} If the truth-assignment
$\tau$ satisfies the clauses $C\cup \{x\}$ and
$D \cup \{\bar{x}\}$, then $\tau$ satisfies the resolvent $C\cup D$.
\end{claim}
\begin{theorem}[Soundness Theorem] If there exists a Resolution
Refutation for $\Gamma$, then $\Gamma$ is unsatisfiable.
\end{theorem}
\begin{proof} Suppose $\Gamma$ has a Resolution
Refutation $C_1, \ldots ,C_n$. If there would exist a truth-assignment
$\tau$ that satisfies $\Gamma$, then by the previous claim $\tau$ would
satisfy all $C_i$. Hence $\tau$ would satisfy the unsatisfiable empty clause
$C_n$, quod non. Thus $\Gamma$ is unsatisfiable.
\end{proof}
\bibliography{logic,cstheory}
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