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\title{\lecnumber. First-Order Theories}
\date{}
\institute[Stanford University]

\begin{document}

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\begin{frame}[c]
\Large
\textsc{The Calculus of Computation:}\\
Decision Procedures with\\
Applications to Verification

\bigskip
\bigskip
by\\
Aaron Bradley\\
Zohar Manna

\bigskip
\bigskip
Springer 2007
\end{frame}

\begin{frame}
	\titlepage
\end{frame}

\begin{frame}
\frametitle{First-Order Theories}
\underline{First-order theory} $T$ defined by
\begin{itemize}
	\item \underline{Signature} $\Sigma$ - set of constant, function, and predicate symbols
	\item Set of \underline{axioms} $A_T$ - set of \underline{closed} (no free variables) $\Sigma$-formulae
\end{itemize}
\bigskip
\underline{$\Sigma$-formula} constructed of constants, functions, and predicate symbols from $\Sigma$,
and variables, logical connectives, and quantifiers

\bigskip
The symbols of $\Sigma$ are \underline{just symbols} without prior meaning --- the axioms of $T$ provide their meaning

\bigskip
A $\Sigma$-formula $F$ is \underline{valid in theory $T$} (\underline{$T$-valid}, also $T \Models F$),\\
if every interpretation $I$ that satisfies the axioms of $T$,\\
\quad i.e. $I\Models A$ for every $A \in A_T$ ($T$-interpretation)\\
also satisfies $F$,\\
\quad i.e. $I\Models F$
\end{frame}

\begin{frame}
A $\Sigma$-formula $F$ is \underline{satisfiable in $T$ ($T$-satisfiable)},
if there is a $T$-interpretation (i.e. satisfies all the axioms of $T$)
that satisfies $F$

\bigskip
Two formulae $F_1$ and $F_2$ are \underline{equivalent in $T$ ($T$-equivalent)},
if $T \Models F_1 \biff F_2$, \\
	\quad i.e. if for every $T$-interpretation $I$, $I\Models F_1$ iff $I\Models F_2$

\bigskip
A \underline{fragment of theory $T$} is a syntactically-restricted subset of formulae of the theory.

\smallskip
\quad \underline{Example}: \underline{quantifier-free segment} of theory $T$ is the set of\\
\quad quantifier-free formulae in $T$.

\bigskip
A theory $T$ is \underline{decidable} if $T\Models F$ ($T$-validity) is decidable for every $\Sigma$-formula $F$,\\
\quad i.e., there is an algorithm that always terminate with ``yes'',\\
\quad if $F$ is $T$-valid, and ``no'', if $F$ is $T$-invalid.

\smallskip
A fragment of $T$ is \underline{decidable} if $T\Models F$ is decidable for every $\Sigma$-formula $F$ in the fragment.
\end{frame}

\begin{frame}[shrink=10]
\frametitle{Theory of Equality $T_E$}
\underline{Signature}\\
\qquad \qquad
$\Sigma_= : \set{=,a,b,c,\cdots,f,g,h,\cdots,p,q,r,\cdots}$\\
consists of
\begin{itemize}
	\item $=$, a binary predicate, \underline{interpreted} by axioms.
	\item all constant, function, and predicate symbols.
\end{itemize}

\smallskip
\underline{Axioms of $T_E$}
\begin{enumerate}
\item $\forall x.\ x = x$ \naxiom{reflexivity}
\item $\forall x,y.\ x = y \bimplies y = x$ \naxiom{symmetry}
\item $\forall x,y,z.\ x = y \band y = z \bimplies x = z$ \naxiom{transitivity}
\item for each positive integer $n$ and $n$-ary function symbol $f$,
$ \forall x_1,\ldots,x_n,y_1,\ldots,y_n.\ \bigwedge_i x_i = y_i \bimplies f(x_1,\ldots,x_n) = f(y_1,\ldots,y_n) $
\naxiom{congruence}
\item for each positive integer $n$ and $n$-ary predicate symbol $p$,
$ \forall x_1,\ldots,x_n,y_1,\ldots,y_n.\ \bigwedge_i x_i = y_i \bimplies (p(x_1,\ldots,x_n) \sbiff p(y_1,\ldots,y_n)) $
\naxiom{equivalence}
\end{enumerate}
Congruence and Equivalence are \underline{axiom schemata}.
For example, Congruence for binary function $f_2$ for $n=2$:\\
\qquad $ \forall x_1, x_2, y_1, y_2.\ x_1=y_1 \band x_2=y_2 \bimplies f_2(x_1,x_2)=f_2(y_1,y_2) $
\end{frame}

\begin{frame}[shrink=5]
\fbox{\begin{minipage}{.9\textwidth}
$T_E$ is undecidable.

\smallskip
The quantifier-free fragment of $T_E$ is decidable. 
Very efficient algorithm.
\end{minipage}}

\smallskip
Semantic argument method can be used for $T_E$\\
\underline{Example}: Prove\\
\smallskip
\qquad $F:\ a = b \band b = c \bimplies g(f(a), b) = g(f(c), a)$ \quad $T_E$-valid.\\
\smallskip
Suppose not; then there exists a $\teuf$-interpretation $I$
such that $I \Nmodels F$.  Then,
\[
\begin{nsemantics}
1. & I & \Nmodels & F & \mbox{assumption} \\
2. & I & \Models & a = b \band b = c & \mbox{1, $\bimplies$} \\
3. & I & \Nmodels & g(f(a), b) = g(f(c), a) & \mbox{1, $\bimplies$} \\
4. & I & \Models & a = b & \mbox{2, $\band$} \\
5. & I & \Models & b = c & \mbox{2, $\band$} \\
6. & I & \Models & a = c & \mbox{4, 5, \axiom{transitivity}} \\
7. & I & \Models & f(a) = f(c) & \mbox{6, \axiom{congruence}} \\
8. & I & \Models & g(f(a), b) = g(f(c), a) & \mbox{4, 7, \axiom{congruence}, \axiom{symmetry}}
\end{nsemantics}
\]
3 and 8 are contradictory $\Bimplies$ $F$ is $\teuf$-valid
\end{frame}

\begin{frame}
\frametitle{Natural Numbers and Integers}
\begin{tabular}{ll}
Natural numbers & $\mathbb{N} = \set{0,1,2,\cdots}$ \\
Integers & $\mathbb{Z} = \set{\cdots,-2,-1,0,1,2,\cdots}$
\end{tabular}

\bigskip
Three variations:
\begin{itemize}
	\item \underline{Peano arithmetic $\tpeano$}: natural numbers with addition and multiplication
	\item \underline{Presburger arithmetic $\tpresb$}: natural numbers with addtion
	\item \underline{Theory of integers $\tint$}: integers with $+, -, >$
\end{itemize}
\end{frame}

\begin{frame}
1. \underline{Peano Arithmetic $T_{PA}$ (first-order arithmetic)}\\
\bigskip
\qquad $\Sigma_\peano:\ \set{0,\ 1,\ +,\ \cdot,\ =}$\\
\bigskip
The axioms:
\begin{enumerate}
\item $\forall x.\ \bnot (x + 1 = 0)$ \naxiom{zero}
\item $\forall x,y.\ x + 1 = y + 1 \bimplies x = y$ \naxiom{successor}
\item $F[0] \band (\forall x.\ F[x] \bimplies F[x + 1]) \bimplies \forall x.\ F[x]$
  \naxiom{induction}
\item $\forall x.\ x + 0 = x$ \naxiom{plus zero}
\item $\forall x, y.\ x + (y + 1) = (x + y) + 1$ \naxiom{plus successor}
\item $\forall x.\ x \cdot 0 = 0$ \naxiom{times zero}
\item $\forall x, y.\ x \cdot (y + 1) = x \cdot y + x$ \naxiom{times successor}
\end{enumerate}
Line 3 is an axiom schema.

\bigskip
\underline{Example}: $3 x + 5 = 2 y$ can be written using $\Sigma_\peano$ as
\[
x + x + x + 1 + 1 + 1 + 1 + 1 = y + y
\]

\end{frame}

\begin{frame}
We have $>$ and $\ge$ since\\
\qquad $3x + 5 > 2y$ \quad write as \quad $\exists z.\ z \ne 0 \band 3x + 5 = 2y + z$\\
\qquad $3x + 5 \ge 2y$ \quad write as \quad $\exists z.\ 3x + 5 = 2y + z$

\bigskip
\underline{Example}:
\begin{itemize}
\item Pythagorean Theorem is $\tpeano$-valid\\
\qquad $\exists x, y, z.\ x \ne 0 \band y \ne 0 \band z \ne 0 \band x x + y y = z z$

\item Fermat's Last Theorem is $\tpeano$-invalid (Andrew Wiles, 1994)\\
$\exists n.\,n>2 \bimplies \exists x, y, z.\,x \ne 0 \sband y \ne 0 \sband z \ne 0 \sband x^n + y^n = z^n$
\end{itemize}

\underline{Remark} (\godel 's first incompleteness theorem)\\
Peano arithmetic $T_{PA}$ does not capture true arithmetic:\\
There exist closed $\Sigma_{PA}$-formulae representing valid propositions
of number theory that are not $T_{PA}$-valid.\\
The reason: $T_{PA}$ actually admits \underline{nonstandard interpretations}\\
\begin{center}
\begin{tabular}{|c|}
\hline
Satisfiability and validity in $T_{PA}$ is undecidable.\\
Restricted theory -- no multiplication\\
\hline
\end{tabular}
\end{center}
\end{frame}

\begin{frame}
2. \underline{Presburger Arithmetic $T_\mathbb{N}$}\\
\bigskip
\qquad $\Sigma_\presb:\ \set{0,\ 1,\ +,\ =}$ \qquad \qquad no multiplication!\\
\bigskip
Axioms $\tpresb$:
\begin{enumerate}
\item $\forall x.\ \bnot (x + 1 = 0)$ \naxiom{zero}
\item $\forall x,y.\ x + 1 = y + 1 \bimplies x = y$ \naxiom{successor}
\item $F[0] \band (\forall x.\ F[x] \bimplies F[x + 1]) \bimplies \forall x.\ F[x]$
  \naxiom{induction}
\item $\forall x.\ x + 0 = x$ \naxiom{plus zero}
\item $\forall x, y.\ x + (y + 1) = (x + y) + 1$ \naxiom{plus successor}
\end{enumerate}

\smallskip
3 is an axiom schema.

\bigskip
\begin{center}
\begin{tabular}{|l|}
\hline
$\tpresb$-satisfiability and $\tpresb$-validity are decidable\\
(Presburger, 1929)\\
\hline
\end{tabular}
\end{center}
\end{frame}

\begin{frame}
3. \underline{Theory of Integers $T_\mathbb{Z}$}\\
\bigskip
$\Sigma_\integers:\ \set{\ldots, -2, -1, 0,\ 1,\ 2,\ \ldots, -3 \cdot, -2 \cdot,\ 2 \cdot,\ 3 \cdot,\ \ldots,\ +,\ -,\ =,\ >}$\\
\smallskip
where
\begin{itemize}
\item $\ldots, -2, -1, 0,\ 1,\ 2,\ \ldots$ are constants
\item $\ldots, -3 \cdot, -2 \cdot, 2 \cdot,\ 3 \cdot,\
\ldots$ are unary functions\\
\qquad (intended $2\cdot x$ is $2x$)
\item $+, -, =, >$
\end{itemize}
\begin{center}
\fbox{$\tint$ and $\tpresb$ have the same expressiveness}
\end{center}
$\bullet$ \underline{Every $\tint$-formula can be reduced to $\spresb$-formula.}

\bigskip
\underline{Example}: Consider the $\tint$-formula\\
\smallskip
$F_0:\ \forall w, x.\ \exists y, z.\ x + 2y - z - 13 > -3w + 5$

\bigskip
Introduce two variables, $v_p$ and $v_n$ (range over the nonnegative integers)
for each variable $v$ (range over the integers) of $F_0$
\end{frame}

\begin{frame}
\[
F_1:\
\renewcommand{\arraystretch}{1.2}
\begin{array}{l}
\forall w_p, w_n, x_p, x_n.\ \exists y_p, y_n, z_p, z_n. \\
\hspace*{1em}
(x_p - x_n) + 2 (y_p - y_n) - (z_p - z_n) - 13 > -3(w_p - w_n) + 5
\end{array}
\]
Eliminate $-$ by moving to the other side of $>$
\[
F_2:\
\renewcommand{\arraystretch}{1.2}
\begin{array}{l}
\forall w_p, w_n, x_p, x_n.\ \exists y_p, y_n, z_p, z_n. \\
\hspace*{1em}
x_p + 2y_p + z_n + 3w_p >
x_n + 2y_n + z_p + 13 + 3w_n + 5
\end{array}
\]
Eliminate $>$
\[
F_3:\
\begin{array}{l}
\forall w_p, w_n, x_p, x_n.\ \exists y_p, y_n, z_p, z_n.\ \exists u. \\
\hspace*{1em}
\bnot (u = 0) \band \\
\hspace*{1em}
x_p + y_p + y_p + z_n + w_p + w_p + w_p \\
\hspace*{2em}
= x_n + y_n + y_n + z_p + w_n + w_n + w_n + u \\
\hspace*{3em}
+ 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 \\
\hspace*{3em}
+ 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1~.
\end{array}
\]
which is a $\tpresb$-formula equivalent to $F_0$.
\end{frame}

\begin{frame}
$\bullet$ \underline{Every $\tpresb$-formula can be reduced to $\sint$-formula.}\\
\medskip
\underline{Example}: 
To decide the $\tpresb$-validity of the $\tpresb$-formula
\[
\forall x.\ \exists y.\ x = y + 1
\]
decide the $\tint$-validity of the $\tint$-formula
\[
\forall x.\ x \ge 0 \bimplies \exists y.\ y \ge 0 \band x = y + 1~,
\]
where $t_1 \ge t_2$ expands to $t_1 = t_2 \bor t_1 > t_2$

\medskip
\begin{center}
\fbox{$\tint$-satisfiability and $\tpresb$-validity is decidable}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Rationals and Reals}
\qquad$\Sigma = \set{0,\ 1,\ +,\ -,\ =,\ \ge}$

\bigskip
\begin{itemize}
\item Theory of Reals $\talg$ (with multiplication)
\[ x^2=2 \quad\Bimplies\quad x = \pm \sqrt{2} \]

\item Theory of Rationals $\trat$ (no multiplication)
\[ \underbrace{2x}_{x+x} = 7 \quad\Bimplies\quad x = \frac{2}{7} \]

\end{itemize}

\bigskip
\underline{Note}: Strict inequality OK\\
\smallskip
\qquad\qquad $\forall x,y.\ \exists z.\ x+y>z$\\
\smallskip
\qquad rewrite as\\
\smallskip
\qquad\qquad $\forall x,y.\ \exists z.\ \neg (x+y=z) \band x+y \ge z$
\end{frame}

\begin{frame}
\underline{1. Theory of Reals $\talg$}\\
\bigskip
\qquad $\Sigma_\alg:\ \set{0,\ 1,\ +,\ -,\ \cdot,\ =,\ \ge}$

\smallskip
with multiplication.\\
Axioms in text.

\bigskip
\underline{Example}:
\[
\forall a,b,c.\ b^2 - 4ac \ge 0 \biff \exists x.\ a x^2 + b x + c = 0
\]
is $\talg$-valid.

\begin{center}
\begin{tabular}{|l|}
\hline
$\talg$ is decidable (Tarski, 1930)\\
High time complexity\\
\hline
\end{tabular}
\end{center}
\end{frame}

\begin{frame}
\underline{2. Theory of Rationals $\trat$}\\
\bigskip
\qquad $\Sigma_\rat:\ \set{0,\ 1,\ +,\ -,\ =,\ \ge}$

\smallskip
without multiplication.\\
Axioms in text.

\bigskip
Rational coefficients are simple to express in $\trat$

\bigskip
\underline{Example}: Rewrite
\[
\frac{1}{2} x + \frac{2}{3} y \ge 4
\]
as the $\srat$-formula
\[
3 x + 4 y \ge 24
\]

\begin{center}
\begin{tabular}{|l|}
\hline
$\trat$ is decidable\\
Quantifier-free fragment of $\trat$ is efficiently decidable\\
\hline
\end{tabular}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Recursive Data Structures (RDS)}
\underline{1. RDS theory of LISP-like lists, $\tcons$}
\[
\Sigma_\cons:\ \set{\cons,\ \car,\ \cdr,\ \atom,\ =}
\]
where\\
\smallskip
\begin{tabular}{l@{\ --\ }l}
$\cons(a, b)$ & list constructed by concatenating $a$ and $b$\\
$\car(x)$ & left projector of $x$: $\car(\cons(a, b)) = a$\\
$\cdr(x)$ & right projector of $x$: $\cdr(\cons(a, b)) = b$\\
$\atom(x)$ & $\truev$ iff $x$ is a single-element list
\end{tabular}

\bigskip
\underline{Axioms}:
\begin{enumerate}
\item The axioms of \underline{reflexivity}, \underline{symmetry}, and \underline{transitivity} of $=$
\item \underline{Congruence} axioms
  \[
  \begin{array}{l}
    \forall x_1,x_2,y_1,y_2.\ x_1 = x_2 \band y_1 = y_2 \bimplies \cons(x_1,y_1) = \cons(x_2,y_2) \\
    \forall x,y.\ x = y \bimplies \car(x) = \car(y) \\
    \forall x,y.\ x = y \bimplies \cdr(x) = \cdr(y)
  \end{array}
  \]
\end{enumerate}
\end{frame}

\begin{frame}
\begin{enumerate}
\setcounter{enumi}{2}
\item \underline{Equivalence} axiom
  \[
  \forall x,y.\ x = y \bimplies (\atom(x) \biff \atom(y))
  \]
\item $\forall x, y.\ \car(\cons(x,y)) = x$ \naxiom{left projection}
\item $\forall x, y.\ \cdr(\cons(x,y)) = y$ \naxiom{right projection}
\item $\forall x.\ \bnot \atom(x) \bimplies \cons(\car(x), \cdr(x)) = x$
  \naxiom{construction}
\item $\forall x, y.\ \bnot \atom(\cons(x, y))$ \naxiom{atom}
\end{enumerate}

\begin{center}
\begin{tabular}{|l|}
\hline
$\tcons$ is undecidable\\
Quantifier-free fragment of $\tcons$ is efficiently decidable\\
\hline
\end{tabular}
\end{center}
\end{frame}

\begin{frame}
\underline{2. Lists + equality}\\
\medskip
\qquad $\tconse \quad = \quad \teuf \ \cup\ \tcons$

\bigskip
\underline{Signature}: \qquad $\Sigma_\euf \ \cup\ \scons$\\
\smallskip
(this includes uninterpreted constants, functions, and predicates)

\bigskip
\underline{Axioms}: union of the axioms of $\teuf$ and $\tcons$

\begin{center}
\begin{tabular}{|l|}
\hline
$\tconse$ is undecidable\\
Quantifier-free fragment of $\tconse$ is efficiently decidable\\
\hline
\end{tabular}
\end{center}

\underline{Example}:
We argue that the $\sconse$-formula
\[
F:\ \begin{array}{l}
\car(a) = \car(b) \band \cdr(a) = \cdr(b) \band \bnot \atom(a)
\band \bnot \atom(b) \\
\hspace*{1em} \bimplies f(a) = f(b)
\end{array}
\]
is $\tconse$-valid.  \end{frame}

\begin{frame}
Suppose not; then there exists a $\tconse$-interpretation $I$ such that $I \Nmodels F$.
Then,
\[
\begin{nsemantics}
1. & I & \Nmodels & F & \mbox{assumption} \\
2. & I & \Models & \car(a) = \car(b) & \mbox{1, $\bimplies$, $\band$} \\
3. & I & \Models & \cdr(a) = \cdr(b) & \mbox{1, $\bimplies$, $\band$} \\
4. & I & \Models & \bnot \atom(a) & \mbox{1, $\bimplies$, $\band$} \\
5. & I & \Models & \bnot \atom(b) & \mbox{1, $\bimplies$, $\band$} \\
6. & I & \Nmodels & f(a) = f(b) & \mbox{1, $\bimplies$} \\
7. & I & \Models & \multicolumn{2}{l}{\cons(\car(a), \cdr(a)) = \cons(\car(b), \cdr(b))} \\
   &   &         & & \mbox{2, 3, \axiom{congruence}} \\
8. & I & \Models & \cons(\car(a), \cdr(a)) = a & \mbox{4, \axiom{construction}} \\
9. & I & \Models & \cons(\car(b), \cdr(b)) = b & \mbox{5, \axiom{construction}} \\
10. & I & \Models & a = b & \mbox{7, 8, 9, \axiom{transitivity}} \\
11. & I & \Models & f(a) = f(b) & \mbox{10, \axiom{congruence}}
\end{nsemantics}
\]
Lines 6 and 11 are contradictory, so our assumption that $I
\Nmodels F$ must be wrong.  Therefore, $F$ is $\tconse$-valid.

\end{frame}

\begin{frame}
\frametitle{Theory of Arrays}
\underline{1. Theory of Arrays $\tarr$}\\
\medskip
\underline{Signature}
\[
\Sigma_\arr:\ \set{\ard{\cdot}{\cdot},\ \awt{\cdot}{\cdot}{\cdot},\ =}
\]
where
\begin{itemize}
\item $\ard{a}{i}$ \quad binary function --\\
 read array $a$ at index $i$ (``read($a$,$i$)'')
\item $\awt{a}{i}{v}$ \quad ternary function --\\
 write value $v$ to index $i$ of array $a$ (``write($a$,$i$,$e$)'')
\end{itemize}

\medskip
\underline{Axioms}
\begin{enumerate}
\item the axioms of \axiom{reflexivity}, \axiom{symmetry}, and
  \axiom{transitivity} of $\tequ$
\item $\forall a,i,j.\ i = j \bimplies \ard{a}{i} = \ard{a}{j}$
  \naxiom{array congruence}
\item $\forall a,v,i,j.\ i = j \bimplies \ard{\awt{a}{i}{v}}{j} = v$
  \naxiom{read-over-write 1}
\item $\forall a,v,i,j.\ i \ne j \bimplies \ard{\awt{a}{i}{v}}{j} = \ard{a}{j}$
  \naxiom{read-over-write 2}
\end{enumerate}

\end{frame}

\begin{frame}
\underline{Note}: $=$ is only defined for array elements
\[
F:\ \ard{a}{i} = e \bimplies \awt{a}{i}{e} = a
\]
not $\tarr$-valid, but
\[
F':\ \ard{a}{i} = e \bimplies \forall j.\ \ard{\awt{a}{i}{e}}{j} = \ard{a}{j}~,
\]
is $\tarr$-valid.

\begin{center}
\begin{tabular}{|l|}
\hline
$\tarr$ is undecidable\\
Quantifier-free fragment of $\tarr$ is decidable\\
\hline
\end{tabular}
\end{center}
\end{frame}

\begin{frame}
\underline{2. Theory of Arrays $\tarre$ (with extensionality)}\\
\smallskip
Signature and axioms of $\tarre$ are the same as $\tarr$, with one additional axiom
\[
\forall a, b.\
(\forall i.\ \ard{a}{i} = \ard{b}{i}) \biff a = b \quad
\naxiom{extensionality}
\]
\underline{Example}:
\[
F:\ \ard{a}{i} = e \bimplies \awt{a}{i}{e} = a
\]
is $\tarre$-valid.

\begin{center}
\begin{tabular}{|l|}
\hline
$\tarre$ is undecidable\\
Quantifier-free fragment of $\tarre$ is decidable\\
\hline
\end{tabular}
\end{center}
\end{frame}

\begin{frame}
\frametitle{Combination of Theories}
How do we show that
\[ 1 \le x \band x \le 2 \band f(x) \neq f(1) \band f(x) \neq f(2) \]
is $(\teuf\ \cup\ \tint)$-unsatisfiable?

\smallskip
Or how do we prove properties about\\
\qquad an array of integers, or\\
\qquad a list of reals \ldots?

\bigskip
Given theories $T_1$ and $T_2$ such that
\[ \Sigma_1 \ \cap \ \Sigma_2 \quad = \quad \set{=} \]
The \underline{combined theory} $T_1\ \cup\ T_2$ has
\begin{itemize}
\item signature $\Sigma_1\ \cup \ \Sigma_2$
\item axioms $A_1 \ \cup \ A_2$
\end{itemize}
\end{frame}

\begin{frame}
\begin{flushright}
\fbox{qff = quantifier-free fragment}
\end{flushright}

Nelson \& Oppen showed that\\
\smallskip
\qquad if satisfiability of qff of $T_1$ is decidable,\\
\qquad \quad satisfiability of qff of $T_2$ is decidable, and\\
\qquad \quad certain technical simple requirements are met\\
\qquad then satisfiability of qff of $T_1 \ \cup \ T_2$ is decidable.
\end{frame}
\end{document}
