\documentclass[12pt]{article} \usepackage{amssymb, amsmath} \textheight 23cm \textwidth 6.5in \oddsidemargin -.1in \evensidemargin -.1in \topmargin -1.5cm \input{prooftree} \renewcommand{\theenumi}{\arabic{enumi}} \renewcommand{\labelenumi}{\theenumi)} \newtheorem{exer}{Exercise} \newcommand{\eqsyn}{=_{syn}} \newcommand{\ra}{\rightarrow} \newcommand{\SL}{{\cal L}} \newcommand{\N}{\mathbb{N}} \newcommand{\underN}{\underline{\mathbb{N}}} \newcommand{\calM}{{\cal M}} \def\Equiv{\Longleftrightarrow} \def\Konig{K\"{o}nig} \parskip 2ex \parindent 0ex \setcounter{page}{18} \begin{document} {\large\bf CS 4995}\hspace{.8in}{\large\bf Notes} (S. Cook and T. Pitassi) \hfill {\large\bf Fall, 2022} \begin{center} {\Large\bf Predicate Calculus}\\ (First-Order Logic) \end{center} \bigskip {\large\bf Syntax} \bigskip A {\em first-order vocabulary} (or just {\em vocabulary} or {\em language}) $\cal L$ is specified by the following: \begin{enumerate} \item For each $n \in {\mathbb N}$ a set of $n$-ary function symbols (possibly empty). We use $f,g,h,...$ and also $+,\cdot, s$ as metasymbols for function symbols. A zero-ary function symbol is called a constant symbol. \item For each $n \geq 0$, a set of $n$-ary predicate symbols (must be non-empty for some $n$). We use $P,Q,R,...$ and also $<, \le, =$ as metasymbols for predicate symbols. A zero-ary predicate symbol is the same as a propositional atom. \end{enumerate} In addition, the following symbols are available to build first-order formulas: \begin{enumerate} \item An infinite set of variables. We use $x,y,z,...$ and sometimes $a,b,c,...$ as metasymbols for variables. (Generally distinct letters $x,y,z$ stand for distinct variables.) \item connectives $\neg, \wedge,\vee $ (not, and, or) \item quantifiers $\forall, \exists$ (for all, there exists) \item (,) (parentheses) \end{enumerate} Terms and Formulas are built from these together with the function and predicate symbols from ${\cal L}$, as described below. The standard vocabulary of arithmetic is $${\cal L}_A = [0, s, +, \cdot \ ; =]$$ \begin{tabbing} $+, \ast$ \= binary function symbol \kill 0 \> constant (zero-ary function symbol)\\ $s$ \> unary function symbol\\ $+, \cdot$ \> binary function symbols\\ $=$ \> binary predicate symbol \end{tabbing} \newpage {\em Terms} (or {\em expressions}) are certain strings built from variables and function symbols, and are intended to represent objects in the universe of discourse. {\bf Definition of an ${\cal L}$-term} \label{Lterm} (Here ${\cal L}$ is a first-order vocabulary): \begin{enumerate} \item Every variable is a term. \item If $f$ is an $n$-ary function symbol of ${\cal L}$ and $t_1, \ldots, t_n$ are $\SL$-terms then $ft_1 \dots t_n$ is an $\SL$-term. \end{enumerate} We will drop mention of $\SL$ when it is not important, or clear from context. Recall that a 0-ary function symbol is called a constant symbol (or sometimes just a {\em constant}). We use $e$ as a metasymbol for constants. Also 0 and 1 are constants. Note that all constants in $\SL$ are $\SL$-terms. {\bf Examples} of ${\cal L}$-terms (where $f$ is binary and $g$ is unary):\\ $fgex, fxy, gfege$. These are parsed $f(g(e),x), f(x,y), g(f(e,g(e))$ respectively. {\bf Unique Readability Theorem for Terms:} If terms $ft_1\cdots t_k$ and $fu_1\cdots u_\ell$ are syntactically equal, then $k=\ell$ and $t_i =_{syn} u_i$, $1\le i\le k$. {\bf Proof:} Similar to the Unique Readability Theorem for propositional formulas (see page 2). To prove the lemma on weights, we assign a weight of $n-1$ to each $n$-ary function symbol, and -1 to each variable. \hfill $\Box$ \begin{exer} Carry out the details in the above argument. \end{exer} {\bf Notation:} We use $r,s,t,...$ to denote terms. In the vocabulary for arithmetic ${\cal L}_A$, in practice we write $+, \cdot$ as though they were infix operators, even though officially they are prefix operators. Thus {\bf Notation} $(t_1 \cdot t_2) =_{syn} \cdot t_1 t_2$\\ $\mbox{}\hspace{.51in}(t_1 + t_2) =_{syn} + t_1 t_2$ Thus examples of our way of writing ${\cal L}_A$ terms are $sss0, ((x+sy)\cdot(ssz+s0))$ {\bf Definition of first-order formula in the vocabulary $\SL$ } (or $\SL$-{\em formula}, or just {\em formula}): \begin{enumerate} \item $Pt_1 \cdots t_n$ is an {\em atomic} $\SL$-formula, where $P$ is an $n$-ary predicate symbol in $\SL$ and $t_1, \cdots, t_n$ are $\SL$-terms. \item If $A$ and $B$ are $\SL$-formulas, so are $\neg A$, $(A \wedge B)$, and $(A \vee B)$ \item If $A$ is an $\SL$-formula and $x$ is a variable, then $\forall x A$ and $\exists x A$ are $\SL$-formulas. \end{enumerate} As in the case of propositional formulas, we use the notation\\ $(A \supset B)$ for $(\neg A \vee B)$\\ $(A \leftrightarrow B)$ for $(A \supset B)\wedge (B \supset A)$ Examples of formulas: $(\neg \forall x Px \vee \exists x \neg Px)$ (Here $P$ is a unary predicate symbol.)\\ $(\forall x \neg Qxy \wedge \neg \forall z Q fyz)$. (Here $Q$ is a binary predicate symbol and $f$ is a unary function symbol.) The Unique Readability Theorem holds for first-order formulas. {\bf Notation} $r=s$ stands for $= rs$\\ $\mbox{}\hspace{.51in}r \neq s$ stands for $ \neg(r=s) $ \bigskip {\bf Example:} Goldbach's conjecture: Every even integer greater than 2 is the sum of two primes. $\forall x ((\mbox{Even}(x) \wedge x>2) \supset \exists y \exists z (\mbox{Prime}(y) \wedge \mbox{Prime}(z) \wedge x=y+z))$ Here Even, Prime are unary predicate symbols. \\ $>$ is a binary predicate symbol (we use infix notation).\\ 2 is a constant symbol.\\ + is a binary function symbol. This can also be stated as a formula in the vocabulary ${\cal L}_A$, since the predicates Even, Prime, and $>$ can be defined in terms of $s,+,\cdot,$ and =. For example, Even$(x)$ can be defined by the formula $\exists y(x=y+y)$. {\bf Free and Bound Variables} {\bf Definition}: An occurrence of $x$ in $A$ is {\em bound} iff it is in a subformula of $A$ of the form $\forall x B$ or $\exists x B$. Otherwise the occurrence is {\em free}. For example, in the formula $ \exists y(x=y+y)$ (which defines Even$(x)$ as above) the occurrence of $x$ is free, while the occurrences of $y$ are bound. Intuitively the meaning of a formula depends on the values assigned to its free variables, but no value need be assigned to a bound variable to give the formula meaning. Notice that a variable can have both free and bound occurrences in one formula. For example, in $Px \wedge \forall x Qx$, the first occurrence of $x$ is free, and the second occurrence is bound. {\bf Definition}: A formula $A$ or a term $t$ is {\em closed} if it contains no free occurrence of a variable. A closed formula is called a {\em sentence}. \bigskip {\large\bf Semantics of Predicate Calculus} \bigskip In the propositional calculus, a truth assignment provides meaning to a formula. In the predicate calculus, we need a more complicated object, called a {\em structure} (or {\em interpretation}) to give meaning to formulas and terms. If $\cal L$ is a first-order vocabulary, then an $\cal L$-structure $\cal M$ consists of the following: \begin{enumerate} \item A nonempty set $M$ called the {\em universe of discourse} (or just {\em universe}). Variables in an $\cal L$-formula range over $M$. \item For each $n$-ary function symbol $f$ in $\cal L$, an associated function $f^{\cal M} : M^n \mapsto M$. \item For each $n$-ary predicate symbol in $\cal L$, an associated relation $P^{\cal M} \subseteq M^n$. If $\cal L$ contains =, then $=^{\cal M}$ must be the true equality relation on $M$. \end{enumerate} Notice that the predicate symbol = gets special treatment in the above definition, in that $=^{\cal M}$ must always be the true equality relation. Other predicate symbols may be interpreted by arbitrary relations of the appropriate arity. For example, if $\cal L$ contains the binary predicate symbol $<$, then $<^{\cal M}$ can be any binary relation on the universe $M$, and is not necessarily an order relation. Every $\cal L$-sentence becomes either true or false when interpreted by an $\cal L$-structure $\cal M$, as explained below. If a sentence $A$ becomes true under $\cal M$, then we say $\cal M$ {\em satisfies} $A$, or $\cal M$ is a {\em model} for $A$, and write ${\cal M} \models A$. {\bf Definition:} We say that a structure $\cal M$ is {\em finite} if the universe $M$ of $\cal M$ is finite. Otherwise $\cal M$ is infinite. If $A$ has free variables, then these variables must be interpreted as specific elements in the universe $M$ before $A$ gets a truth value under the structure $\cal M$. For this we need the following: {\bf Definition:} An {\em object assignment} $\sigma$ for a structure ${\cal M}$ is a mapping from variables to the universe $M$. Below we give the formal definition of notion $\cal M \models A[\sigma]$, which is intended to mean that the structure $\cal M$ satisfies the formula $A$ when the free variables of $A$ are interpreted according to the object assignment $\sigma$. First it is necessary to define the notation $t^{\cal M}[\sigma]$, which is the element of universe $M$ assigned to the term $t$ by the structure $\cal M$ when the variables of $t$ are interpreted according to $\sigma$. \newpage {\large\bf Basic Semantic Definition} Let $\cal L$ be a vocabulary, let $\cal M$ be an $\cal L$-structure, and let $\sigma$ be an object assignment for $\cal M$. Each $\cal L$-term $t$ is assigned an element $t^{\cal M}[\sigma]$ in $M$, defined by structural induction on terms $t$, as follows (refer to the definition of $\cal L$-term, page \pageref{Lterm}): \begin{itemize} \item[a)] $x^{\cal M}[\sigma]$ is $\sigma(x)$, for each variable $x$ \item[b)] $(ft_1 \cdots t_n)^{\cal M}[\sigma] = f^{\cal M}(t_1^{\cal M}[\sigma], \ldots, t_n^{\cal M}[\sigma])$ \end{itemize} {\em Notation}: If $x$ is a variable and $m\in M$, then the object assignment $\sigma(m/x)$ is the same as $\sigma$ except $\sigma(m/x)(x)=m$. For $A$ an $\cal L$-formula, the notion $\cal M \models A[\sigma]$ ($\cal M$ {\em satisfies} $A$ {\em under} $\sigma$) is defined by structural induction on formulas $A$ as follows (refer to the definition of formula): \begin{itemize} \item[a)] ${\cal M} \models (Pt_1 \cdots t_n)[\sigma]$ iff $\langle t_1^{\cal M}[\sigma], \ldots, t_n^{\cal M}[\sigma] \rangle \in P^{\cal M}$ \item[b)] ${\cal M} \models (s=t)[\sigma]$ iff $s^{\cal M}[\sigma] = t^{\cal M}[\sigma]$ \item[c)] $\calM\models\neg A[\sigma]$ iff not $\calM\models A[\sigma]$. \item[d)] $\calM\models (A\vee B)[\sigma]$ iff $\calM\models A[\sigma]$ or $\calM\models B[\sigma]$. \item[e)] $\calM\models (A\wedge B)[\sigma]$ iff $\calM\models A[\sigma]$ and $\calM\models B[\sigma]$. \item[f)] ${\cal M} \models (\forall x A)[\sigma]$ iff $\calM \models A[\sigma(m/x)]$ for all $m \in M$ \item[g)] $\calM \models (\exists x A)[\sigma]$ iff $\calM \models A[\sigma(m/x)]$ for some $m \in M$ \end{itemize} This method of giving meaning is sometimes called Tarski semantics, named after the important logician Alfred Tarski. Note that item b) in the definition of $\calM \models A[\sigma]$ follows from a) and the fact that $=^{\cal M}$ is always the equality relation. If $t$ is a closed term (i.e. contains no variables), then $t^{\cal M}[\sigma]$ is independent of $\sigma$, and so we sometimes just write $t^{\cal M}$. Similarly, if $A$ is a sentence, then we sometimes write $\calM \models A$ instead of $\calM\models A[\sigma]$, since $\sigma$ does not matter. (See the Corollary on the next page.) {\bf Example:} Let $\SL$ be the vocabulary $\{;R,=\}$ and let ${\cal M}$ be the $\SL$-structure whose universe $M=\N$ and such that $R^{\cal M}(m,n)$ holds iff $m\leq n$. Then $ {\cal M} \models \exists x\forall y R(x,y)$ (since 0 is the least element of $\N$) but ${\cal M} \not\models \exists y\forall x R(x,y)$ since there is no largest natural number. {\bf Standard Structure:} The {\em standard structure} $\underN$ for the vocabulary ${\cal L}_A$ has universe $M = \N$ = $\{ 0,1,2,...,\}$, $s^{\underN}(n)=n+1$, and $0,+,\cdot, =$ get their usual meanings on the natural numbers. {\bf Example:} $\underN \models \forall x\forall y \exists z(x+z=y\vee y+z=x)$ (since either $y-x$ or $x-y$ exists) but $\underN \not\models \forall x \exists y(y+y=x)$ since not all natural numbers are even. In the future we sometimes assume that there is some first-order vocabulary $\cal L$ in the background, and do not necessarily mention it explicitly. {\bf Notation:} In general, $\Phi$ denotes a set of formulas, $A,B,C,...$ denote formulas, $\cal M$ denotes a structure, and $\sigma$ denotes an object assignment. {\bf Lemma:} If $\sigma$ and $\sigma'$ agree on the free variables of $A$, then $\calM\models A[\sigma]$ iff $\calM\models A[\sigma']$. {\bf Proof:} Structural induction on formulas $A$. {\bf Corollary:} If $A$ is a sentence, then for any object assignments $\sigma,\sigma'$, $\calM\models A[\sigma]$ iff $\calM\models A[\sigma']$. In view of the Corollary, if $A$ is a sentence, then $\sigma$ is irrelevant, so we omit mention of $\sigma$ and simply write $\calM \models A$. {\bf Definition:} \begin{itemize} \item[a)] $A$ is {\em satisfiable} iff $\calM\models A[\sigma]$ for some $\cal M$ and $\sigma$. \item[b)] $\calM \models \Phi[\sigma]$ iff $\calM\models A[\sigma]$ for all $A \in \Phi$. (We may omit mention of $\sigma$ if $\Phi$ is a set of sentences.) We say $\Phi$ is {\em satisfiable} if $\calM \models \Phi[\sigma]$ for some $\cal M$ and $\sigma$. \item[c)] $\Phi \models A$ iff for all $\cal M$ and all $\sigma$, if $\calM \models \Phi[\sigma]$ then $\calM\models A[\sigma]$. \item[d)] $\models A$ ($A$ is {\em valid}) iff $\calM\models A[\sigma]$ for all $\cal M$ and $\sigma$. \item[e)] $A\Longleftrightarrow B$ ($A$ and $B$ are {\em logically equivalent}, or just {\em equivalent}) iff for all $\cal M$ and all $\sigma$, $\calM\models A[\sigma]$ iff $\calM\models B[\sigma]$. \end{itemize} $\Phi \models A$ is read ``$A$ is a logical consequence of $\Phi$''. This relation is of FUNDAMENTAL IMPORTANCE. Do not confuse this with our other use of the symbol $\models$, as in $\calM\models A$ ($\calM$ satisfies $A$). In the latter, $\cal M$ is a structure, rather than a set of formulas. Note that $\models$ is a symbol of the ``meta language'' (English), as opposed to $\neg,\vee,\wedge,\forall,\exists$, which are symbols of the ``object language''. As in the propositional case, if $\Phi=\{B_1,\ldots,B_n\}$, then we sometimes write $B_1,\ldots,B_n\models A$ instead of $\{B_1,\ldots,B_n\}\models A$. \bigskip {\bf Examples:} \begin{itemize} \item[1] $ (\forall x A \vee \forall x B) \models \forall x (A \vee B)$, for all formulas $A$ and $B$. Proof: We follow the definition of $\Phi\models A$ above. Let $\calM$ be any structure and let $\sigma$ be any object assignment. Assume L.H.S. is true, i.e. $\calM \models (\forall x A \vee \forall x B)[\sigma]$. Then following the Basic Semantic Definition, $\calM\models(\forall x A)[\sigma]$ or $\calM\models(\forall x B)[\sigma]$. Say $\calM\models(\forall x A)[\sigma]$. Then $\calM\models A[\sigma(m/x)] $ for all $m \in M$. Then $\calM\models(A \vee B)[\sigma(m/x)]$ for all $m \in M$. Therefore $\calM\models\forall x (A \vee B)[\sigma] $. Similarly for the case $\calM\models(\forall x B)[\sigma]$. $\Box$ \item[2] $\forall x (A \vee B) \models (\forall x A \vee \forall x B)$? No, not necessarily. Take $A =_{syn} Px$, $B =_{syn} Qx$, define the structure $\calM$ to have universe $M = \N$, define $P^\calM$ to be the set of even natural numbers, and $Q^\calM$ to be the set of odd natural numbers. Then $\calM\models\forall x(Px\vee Qx)$ (every number is even or odd), but not $\calM\models (\forall x Px \vee \forall x Qx)$ (it is not the case that either all numbers are even or all numbers are odd). \item[3] $\neg \forall x A \Equiv \exists x\neg A$, for all formulas $A$. $\neg \exists xA\Equiv\forall x\neg A$, for all formulas $A$. $(\forall x A \wedge \forall x B) \Longleftrightarrow \forall x(A \wedge B)$, for all formulas $A$, $B$. $ \exists x (A \vee B) \Longleftrightarrow (\exists x A \vee \exists x B)$, for all formulas $A$, $B$. $\exists x (A \wedge B) \models (\exists x A \wedge \exists x B)$, for all formulas $A$, $B$. NOT $(\exists x A \wedge \exists x B) \models \exists x(A \wedge B)$ in general $\forall x\forall yA \Equiv \forall y\forall x A$ $\exists x\exists yA \Equiv \exists y\exists x A$ $ \exists y \forall x A \models \forall x \exists y A$, for all formulas $A$. NOT $\forall x \exists y A \models \exists y \forall x A$ in general $ \forall x A \models \exists x A$, because of our requirement that every universe $M$ must be nonempty. $\forall x\forall y(x=y\supset fx=fy)$ is valid. $\forall x\forall y(fx=fy\supset x=y)$ is NOT valid. \end{itemize} \begin{exer} Verify each line in item 3 above. For the two lines beginning NOT give specific formulas $A$ (and $B$) for which the relation is false, and show it is false by giving a specific structure which satisfies the left hand side but not the right hand side. For the last line, give a structure which does not satisfy the formula. \end{exer} \begin{exer} Show that $\{P0,Ps0,Pss0,...\} \not\models \forall xPx$ by giving a specific structure. \end{exer} \begin{exer} Consider the following four formulas over the vocabulary ${\cal L}_A$: P1: $\forall x(sx \ne 0)$ \\ P2: $\forall x\forall y (sx=sy \supset x=y)$\\ P3: $\forall x(x+0 = x)$ \\ P4: $\forall x\forall y(x+sy = s(x+y))$ Prove from the definition of $\models$ that $$ \mbox{P1,P2,P3,P4 $\not\models \forall x\forall y (x+y = y+x)$} $$ {\bf Hint:} Think of + as string concatenation. \end{exer} \begin{exer} Show that $\forall x(gfx=x)$ is NOT a logical consequence of $\forall x(fgx=x)$. \end{exer} \begin{exer} Let ${\cal M}$ be a structure and let $\Phi$ be the set of all sentences $A$ satisfied by ${\cal M}$. Show that $\Phi$ is closed under $\models$. That is, show that if $\Phi\models A$ then $A\in\Phi$. \end{exer} \begin{exer} Give a sentence in the vocabulary ${\cal L} = \{;=\}$ which is satisfied by a structure iff the universe has exactly three elements. \end{exer} \begin{exer} Give a satisfiable sentence $A$ in the vocabulary ${\cal L} = \{;R\}$, where $R$ is a binary predicate symbol, such that $A$ has no finite model. (Hint: Think of $R$ as an order relation.) \end{exer} \begin{exer} Give a sentence $A$ in the vocabulary ${\cal L} = \{;R,=\}$, where $R$ is a binary predicate symbol, such that for all $n\in \N, n>0$, $A$ has a model whose universe has $n$ elements iff $n$ is even. (Hint: Think of $R$ as a pairing relation.) \end{exer} \begin{exer} Give a sentence $A$ of the predicate calculus with the vocabulary ${\cal{L}}=\{;R,=\}$, where $R$ is a binary predicate symbol, such that a finite $\cal{L}$-structure (thought of as a directed graph with edge relation $R$) is a model for $A$ iff it is a disjoint union of directed cycles. Now give an infinite model for $A$. \end{exer} \bigskip Recall that a sentence is a formula with no free variables. Each sentence in the vocabulary $\cal L_A$ (the vocabulary of arithmetic) is either true or false in the standard structure $\underline{\Bbb N}$. Thus $\forall x \forall y(x+y=y+x)$ and Fermat's Last Theorem are true, while $\forall x \neg (0 = x+x)$ is false, and no one knows the truth value of Goldbach's conjecture. On the other hand, a formula such as $\forall y \neg(x=y+y)$ (``$x$ is odd'') has no truth value under any structure, since it has a free variable. Of course it gets a truth value in a structure when an object assignment $\sigma$ is specified. \bigskip {\large\bf Substitution} {\bf Syntactic Definition:} ($s, t$ are terms)\\ \mbox{}\hspace{.5in}$t(s/x)$ is the result of replacing all occurrences of $x$ in $t$ by $s$.\\ \mbox{}\hspace{.5in}$A(s/x)$ is the result of replacing all {\em free} occurrences of $x$ in $A$ by $s$. {\bf Semantics}: {\bf Lemma} For each structure $\calM$ and each object assignment $\sigma$, $$(t(s/x))^\calM[\sigma] = t^\calM[\sigma(m/x)]$$ where $m=s^\calM[\sigma]$. {\bf Example:} Let $\cal M$ be the standard structure $\underN$ for the vocabulary $\SL_A$ of arithmetic. Suppose $\sigma(x)=5$ and $\sigma(y)=7$. Let $s$ be the term $x+y$ and let $t$ be the term $ss0$ (here $s$ is the successor function in $\SL$). Then $s(t/x))$ is $ss0+y$ and so $(s(t/x))^{\underN}[\sigma]=2+7=9$. On the other hand, $m=t^{\underN}=2$, so $s^{\underN}[\sigma(m/x)]=2+7=9$, and the Lemma is verified for this case. {\bf Proof if the Lemma}: Structural induction on $t$. Base case: $t$ is a variable. If the variable is $x$, then both sides of the equation are the same, namely $s^\calM[\sigma]$. If $t$ is a variable $y$ other than $x$, then again both sides are the same, namely $\sigma(y)$. The induction step is straightforward from the Basic Semantic Definition. $\Box$ \begin{exer} Carry out the induction step in detail. \end{exer} Question: Does the above lemma apply to formulas $A$? I.e. can we say $\calM\models A(t/x)[\sigma]$ iff $\calM\models A[\sigma(m/x)]$, where $m=t^\calM[\sigma]$? Something can go wrong. Example: Suppose $A$ is $\forall y \neg (x=y+y)$. This says ``$x$ is odd''. But $A(x+y/x)$ is $\forall y \neg (x+y=y+y)$, which does not say ``$x+y$ is odd'' as desired, but instead it is always false. The problem is that $y$ in the term $x+y$ got ``caught'' by the quantifier $\forall y$. {\bf Definition} A term $t$ is {\em freely substitutable for $x$ in} $A$ iff no free occurrence of $x$ in $A$ is in a subformula of $A$ of the form $\forall y B$ or $\exists y B$, where $y$ occurs in $t$. {\bf Substitution Theorem}: If $t$ is freely substitutable for $x$ in $A$ then for all structures $\calM$ and all object assignments $\sigma$, $\calM\models A(t/x)[\sigma]$ iff $\calM\models A[\sigma(m/x)]$, where $m=t^\calM[\sigma]$. {\bf Proof:} Structural induction on $A$. The interesting case is when $A$ is $\forall y B$. (The case when $A$ is $\exists yB$ is similar). Then we are to prove \begin{equation} \label{subst} \calM\models (\forall y B)(t/x)[\sigma] \mbox{ iff } \calM\models (\forall y B)[\sigma(m/x)] \end{equation} where $m=t^\calM[\sigma]$. If $x$ does not occur free in $\forall yB$, then no substitution is done, so the result is easy. (If $x,y$ are the same variable, then $x$ does not occur free in $\forall yB$.) Hence we may assume that $x,y$ are distinct variables and $x$ occurs free in $B$. Since $t$ is freely substitutible for $x$ in $\forall y B$, $y$ does not occur in $t$. Following the Basic Semantic Definition, the LHS of (\ref{subst}) holds iff $\calM\models B(t/x)[\sigma(n/y)]$ for all $n \in M$. Apply the induction hypothesis to $B$ to obtain $$\calM\models B(t/x)[\sigma(n/y)] \mbox{ iff } \calM\models B[\sigma(n/y)(m'/x)]$$ where now $m'=t^\calM[\sigma(n/y)]$. But note that $m'=t^\calM[\sigma(n/y)]=t^\calM[\sigma] = m$ because $y$ does not occur in $t $. Hence $$\calM\models B(t/x)[\sigma(n/y)] \mbox{ iff } \calM\models B[\sigma(n/y)(m/x)]$$ Now the RHS of (\ref{subst}) holds iff $\calM\models B[\sigma(m/x)(n/y)]$ for all $n\in M$. But $\sigma(n/y)(m/x)=\sigma(m/x)(n/y)$, since $x$ and $y$ are distinct. Hence the LHS holds iff the RHS holds. $\Box$ {\bf Change of Bound Variable} If a term $t$ is not freely substitutible for $x$ in $A$, it is because some variable $y$ in $t$ gets caught by a quantifier $\forall y$ or $\exists y$ in $A$. One way to fix this is simply rename the bound variable $y$ in $A$ to some new variable $z$. It should be intuitively clear that this renaming does not change the meaning of $A$. The definition and lemmas below formalize this process. {\bf Definition}: $\forall z A(z/y)$ results from $\forall y A$ by {\em change of bound variable} provided $z$ does not occur in $A$. Similarly for $\exists z A(z/y)$. {\bf Lemma}: If $z$ does not occur in $A$, then $\forall z A(z/y)$ and $\forall y A$ are logically equivalent. Also $\exists zA(z/y)$ and $\exists y A$ are equivalent. {\bf Proof}: This follows from the Basic Semantic Definition and the Substitution Theorem. (Verify this). $\Box$ {\bf Definition} $A'$ is a {\em variant} of $A$ if $A'$ results by a sequence of changes of bound variables to subformulas of $A$. {\bf Theorem}: If $A'$ is a variant of $A$ then $A$ and $A'$ are equivalent. This follows from the preceding Lemma and the following general result: {\bf Replacement Theorem}: If $B$ and $B'$ are equivalent formulas and $A'$ results from $A$ by replacing some occurrence of $B$ in $A$ by $B'$, then $A$ and $A'$ are equivalent. \begin{exer} Prove the Replacement Theorem, by structural induction on $A$ (relative to $B$). The base case is when $A$ and $B$ coincide. \end{exer} {\bf Example:} $B$ is $\neg \forall xPxy$, $B'$ is $\exists z\neg Pzy$, $A$ is $\forall y (\neg\forall x Pxy\supset Qy)$. Note that $B$ has a free variable that is bound in $A$. $A'$ is $\forall y (\exists z\neg Pzy \supset Qy)$. By the Replacement Theorem, $A$ and $A'$ are equivalent, even though the quantifier $\forall y$ in $A$ catches a variable in $B$. {\large\bf A First-Order Gentzen System} We now extend the propositional proof system $PK$ to the first-order sequent proof system $LK$. For this it is convenient to introduce two kinds of variables: \begin{itemize} \item type ``free'': $a,b,c,...$ \item type ``bound'': $x,y,z,...$ \end{itemize} A first-order formula $A$ is called a {\em proper formula} if it satisfies the restriction that every variable that occurs free has type free, and every variable that occurs bound has type bound. Similarly a {\em proper term} has no variable of type bound. Notice that a subformula of a proper formula is not necessarily proper, and a proper formula may contain terms which are not proper. The sequent system $LK$ is an extension of the propositional system $PK$, where now all formulas $A_1,...,A_k,B_1,...,B_\ell$ in a sequent $A_1,...,A_k\ra B_1,...,B_\ell$ must be proper formulas. In addition to the rules given for $PK$, the system $LK$ has four rules for introducing the quantifiers. {\bf Notation}: In the rules below, $t$ is any proper term and $A(t)$ is the result of substituting $t$ for all free occurrences of $x$ in $A(x)$. Similarly $A(b)$ is the result of substituting $b$ for all free occurrences of $x$ in $A(x)$. Note that $t$ and $b$ can always be freely substituted for $x$ in $A(x)$ because $\forall x A(x)$ and $\exists xA(x)$ are proper formulas. $\forall$ introduction rules $$ {\mbox{\bf left}} \; {\large \frac{A(t),\Gamma\ra\Delta} {\forall xA(x),\Gamma\ra\Delta}} \hspace{2cm} {\mbox{\bf right}}\; {\large \frac{\Gamma\ra\Delta,A(b)} {\Gamma\ra\Delta,\forall xA(x)}} $$ $\exists$ introduction rules $$ {\mbox{\bf left}} \; {\large \frac{A(b),\Gamma\ra\Delta} {\exists xA(x),\Gamma\ra\Delta}} \hspace{2cm} {\mbox{\bf right}}\; {\large \frac{\Gamma\ra\Delta,A(t)} {\Gamma\ra\Delta,\exists xA(x)}} $$ {\bf Restriction}: The free variable $b$ must not occur in the conclusion in $\forall$ {\bf right} and $\exists$ {\bf left}. {\bf Example:} An instance of $\forall$-{\bf left} is $$ {\large \frac{Pbb\ra Pbb} {\forall yPby\ra Pbb}} $$ What is the formula $A(y)$ in this case? {\bf Semantics of first-order sequents} The semantics of first-order sequents is a natural generalization of the semantics of propositional sequents given on page 10. Again a sequent $S =_{syn}$ $$ A_1,...,A_k\ra B_1,...,B_\ell $$ has the same meaning as its associated formula $A_S =_{syn}$ \begin{equation}\label{e:meaning} (A_1\wedge A_2\wedge ... \wedge A_k)\supset (B_1\vee B_2\vee...\vee B_\ell) \end{equation} In particular, we say that the sequent is {\em valid} iff its associated formula is valid. {\bf Definition:} [Universal Closure] Suppose that $A$ is a formula whose free variables comprise the list $a_1,...,a_n$. Then the {\em universal closure} of $A$, written $\forall A$, is the sentence $\forall x_1...\forall x_n A(x_1/a_1,...,x_n/a_n)$, where $x_1,...,x_n$ are new (bound) variables. If $\Phi$ is a set of formulas, then $\forall \Phi$ is the set of all sentences $\forall A$, for $A$ in $\Phi$. Note that every formula $A$ is valid iff its universal closure $\forall A$ is valid. Also $A$ is a logical consequence of its universal closure $\forall A$, but $\forall A$ is not necessarily a logical consequence of $A$ (for example take $A =_{syn} Pa$). Recall that for the propositional system $PK$, for each rule the bottom sequent is a logical consequence of the top sequent(s). This remains true for $LK$, with the exception of the rules $\forall$-{\bf right} and $\exists$-{\bf left}. For these rules we can make a weaker statement: the universal closure of (the meaning of) the bottom sequent is a logical consequence of the universal closure of (the meaning of) the top sequent. The following proposition makes this weaker statement for all the $PK$ rules. (The statement is weaker, because for any formulas $A$ and $B$, if $A\models B$, then $\forall A\models \forall B$). {\bf Lemma} For each $PK$ rule, the universal closure of the meaning of the bottom sequent is a logical consequence of the universal closure(s) of the meaning(s) of the top sequent(s). Here the {\em meaning} of a sequent $S$ is the formula $A_S$ given in (\ref{e:meaning}). {\bf Proof:} The argument for the propositional rules is essentially the same as for the system $PK$. The arguments for $\forall$-{\bf left} and $\exists$-{\bf right} are easy; and in fact in these cases it is not necessary to take universal closures. We illustrate the remaining two rules by considering the case of of $\forall$-{\bf right}. Note that because of the {\bf Restriction} for this rule, the variable $b$ cannot occur in $\Gamma$ or $\Delta$. Hence it suffices to verify that $$ \forall x(\bigwedge\Gamma\supset (\bigvee \Delta \vee A(x))) \ \ \models \ \ \bigwedge\Gamma\supset(\bigvee\Delta\vee\forall xA(x)) $$ To see that this logical consequence holds, suppose that $\calM$ is a structure and $\sigma$ is an object assignment. Suppose that $\calM$ satisfies the left hand side under $\sigma$, i.e. $$ \calM \models \forall x(\bigwedge\Gamma\supset (\bigvee \Delta \vee A(x))) [\sigma] $$ Either $\calM$ satisfies $\forall xA(x)$ under $\sigma$ or not. In the first case it follows immediately that $\calM$ satisfies the right hand side under $\sigma$. In the second case, it must be that $$ \calM \models \forall x(\bigwedge\Gamma\supset \bigvee \Delta ) [\sigma] $$ and hence again $\calM$ satisfies the right hand side under $\sigma$. \hfill $\Box$ \begin{exer} Give the argument for the other three quantifier rules. \end{exer} {\bf Soundness Theorem for $LK$}: Every sequent provable in $LK$ is valid. {\bf Proof:} This is proved by induction on the number of sequents in the $LK$ proof. For the base case, obviously each axiom $A\ra A$ is valid. For the induction step, it follows from the above lemma that for each rule, if all sequents on top are valid, then the sequent on the bottom is valid. \hfill $\Box$ \begin{exer} Give a specific example of a sequent $\Gamma\ra\Delta,A(b)$ which is valid, but the bottom sequent $\Gamma\ra\Delta,\forall xA(x)$ is not valid, because the restriction for the $\forall$ {\bf right} rule is violated (i.e. $b$ occurs in $\Gamma$ or $\Delta$ or $\forall xA(x)$). Do the same for the $\exists$ {\bf left} rule. \end{exer} An $LK$ proof of a valid first-order sequent can be obtained using the same method as in the propositional case: Write the goal sequent at at the bottom, and move up by using the introduction rules in reverse. A good heuristic is: if there is a choice about which quantifier to remove next, choose $\forall$ {\bf right} and $\exists$ {\bf left} first (working backwards), since these rules carry a restriction. Here is an $LK$ proof of the sequent $(\forall xPx\vee \forall xQx)\ra\forall x(Px\vee Qx)$. $$ \begin{prooftree} \[ \[ \[ \[ Pb\ra Pb \justifies Pb\ra Pb, Qb \using (weakening) \] \justifies Pb\ra (Pb\vee Qb) \using (\vee \ right) \] \justifies \forall xPx\ra (Pb\vee Qb) \using (\forall \ left) \] \qquad \[ \[ \[ Qb\ra Qb\ \justifies Qb\ra Pb,Qb \using (weakening) \] \justifies Qb \ra (Pb\vee Qb) \using (\vee \ right) \] \justifies \forall xQx \ra (Pb\vee Qb) \using (\forall \ left) \] \justifies (\forall xPx\vee\forall xQx)\ra (Pb\vee Qb) \using (\vee \ left) \] \justifies (\forall xPx\vee\forall xQx)\ra \forall x(Px\vee Qx) \using (\forall \ right) \end{prooftree} $$ \begin{exer} Give $LK$ proofs for the following valid sequents: $\forall xPx\wedge\forall xQx \ra \forall x(Px\wedge Qx)$\\ $\forall x(Px\wedge Qx)\ra \forall xPx\wedge \forall xQx$\\ $\exists x (Px\vee Qx)\ra \exists xPx\vee \exists xQx$\\ $\exists x Px\vee \exists xQx\ra \exists x(Px\vee Qx)$\\ $\exists x(Px\wedge Qx)\ra \exists xPx\wedge\exists xQx$\\ $\exists y\forall xPxy\ra\forall x\exists yPxy$\\ $\forall xPx\ra \exists xPx$ Check that the rule restrictions seem to prevent generating $LK$ proofs for the following invalid sequents: $\exists xPx\wedge\exists xQx\ra\exists x(Px\wedge Qx)$\\ $\forall x\exists yPxy\ra \exists y\forall xPxy$ \end{exer} \end{document}