Section 5: Truth in QL

For SL, we split the definition of truth into two parts: a truth value assignment (\(a\)) for sentence letters and a truth function (\(v\)) for all sentences. The truth function covered the way that complex sentences could be built out of sentence letters and connectives.

In the same way that truth for SL is always truth given a truth value assignment, truth for QL is truth in a model. The simplest atomic sentence of QL consists of a one-place predicate followed by a constant, like \(Pj\). It is true in a model \(\mathbb{M}\) if and only if the referent of \(j\) is in the extension of \(P\) in \(\mathbb{M}\).

We could go on in this way to define truth for all atomic sentences that contain only predicates and constants: Consider any sentence of the form \(\mathcal{RC}\)₁ ...\(\mathcal{c}\)_n where \(\mathcal{R}\) is an n-place predicate and the \(\mathcal{c}\)s are constants. It is true in \(\mathbb{M}\) if and only if hreferent(\(\mathcal{c}\)₁),...,referent(\(\mathcal{c}\)_n)i is in extension(\(\mathcal{R}\)) in \(\mathbb{M}\).

We could then define truth for sentences built up with sentential connectives in the same way we did for SL. For example, the sentence (\(Pj\) → \(Mda\)) is true in \(\mathbb{M}\) if either \(Pj\) is false in \(\mathbb{M}\) or \(Mda\) is true in \(\mathbb{M}\).

Unfortunately, this approach will fail when we consider sentences containing quantifiers. Consider ∀\(xPx\). When is it true in a model \(\mathbb{M}\)? The answer cannot depend on whether \(Px\) is true or false in \(\mathbb{M}\), because the \(x\) in \(Px\) is a free variable. \(Px\) is not a sentence. It is neither true nor false.

We were able to give a recursive definition of truth for SL because every wellformed formula of SL has a truth value. This is not true in QL, so we cannot define truth by starting with the truth of atomic sentences and building up. We also need to consider the atomic formulae which are not sentences. In order to do this we will define satisfaction; every well-formed formula of QL will be satisfied or not satisfied, even if it does not have a truth value. We will then be able to define truth for sentences of QL in terms of satisfaction.

Satisfaction

The formula \(Px\) says, roughly, that \(x\) is one of the \(Ps\). This cannot be quite right, however, because \(x\) is a variable and not a constant. It does not name any particular member of the UD. Instead, its meaning in a sentence is determined by the quantifier that binds it. The variable \(x\) must stand-in for every member of the UD in the sentence ∀\(xPx\), but it only needs to stand-in for one member in ∃\(xPx\). Since we want the definition of satisfaction to cover \(Px\) without any quantifier whatsoever, we will start by saying how to interpret a free variable like the \(x\) in \(Px\).

We do this by introducing a variable assignment. Formally, this is a function that matches up each variable with a member of the UD. Call this function ‘\(\a\).’ (The ‘\(a\)’ is for ‘assignment’, but this is not the same as the truth value assignment that we used in defining truth for SL.)

The formula \(Px\) is satisfied in a model \(\mathbb{M}\) by a variable assignment \(a\) if and only if \(a\)(\(x\)), the object that \(a\) assigns to \(x\), is in the the extension of \(P\) in \(\mathbb{M}\).

When is ∀\(xPx\) satisfied? It is not enough if \(Px\) is satisfied in \(\mathbb{M}\) by \(a\), because that just means that \(a\)(\(x\)) is in extension(\(P\)). ∀\(xPx\) requires that every other member of the UD be in extension(\(P\)) as well.

So we need another bit of technical notation: For any member Ω of the UD and any variable \(x\), let a[Ω|\(x\)] be the variable assignment that assigns Ω to \(x\) but agrees with \(a\) in all other respects. We have used Ω, the Greek letter Omega, to underscore the fact that it is some member of the UD and not some symbol of QL. Suppose, for example, that the UD is presidents of the United States. The function \(a\)[Grover Cleveland|\(x\)] assigns Grover Cleveland to the variable \(x\), regardless of what \(a\) assigns to \(x\); for any other variable, \(a\)[Grover Cleveland|\(x\)] agrees with \(a\).

We can now say concisely that ∀\(xPx\) is satisfied in a model \(\mathbb{M}\) by a variable assignment \(a\) if and only if, for every object Ω in the UD of \(\mathbb{M}\), \(Px\) is satisfied in \(\mathbb{M}\) by \(a\)[Ω|\(x\)].

You may worry that this is circular, because it gives the satisfaction conditions for the sentence ∀\(xPx\) using the phrase ‘for every object.’ However, it is important to remember the difference between a logical symbol like ‘∀’ and an English language word like ‘every.’ The word is part of the metalanguage that we use in defining satisfaction conditions for object language sentences that contain the symbol.

We can now give a general definition of satisfaction, extending from the cases we have already discussed. We define a function \(s\) (for ‘satisfaction’) in a model \(\mathbb{M}\) such that for any wff \(\mathcal{A}\) and variable assignment \(a\), s(\(\mathcal{A}\),\(a\)) = 1 if \(\mathcal{A}\) is satisfied in \(\mathbb{M}\) by \(a\); otherwise s(\(\mathcal{A}\),\(a\)) = 0.

1. If \(\mathcal{A}\) is an atomic wff of the form \(\mathcal{Pt}\)₁ ...\(\mathcal{t}\)_n and Ω_i is the object picked out by \(t\)_i, then

\( s ( \mathcal { A } , a ) = \left\{ \begin{array} { l l } { 1 } & { \text { if } \left\langle \Omega _ { 1 } \ldots \Omega _ { n } \right\rangle \text { is in extension } ( \mathcal { P } ) \text { in } \mathbb { M } } \\ { 0 } & { \text { otherwise. } } \end{array} \right. \)

For each term \(t\)_i: If \(t\)_i is a constant, then Ω_i = referent(\(t\)_i). If \(t\)_i is a variable, then Ω_i = \(a\)(\(t\)_i).

2. If \(\mathcal{A}\) is ¬\(\mathcal{B}\) for some wff \(\mathcal{B}\), then

\( s ( \mathcal { A } , a ) = \left\{ \begin{array} { l l } { 1 } & { \text { if } s ( \mathcal { B } , a ) = 0 } \\ { 0 } & { \text { otherwise } } \end{array} \right. \)

3. If \(\mathcal{A}\) is (\(\mathcal{B}\) & \(\mathcal{C}\)) for some wffs \(\mathcal{B}\),\(\mathcal{C}\), then

\( s ( \mathcal { A } , a ) = \left\{ \begin{array} { l l } { 1 } & { \text { if } s ( \mathcal { B } , a ) = 1 \text { and } s ( C , a ) = 1 } \\ { 0 } & { \text { otherwise } } \end{array} \right. \)

4. If \(\mathcal{A}\) is (\(\mathcal{B}\)∨\(\mathcal{C}\)) for some wffs \(\mathcal{B}\),\(\mathcal{C}\), then

\( s ( \mathcal { A } , a ) = \left\{ \begin{array} { l l } { 0 } & { \text { if } s ( \mathcal { B } , a ) = 0 \text { and } s ( C , a ) = 0 } \\ { 1 } & { \text { otherwise } } \end{array} \right. \)

5. If \(\mathcal{A}\) is (\(\mathcal{B}\) → \(\mathcal{C}\)) for some wffs \(\mathcal{B}\),\(\mathcal{C}\), then

\( s ( \mathcal { A } , a ) = \left\{ \begin{array} { l l } { 0 } & { \text { if } s ( \mathcal { B } , a ) = 1 \text { and } s ( C , a ) = 0 } \\ { 1 } & { \text { otherwise } } \end{array} \right. \)

6. If \(\mathcal{A}\) is (\(\mathcal{B}\) ↔ \(\mathcal{C}\)) for some sentences \(\mathcal{B}\),\(\mathcal{C}\), then

\( s ( \mathcal { A } , a ) = \left\{ \begin{array} { l l } { 1 } & { \text { if } s ( \mathcal { B } , a ) = s ( C , a ) } \\ { 0 } & { \text { otherwise } } \end{array} \right. \)

7. If \(\mathcal{A}\) is ∀\(\mathcal{xB}\) for some wff \(\mathcal{B}\) and some variable \(\mathcal{x}\), then

\( s ( \mathcal { A } , a ) = \left\{ \begin{array} { l l } { 1 } & { \text { if } s ( \mathcal { B } , a [ \Omega | x ] ) = 1 \text { for every member } \Omega \text { of the UD, } } \\ { 0 } & { \text { otherwise. } } \end{array} \right. \)

8. If \(\mathcal{A}\) is ∃\(\mathcal{xB}\) for some wff \(\mathcal{B}\) and some variable \(\mathcal{x}\), then

\( s ( \mathcal { A } , a ) = \left\{ \begin{array} { l l } { 1 } & { \text { if } s ( \mathcal { B } , a [ \Omega | x ] ) = 1 \text { for at least one member } \Omega \text { of the UD, } } \\ { 0 } & { \text { otherwise. } } \end{array} \right. \)

This definition follows the same structure as the definition of a wff for QL, so we know that every wff of QL will be covered by this definition. For a model \(\mathbb{M}\) and a variable assignment \(a\), any wff will either be satisfied or not. No wffs are left out or assigned conflicting values.

Truth

Consider a simple sentence like∀\(xPx\). By part 7 in the definition of satisfaction, this sentence is satisfied if \(a\)[Ω|\(x\)] satisfies \(Px\) in \(\mathbb{M}\) for every Ω in the UD. By part 1 of the definition, this will be the case if every Ω is in the extension of \(P\). Whether ∀\(xPx\) is satisfied does not depend on the particular variable assignment \(a\). If this sentence is satisfied, then it is true. This is a formalization of what we have said all along: ∀\(xPx\) is true if everything in the UD is in the extension of \(P\).

The same thing holds for any sentence of QL. Because all of the variables are bound, a sentence is satisfied or not regardless of the details of the variable assignment. So we can define truth in this way: A sentence \(\mathcal{A}\) is true in \(\mathbb{M}\) if and only if some variable assignment satisfies \(\mathcal{A}\) in \(M\); \(\mathcal{A}\) is false in \(\mathbb{M}\) otherwise.

Truth in QL is truth in a model. Sentences of QL are not flat-footedly true or false as mere symbols, but only relative to a model. A model provides the meaning of the symbols, insofar as it makes any difference to truth and falsity.

Reasoning about all models (reprise)

At the end of section 5.4, we were stymied when we tried to show that∀\(x\)(\(Rxx\) → \(Rxx\)) is a tautology. Having defined satisfaction, we can now reason in this way:

Consider some arbitrary model \(\mathbb{M}\). Now consider an arbitrary member of the UD; for the sake of convenience, call it Ω. It must be the case either that <Ω,Ω> is in the extension of \(R\) or that it is not. If <Ω,Ω> is in the extension of \(R\), then \(Rxx\) is satisfied by a variable assignment that assigns Ω to \(x\) (by part 1 of the definition of satisfaction); since the consequent of \(Rxx\) → \(Rxx\) is satisfied, the conditional is satisfied (by part 5). If <Ω,Ω> is not in the extension of \(R\), then \(Rxx\) is not satisfied by a variable assignment that assigns Ω to \(x\) (by part 1); since antecedent of \(Rxx\) → \(Rxx\) is not satisfied, the conditional is satisfied (by part 5). In either case, \(Rxx\) → \(Rxx\) is satisfied. This is true for any member of the UD, so ∀\(x\)(\(Rxx\) → \(Rxx\)) is satisfied by any truth value assignment (by part 7). So ∀\(x\)(\(Rxx\) → \(Rxx\)) is true in \(\mathbb{M}\) (by the definition of truth). This argument holds regardless of the exact UD and regardless of the exact extension of \(R\), so ∀\(x\)(\(Rxx\) → \(Rxx\)) is true in any model. Therefore, it is a tautology.

Giving arguments about all possible models typically requires clever combination of two strategies:

1. Divide cases between two possible kinds, such that every case must be one kind or the other. In the argument on p. 92, for example, we distinguished two kinds of models based on whether or not a specific ordered pair was in extension(\(R\)). In the argument above, we distinguished cases in which an ordered pair was in extension(\(R\)) and cases in which it was not.

2. Consider an arbitrary object as a way of showing something more general. In the argument above, it was crucial that Ω was just some arbitrary member of the UD. We did not assume anything special about it. As such, whatever we could show to hold of Ω must hold of every member of the UD— if we could show it for Ω, we could show it for anything. In the same way, we did not assume anything special about \(\mathbb{M}\), and so whatever we could show about \(\mathbb{M}\) must hold for all models.

Consider one more example. The argument ∀\(x\)(\(Hx\)&\(Jx\)) .˙.∀\(xHx\) is obviously valid. We can only show that the argument is valid by considering what must be true in every model in which the premise is true.

Consider an arbitrary model \(\mathbb{M}\) in which the premise ∀\(x\)(\(Hx\)&\(Jx\)) is true. The conjunction \(Hx\)&\(Jx\) is satisfied regardless of what is assigned to \(x\), so \(Hx\) must be also (by part 3 of the definition of satisfaction). As such, (∀\(x\))\(Hx\) is satisfied by any variable assignment (by part 7 of the definition of satisfaction) and true in \(\mathbb{M}\) (by the definition of truth). Since we did not assume anything about \(\mathbb{M}\) besides ∀\(x\)(\(Hx\)&\(Jx\)) being true, (∀\(x\))\(Hx\) must be true in any model in which ∀\(x\)(\(Hx\)&\(Jx\)) is true. So ∀\(x\)(\(Hx\)&\(Jx\)) |= ∀\(xHx\).

Even for a simple argument like this one, the reasoning is somewhat complicated. For longer arguments, the reasoning can be insufferable. The problem arises because talking about an infinity of models requires reasoning things out in English. What are we to do?

We might try to formalize our reasoning about models, codifying the divide-andconquer strategies that we used above. This approach, originally called semantic tableaux, was developed in the 1950s by Evert Beth and Jaakko Hintikka. Their tableaux are now more commonly called truth trees.

A more traditional approach is to consider deductive arguments as proofs. A proof system consists of rules that formally distinguish between legitimate and illegitimate arguments— without considering models or the meanings of the symbols. In the next chapter, we develop proof systems for SL and QL.