Section 4: Working with models

We will use the double turnstile symbol for QL much as we did for SL. ‘\(\mathcal{A}\) |= \(\mathcal{B}\)’ means that ‘\(\mathcal{A}\) entails \(\mathcal{B}\)’: When \(\mathcal{A}\) and \(\mathcal{B}\) are two sentences of QL, \(\mathcal{A}\) |= \(\mathcal{B}\) means that there is no model in which \(\mathcal{A}\) is true and \(\mathcal{B}\) is false. |= \(\mathcal{A}\) means that \(\mathcal{A}\) is true in every model.

This allows us to give definitions for various concepts in QL. Because we are using the same symbol, these definitions will look similar to the definitions in SL. Remember, however, that the definitions in QL are in terms of models rather than in terms of truth value assignments.

A tautology in ql is a sentence \(\mathcal{A}\) that is true in every model;
i.e., |= \(\mathcal{A}\).
A contradiction in ql is a sentence \(\mathcal{A}\) that is false in every model;
i.e., |= ¬\(\mathcal{A}\).

A sentence is contingent in ql if and only if it is neither a tautology nor a contradiction.

An argument “ \(\mathcal{P}\)₁,\(\mathcal{P}\)₂,···, .˙. \(\mathcal{C}\) ” is valid in ql if and only if there is no model in which all of the premises are true and the conclusion is false; i.e., {\(\mathcal{P}\)₁,\(\mathcal{P}\)₂,···}|= \(\mathcal{C}\). It is invalid in ql otherwise.

Two sentences \(\mathcal{A}\) and \(\mathcal{B}\) are logically equivalent in ql if and only if both \(\mathcal{A}\) |= \(\mathcal{B}\) and \(\mathcal{B}\) |= \(\mathcal{A}\).

The set {\(\mathcal{A}\)₁,\(\mathcal{A}\)₂,\(\mathcal{A}\)₃,···} is consistent in ql if and only if there is at least one model in which all of the sentences are true. The set is inconsistent in ql if and if only there is no such model.

Constructing models

Suppose we want to show that ∀\(xAxx\) → \(Bd\) is not a tautology. This requires showing that the sentence is not true in every model; i.e., that it is false in some model. If we can provide just one model in which the sentence is false, then we will have shown that the sentence is not a tautology.

What would such a model look like? In order for ∀\(xAxx\) → \(Bd\) to be false, the antecedent (∀\(xAxx\)) must be true, and the consequent (\(Bd\)) must be false.

To construct such a model, we start with a UD. It will be easier to specify extensions for predicates if we have a small UD, so start with a UD that has just one member. Formally, this single member might be anything. Let’s say it is the city of Paris.

We want ∀\(xAxx\) to be true, so we want all members of the UD to be paired with themself in the extension of \(A\); this means that the extension of \(A\) must be {<Paris,Paris>}.

We want \(Bd\) to be false, so the referent of \(d\) must not be in the extension of \(B\).

We give \(B\) an empty extension.

Since Paris is the only member of the UD, it must be the referent of \(d\). The model we have constructed looks like this:

UD = {Paris}
extension(\(A\)) = {<Paris,Paris>}
extension(\(B\)) = ∅
referent(\(d\)) = Paris

Strictly speaking, a model specifies an extension for every predicate of QL and a referent for every constant. As such, it is generally impossible to write down a complete model. That would require writing down infinitely many extensions and infinitely many referents. However, we do not need to consider every predicate in order to show that there are models in which ∀\(xAxx\) → \(Bd\) is false. Predicates like \(H\) and constants like \(f\)₁₃ make no difference to the truth or falsity of this sentence. It is enough to specify extensions for \(A\) and \(B\) and a referent for \(d\), as we have done. This provides a partial model in which the sentence is false.

Perhaps you are wondering: What does the predicate \(A\) mean in English? The partial model could correspond to an interpretation like this one:

UD: Paris
\(Axy\): \(x\) is in the same country as \(y\).
\(Bx\): \(x\) was founded in the 20th century.
\(d\): the City of Lights

However, all that the partial model tells us is that \(A\) is a predicate which is true of Paris and Paris. There are indefinitely many predicates in English that have this extension. \(Axy\) might instead translate ‘\(x\) is the same size as \(y\)’ or ‘\(x\) and \(y\) are both cities.’ Similarly, \(Bx\) is some predicate that does not apply to Paris; it might instead translate ‘\(x\) is on an island’ or ‘\(x\) is a subcompact car.’ When we specify the extensions of \(A\) and \(B\), we do not specify what English predicates \(A\) and \(B\) should be used to translate. We are concerned with whether the ∀\(xAxx\) → \(Bd\) comes out true or false, and all that matters for truth and falsity in QL is the information in the model: the UD, the extensions of predicates, and the referents of constants.

We can just as easily show that ∀\(xAxx\) → \(Bd\) is not a contradiction. We need only specify a model in which ∀\(xAxx\) → \(Bd\) is true; i.e., a model in which either ∀\(xAxx\) is false or \(Bd\) is true. Here is one such partial model:

UD = {Paris}
extension(\(A\)) = {<Paris,Paris>}
extension(\(B\)) = {Paris}
referent(\(d\)) = Paris

We have now shown that ∀\(xAxx\) → \(Bd\) is neither a tautology nor a contradiction. By the definition of ‘contingent in QL,’ this means that ∀\(xAxx \)→ \(Bd\) is contingent. In general, showing that a sentence is contingent will require two models: one in which the sentence is true and another in which the sentence is false.

Suppose we want to show that ∀\(xSx\) and ∃\(xSx\) are not logically equivalent. We need to construct a model in which the two sentences have different truth values; we want one of them to be true and the other to be false. We start by specifying a UD. Again, we make the UD small so that we can specify extensions easily. We will need at least two members. Let the UD be {Duke, Miles}. (If we chose a UD with only one member, the two sentences would end up with the same truth value. In order to see why, try constructing some partial models with one-member UDs.)

We can make ∃\(xSx\) true by including something in the extension of \(S\), and we can make ∀\(xSx\) false by leaving something out of the extension of \(S\). It does not matter which one we include and which one we leave out. Making Duke the only \(S\), we get a partial model that looks like this:

UD = {Duke, Miles}
extension(\(S\)) = {Duke}

This partial model shows that the two sentences are not logically equivalent.

Back on p. 63, we said that this argument would be invalid in QL:

(\(Rc\)&\(K\)₁c)&\(Tc\)
.˙. \(Tc\)&\(K\)₂\(c\)

In order to show that it is invalid, we need to show that there is some model in which the premises are true and the conclusion is false. We can construct such a model deliberately. Here is one way to do it:

UD = {Björk}
extension(\(T\)) = {Björk}
extension(\(K\)₁) = {Björk}
extension(\(K\)₂) = ∅
extension(\(R\)) = {Björk}
referent(\(c\)) = Björk

Similarly, we can show that a set of sentences is consistent by constructing a model in which all of the sentences are true.

Reasoning about all models

We can show that a sentence is not a tautology just by providing one carefully specified model: a model in which the sentence is false. To show that something is a tautology, on the other hand, it would not be enough to construct ten, one hundred, or even a thousand models in which the sentence is true. It is only a tautology if it is true in every model, and there are infinitely many models. This cannot be avoided just by constructing partial models, because there are infinitely many partial models.

Consider, for example, the sentence \(Raa\) ↔ \(Raa\). There are two logically distinct partial models of this sentence that have a 1-member UD. There are 32 distinct partial models that have a 2-member UD. There are 1526 distinct partial models that have a 3-member UD. There are 262,144 distinct partial models that have a 4-member UD. And so on to infinity. In order to show that this sentence is a tautology, we need to show something about all of these models. There is no hope of doing so by dealing with them one at a time.

Nevertheless, \(Raa\) ↔ \(Raa\) is obviously a tautology. We can prove it with a simple argument:

There are two kinds of models: those in which {referent(\(a\)),referent(\(a\))i is in the extension of \(R\) and those in which it is not. In the first kind of model, \(Raa\) is true; by the truth table for the biconditional, \(Raa\) ↔ \(Raa\) is also true. In the second kind of model, \(Raa\) is false; this makes \(Raa\) ↔ \(Raa\) true. Since the sentence is true in both kinds of model, and since every model is one of the two kinds, \(Raa\) ↔ \(Raa\) is true in every model. Therefore, it is a tautology.

This argument is valid, of course, and its conclusion is true. However, it is not an argument in QL. Rather, it is an argument in English about QL; it is an argument in the metalanguage. There is no formal procedure for evaluating or constructing natural language arguments like this one. The imprecision of natural language is the very reason we began thinking about formal languages.

There are further difficulties with this approach.

Consider the sentence ∀\(x\)(\(Rxx\) → \(Rxx\)), another obvious tautology. It might be tempting to reason in this way: ‘\(Rxx\) → \(Rxx\) is true in every model, so ∀\(x\)(\(Rxx\) → \(Rxx\)) must be true.’ The problem is that \(Rxx\) → \(Rxx\) is not true in every model. It is not a sentence, and so it is neither true nor false. We do not yet have the vocabulary to say what we want to say about \(Rxx\) → \(Rxx\). In the next section, we introduce the concept of satisfaction; after doing so, we will be better able to provide an argument that ∀\(x\)(\(Rxx\) → \(Rxx\)) is a tautology.

It is necessary to reason about an infinity of models to show that a sentence is a tautology. Similarly, it is necessary to reason about an infinity of models to show that a sentence is a contradition, that two sentences are equivalent, that a set of sentences is inconsistent, or that an argument is valid. There are other things we can show by carefully constructing a model or two. Table 5.1 summarizes which things are which.

Table 5.1: It is relatively easy to answer a question if you can do it by constructing a model or two. It is much harder if you need to reason about all possible models. This table shows when constructing models is enough.

	YES	NO
Is \(\mathcal{A}\) a tautology?	show that \(\mathcal{A}\) must be true in any model	construct a model in which \(\mathcal{A}\) is false
Is \(\mathcal{A}\) a contradiction?	show that \(\mathcal{A}\) must be false in any model	construct a model in which \(\mathcal{A}\) is true
Is \(\mathcal{A}\) contingent?	construct two models, one in which \(\mathcal{A}\) is true and another in which \(\mathcal{A}\) is false	either show that \(\mathcal{A}\) is a tautology or show that \(\mathcal{A}\) is a contradiction
Are \(\mathcal{A}\) and \(\mathcal{B}\) equivalent?	show that \(\mathcal{A}\) and \(\mathcal{B}\) must have the same truth value in any model	construct a model in which \(\mathcal{A}\) and \(\mathcal{B}\) have different truth values
Is the set \(\mathbb{A}\) consistent?	construct a model in which all the sentences in \(\mathbb{A}\) are true	show that the sentences could not all be true in any model
Is the argument `\(\mathcal{P}\), .˙. \(\mathcal{C}\)' valid?	show that any model in which \(\mathcal{P}\) is true must be a model in which \(\mathcal{C}\) is true	construct a model in which \(\mathcal{P}\) is true and \(\mathcal{C}\) is false