# Chapter 24 Sentences with one quantifier

We now have all of the pieces of FOL. Symbolizing more complicated sentences is just a matter of knowing how to combine predicates, names, quantifiers, and connectives. There is a knack to this, and there is no substitute for practice.

## 24.1 Common quantifier phrases

Consider these sentences:

1. 1.

Every coin in my pocket is a quarter.

2. 2.

Some coin on the table is a dime.

3. 3.

Not all the coins on the table are dimes.

4. 4.

None of the coins in my pocket are dimes.

In providing a symbolization key, we need to specify a domain. Since we are talking about coins in my pocket and on the table, the domain must at least contain all of those coins. Since we are not talking about anything besides coins, we let the domain be all coins. Since we are not talking about any specific coins, we do not need to deal with any names. So here is our key:

domain:

all coins

$P(x)$:

blank${}_{x}$ is in my pocket

$T(x)$:

blank${}_{x}$ is on the table

$Q(x)$:

blank${}_{x}$ is a quarter

$D(x)$:

blank${}_{x}$ is a dime

Sentence 1 is most naturally symbolized using a universal quantifier. The universal quantifier says something about everything in the domain, not just about the coins in my pocket. Sentence 1 can be paraphrased as ‘for any coin, if that coin is in my pocket then it is a quarter’. So we can symbolize it as ‘$\forall x(P(x)\rightarrow Q(x))$’.

Since sentence 1 is about coins that are both in my pocket and that are quarters, it might be tempting to symbolize it using a conjunction. However, the sentence ‘$\forall x(P(x)\wedge Q(x))$’ would symbolize the sentence ‘every coin is both a quarter and in my pocket’. This obviously means something very different than sentence 1. And so we see:

A sentence can be symbolized as $\forall x(\mathscr{F}(x)\rightarrow\mathscr{G}(x))$ if it can be paraphrased in English as ‘every $F$ is $G$’.

Sentence 2 is most naturally symbolized using an existential quantifier. It can be paraphrased as ‘there is some coin which is both on the table and which is a dime’. So we can symbolize it as ‘$\exists x(T(x)\wedge D(x))$’.

Notice that we needed to use a conditional with the universal quantifier, but we used a conjunction with the existential quantifier. Suppose we had instead written ‘$\exists x(T(x)\rightarrow D(x))$’. That would mean that there is some object in the domain of which ‘$(T(x)\rightarrow D(x))$’ is true. Recall that, in TFL, $\mathscr{A}\rightarrow\mathscr{B}$ is logically equivalent (in TFL) to $\neg\mathscr{A}\vee\mathscr{B}$. This equivalence will also hold in FOL. So ‘$\exists x(T(x)\rightarrow D(x))$’ is true if there is some object in the domain, such that ‘$(\neg T(x)\vee D(x))$’ is true of that object. That is, ‘$\exists x(T(x)\rightarrow D(x))$’ is true if some coin is either not on the table or is a dime. Of course there is a coin that is not on the table: there are coins in lots of other places. So it is very easy for ‘$\exists x(T(x)\rightarrow D(x))$’ to be true. A conditional will usually be the natural connective to use with a universal quantifier, but a conditional within the scope of an existential quantifier tends to say something very weak indeed. As a general rule of thumb, do not put conditionals in the scope of existential quantifiers unless you are sure that you need one.

A sentence can be symbolized as $\exists x(\mathscr{F}(x)\wedge\mathscr{G}(x))$ if it can be paraphrased in English as ‘some $F$ is $G$’.

Sentence 3 can be paraphrased as, ‘It is not the case that every coin on the table is a dime’. So we can symbolize it by ‘$\neg\forall x(T(x)\rightarrow D(x))$’. You might look at sentence 3 and paraphrase it instead as, ‘Some coin on the table is not a dime’. You would then symbolize it by ‘$\exists x(T(x)\wedge\neg D(x))$’. Although it is probably not immediately obvious yet, these two sentences are logically equivalent. (This is due to the logical equivalence between $\neg\forall x\,\mathscr{A}$ and $\exists x\,\neg\mathscr{A}$, mentioned in chapter 23, along with the equivalence between $\neg(\mathscr{A}\rightarrow\mathscr{B})$ and $\mathscr{A}\wedge\neg\mathscr{B}$.)

Sentence 4 can be paraphrased as, ‘It is not the case that there is some dime in my pocket’. This can be symbolized by ‘$\neg\exists x(P(x)\wedge D(x))$’. It might also be paraphrased as, ‘Everything in my pocket is a non-dime’, and then could be symbolized by ‘$\forall x(P(x)\rightarrow\neg D(x))$’. Again the two symbolizations are logically equivalent; both are correct symbolizations of sentence 4.

A sentence that can be paraphrased as ‘no $F$ is $G$’ can be symbolized as $\neg\exists x(\mathscr{F}(x)\wedge\mathscr{G}(x))$ and also as $\forall x(\mathscr{F}(x)\rightarrow\neg\mathscr{G}(x))$.

Finally, consider ‘only’, as in:

1. 5.

Only dimes are on the table.

How should we symbolize this? A good strategy is to consider when the sentence would be false. If we are saying that only dimes are on the table, we are excluding all the cases where something on the table is a non-dime. So we can symbolize the sentence the same way we would symbolize ‘No non-dimes are on the table.’ Remembering the lesson we just learned, and symbolizing ‘$x$ is a non-dime’ as ‘$\neg D(x)$’, the possible symbolizations are: ‘$\neg\exists x(T(x)\wedge\neg D(x))$’, or alternatively: ‘$\forall x(T(x)\rightarrow\neg\neg D(x))$’. Since double negations cancel out, the second is just as good as ‘$\forall x(T(x)\rightarrow D(x))$’. In other words, ‘Only dimes are on the table’ and ‘Everything on the table is a dime’ are symbolized the same way.

A sentence that can be paraphrased as ‘only $F$s are $G$s’ can be symbolized as $\neg\exists x(\mathscr{G}(x)\wedge\neg\mathscr{F}(x))$ and also as $\forall x(\mathscr{G}(x)\rightarrow\mathscr{F}(x))$.

## 24.2 Empty predicates

In chapter 23, we emphasized that a name must pick out exactly one object in the domain. However, a predicate need not apply to anything in the domain. A predicate that applies to nothing in the domain is called an empty predicate . This is worth exploring.

Suppose we want to symbolize these two sentences:

1. 6.

Every monkey knows sign language.

2. 7.

Some monkey knows sign language.

It is possible to write the symbolization key for these sentences in this way:

domain:

animals

$M(x)$:

blank${}_{x}$ is a monkey.

$S(x)$:

blank${}_{x}$ knows sign language.

Sentence 6 can now be symbolized by ‘$\forall x(M(x)\rightarrow S(x))$’. Sentence 7 can be symbolized as ‘$\exists x(M(x)\wedge S(x))$’.

It is tempting to say that sentence 6 entails sentence 7. That is, we might think that it is impossible that every monkey knows sign language unless some monkey knows sign language. But this would be a mistake. It is possible for the sentence ‘$\forall x(M(x)\rightarrow S(x))$’ to be true even though the sentence ‘$\exists x(M(x)\wedge S(x))$’ is false.

How can this be? The answer comes from considering whether these sentences would be true or false if there were no monkeys. If there were no monkeys at all (in the domain), then ‘$\forall x(M(x)\rightarrow S(x))$’ would be vacuously true: take any monkey you like—it knows sign language! But if there were no monkeys at all (in the domain), then ‘$\exists x(M(x)\wedge S(x))$’ would be false.

Another example will help to bring this home. Suppose we extend the above symbolization key, by adding:

$R(x)$:

blank${}_{x}$ is a refrigerator

Now consider the sentence ‘$\forall x(R(x)\rightarrow M(x))$’. This symbolizes ‘every refrigerator is a monkey’. This sentence is true, given our symbolization key, which is counterintuitive, since we (presumably) do not want to say that there are a whole bunch of refrigerator monkeys. It is important to remember, though, that ‘$\forall x(R(x)\rightarrow M(x))$’ is true if⁠f any member of the domain that is a refrigerator is a monkey. Since the domain is animals, there are no refrigerators in the domain. Again, then, the sentence is vacuously true.

If you were actually dealing with the sentence ‘All refrigerators are monkeys’, then you would most likely want to include kitchen appliances in the domain. Then the predicate ‘$R$’ would not be empty and the sentence ‘$\forall x(R(x)\rightarrow M(x))$’ would be false.

When $\mathscr{F}$ is an empty predicate, any sentence $\forall x(\mathscr{F}(x)\rightarrow\ldots)$ is vacuously true.

## 24.3 Picking a domain

The appropriate symbolization of an English language sentence in FOL will depend on the symbolization key. Choosing a key can be difficult. Suppose we want to symbolize the English sentence:

1. 8.

Every rose has a thorn.

We might offer this symbolization key:

$R(x)$:

blank${}_{x}$ is a rose

$T(x)$:

blank${}_{x}$ has a thorn

It is tempting to say that sentence 8 should be symbolized as ‘$\forall x(R(x)\rightarrow T(x))$’, but we have not yet chosen a domain. If the domain contains all roses, this would be a good symbolization. Yet if the domain is merely things on my kitchen table, then ‘$\forall x(R(x)\rightarrow T(x))$’ would only come close to covering the fact that every rose on my kitchen table has a thorn. If there are no roses on my kitchen table, the sentence would be trivially true. This is not what we want. To symbolize sentence 8 adequately, we need to include all the roses in the domain, but now we have two options.

First, we can restrict the domain to include all roses but only roses. Then sentence 8 can, if we like, be symbolized with ‘$\forall x\,T(x)$’. This is true if⁠f everything in the domain has a thorn; since the domain is just the roses, this is true if⁠f every rose has a thorn. By restricting the domain, we have been able to symbolize our English sentence with a very short sentence of FOL. So this approach can save us trouble, if every sentence that we want to deal with is about roses.

Second, we can let the domain contain things besides roses: rhododendrons, rats, rifles, whatevers; and we will certainly need to include a more expansive domain if we simultaneously want to symbolize sentences like:

1. 9.

Our domain must now include both all the roses (so that we can symbolize sentence 8) and all the cowboys (so that we can symbolize sentence 9). So we might offer the following symbolization key:

domain:

people and plants

$C(x)$:

blank${}_{x}$ is a cowboy

$S(x)$:

blank${}_{x}$ sings a sad, sad song

$R(x)$:

blank${}_{x}$ is a rose

$T(x)$:

blank${}_{x}$ has a thorn

Now we will have to symbolize sentence 8 with ‘$\forall x(R(x)\rightarrow T(x))$’, since ‘$\forall x\,T(x)$’ would symbolize the sentence ‘every person or plant has a thorn’. Similarly, we will have to symbolize sentence 9 with ‘$\forall x(C(x)\rightarrow S(x))$’.

In general, the universal quantifier can be used to symbolize the English expression ‘everyone’ if the domain only contains people. If there are people and other things in the domain, then ‘everyone’ must be treated as ‘every person’.

## 24.4 The utility of paraphrase

When symbolizing English sentences in FOL, it is important to understand the structure of the sentences you want to symbolize. What matters is the final symbolization in FOL, and sometimes you will be able to move from an English language sentence directly to a sentence of FOL. Other times, it helps to paraphrase the sentence one or more times. Each successive paraphrase should move from the original sentence closer to something that you can easily symbolize directly in FOL.

For the next several examples, we will use this symbolization key:

domain:

people

$B(x)$:

blank${}_{x}$ is a bassist

$R(x)$:

blank${}_{x}$ is a rock star

$k$:

Kim Deal

Now consider these sentences:

1. 10.

If Kim Deal is a bassist, then she is a rock star.

2. 11.

If a person is a bassist, then she is a rock star.

The same words appear as the consequent in sentences 10 and 11 (‘$\ldots$ she is a rock star’), but they mean very different things. To make this clear, it often helps to paraphrase the original sentences, removing pronouns.

Sentence 10 can be paraphrased as, ‘If Kim Deal is a bassist, then Kim Deal is a rockstar’. This can obviously be symbolized as ‘$B(k)\rightarrow R(k)$’.

Sentence 11 must be paraphrased differently: ‘If a person is a bassist, then that person is a rock star’. This sentence is not about any particular person, so we need a variable. As an intermediate step, we can paraphrase this as, ‘For any person x, if x is a bassist, then x is a rockstar’. Now this can be symbolized as ‘$\forall x(B(x)\rightarrow R(x))$’. This is the same sentence we would have used to symbolize ‘Everyone who is a bassist is a rock star’. On reflection, that is surely true if⁠f sentence 11 is true, as we would hope.

Consider these further sentences:

1. 12.

If anyone is a bassist, then Kim Deal is a rock star.

2. 13.

If anyone is a bassist, then she is a rock star.

The same words appear as the antecedent in sentences 12 and 13 (‘If anyone is a bassist$\ldots$’), but it can be tricky to work out how to symbolize these two uses. Again, paraphrase will come to our aid.

Sentence 12 can be paraphrased, ‘If there is at least one bassist, then Kim Deal is a rock star’. It is now clear that this is a conditional whose antecedent is a quantified expression; so we can symbolize the entire sentence with a conditional as the main logical operator: ‘$\exists xB(x)\rightarrow R(k)$’.

Sentence 13 can be paraphrased, ‘For all people $x$, if $x$ is a bassist, then $x$ is a rock star’. Or, in more natural English, it can be paraphrased by ‘All bassists are rock stars’. It is best symbolized as ‘$\forall x(B(x)\rightarrow R(x))$’, just like sentence 11.

The moral is that the English words ‘any’ and ‘anyone’ should typically be symbolized using quantifiers, and if you are having a hard time determining whether to use an existential or a universal quantifier, try paraphrasing the sentence with an English sentence that uses words besides ‘any’ or ‘anyone’.

## 24.5 Quantifiers and scope

Continuing the example, suppose we want to symbolize these sentences:

1. 14.

If everyone is a bassist, then Lars is a bassist.

2. 15.

Everyone is such that, if they are a bassist, then Lars is a bassist.

To symbolize these sentences, we will have to add a new name to the symbolization key, namely:

$l$:

Lars

Sentence 14 is a conditional, whose antecedent is ‘everyone is a bassist’, so we will symbolize it with ‘$\forall x\,B(x)\rightarrow B(l)$’. This sentence is necessarily true: if everyone is indeed a bassist, then take anyone you like—for example Lars—and he will be a bassist.

Sentence 15, by contrast, might best be paraphrased by ‘every person $x$ is such that, if $x$ is a bassist, then Lars is a bassist’. This is symbolized by ‘$\forall x(B(x)\rightarrow B(l))$’. This sentence can well be false. For instance, Kim Deal is a bassist. So ‘$B(k)$’ is true. Suppose that Lars is not a bassist (say, he’s a drummer instead), so ‘$B(l)$’ is false. Accordingly, ‘$B(k)\rightarrow B(l)$’ will be false, so ‘$\forall x(B(x)\rightarrow B(l))$’ will be false as well. This example shows something else: ‘$\forall x(B(x)\rightarrow B(l))$’ is false exactly when $\exists x\,B(x)\rightarrow B(l)$ is false; they are equivalent. Recalling the lesson about how to symbolize ‘any’ in sentence 12 from the previous section, we can conclude that sentence 15 is just a convoluted way of saying “Lars is a bassist, if anyone is”.

In short, ‘$\forall xB(x)\rightarrow B(l)$’ and ‘$\forall x(B(x)\rightarrow B(l))$’ are very different sentences. We can explain the difference in terms of the scope of the quantifier. The scope of quantification is very much like the scope of negation, which we considered when discussing TFL, and it will help to explain it in this way.

In the sentence ‘$\neg B(k)\rightarrow B(l)$’, the scope of ‘$\neg$’ is just the antecedent of the conditional. We are saying something like: if ‘$B(k)$’ is false, then ‘$B(l)$’ is true. Similarly, in the sentence ‘$\forall xB(x)\rightarrow B(l)$’, the scope of ‘$\forall x$’ is just the antecedent of the conditional. We are saying something like: if ‘$B(x)$’ is true of everything, then ‘$B(l)$’ is also true.

In the sentence ‘$\neg(B(k)\rightarrow B(l))$’, the scope of ‘$\neg$’ is the entire sentence. We are saying something like: ‘$(B(k)\rightarrow B(l))$’ is false. Similarly, in the sentence ‘$\forall x(B(x)\rightarrow B(l))$’, the scope of ‘$\forall x$’ is the entire sentence. We are saying something like: ‘$(B(x)\rightarrow B(l))$’ is true of everything.

The moral of the story is simple. When you are using conditionals, be very careful to make sure that you have sorted out the scope correctly.

## 24.6 Ambiguous predicates

Suppose we just want to symbolize this sentence:

1. 16.

Let the domain be people, let $K(x)$ mean ‘$x$ is a skilled surgeon’, and let $a$ mean Adina. Sentence 16 is simply $K(a)$.

Suppose instead that we want to symbolize this argument:

• The hospital will only hire a skilled surgeon.

• All surgeons are greedy.

• Billy is a surgeon, but is not skilled.

• Billy is greedy, but the hospital will not hire him.

We need to distinguish being a skilled surgeon from merely being a surgeon. So we define this symbolization key:

domain:

people

$G(x)$:

blank${}_{x}$ is greedy

$H(x)$:

The hospital will hire blank${}_{x}$

$R(x)$:

blank${}_{x}$ is a surgeon

$K(x)$:

blank${}_{x}$ is skilled

$b$:

Billy

Now the argument can be symbolized in this way:

• $\forall x\bigl{[}\neg(R(x)\wedge K(x))\rightarrow\neg H(x)\bigr{]}$

• $\forall x(R(x)\rightarrow G(x))$

• $R(b)\wedge\neg K(b)$

• $G(b)\wedge\neg H(b)$

Next suppose that we want to symbolize this argument:

• Carol is a skilled surgeon and a tennis player.

• Carol is a skilled tennis player.

If we start with the symbolization key we used for the previous argument, we could add a predicate (let $T(x)$ mean ‘$x$ is a tennis player’) and a name (let ‘$c$’ be a name for Carol). Then the argument becomes:

• $(R(c)\wedge K(c))\wedge T(c)$

• $T(c)\wedge K(c)$

This symbolization is a disaster! It takes what in English is a terrible argument and symbolizes it as a valid argument in FOL. The problem is that there is a difference between being skilled as a surgeon and skilled as a tennis player. Symbolizing this argument correctly requires two separate predicates, one for each type of skill. If we let $K_{1}(x)$ mean ‘$x$ is skilled as a surgeon’ and $K_{2}(x)$ mean ‘$x$ is skilled as a tennis player,’ then we can symbolize the argument in this way:

• $(R(c)\wedge K_{1}(c))\wedge T(c)$

• $T(c)\wedge K_{2}(c)$

Like the English language argument it symbolizes, this is invalid.

The moral of these examples is that you need to be careful of symbolizing predicates in an ambiguous way. Similar problems can arise with predicates like good, bad, big, and small. Just as skilled surgeons and skilled tennis players have different skills, big dogs, big mice, and big problems are big in different ways.

Is it enough to have a predicate that means ‘$x$ is a skilled surgeon’, rather than two predicates ‘$x$ is skilled’ and ‘$x$ is a surgeon’? Sometimes. As sentence 16 shows, sometimes we do not need to distinguish between skilled surgeons and other surgeons.

Must we always distinguish between different ways of being skilled, good, bad, or big? No. As the argument about Billy shows, sometimes we only need to talk about one kind of skill. If you are symbolizing an argument that is just about dogs, it is fine to define a predicate that means ‘$x$ is big.’ If the domain includes dogs and mice, however, it is probably best to make the predicate mean ‘$x$ is big for a dog.’

## Practice exercises

A. Here are the syllogistic figures identified by Aristotle and his successors, along with their medieval names:

1. 1.

Barbara. All G are F. All H are G. So: All H are F.

2. 2.

Celarent. No G are F. All H are G. So: No H are F.

3. 3.

Ferio. No G are F. Some H is G. So: Some H is not F.

4. 4.

Darii. All G are F. Some H is G. So: Some H is F.

5. 5.

Camestres. All F are G. No H are G. So: No H are F.

6. 6.

Cesare. No F are G. All H are G. So: No H are F.

7. 7.

Baroko. All F are G. Some H is not G. So: Some H is not F.

8. 8.

Festino. No F are G. Some H are G. So: Some H is not F.

9. 9.

Datisi. All G are F. Some G is H. So: Some H is F.

10. 10.

Disamis. Some G is F. All G are H. So: Some H is F.

11. 11.

Ferison. No G are F. Some G is H. So: Some H is not F.

12. 12.

Bokardo. Some G is not F. All G are H. So: Some H is not F.

13. 13.

Camenes. All F are G. No G are H. So: No H is F.

14. 14.

Dimaris. Some F is G. All G are H. So: Some H is F.

15. 15.

Fresison. No F are G. Some G is H. So: Some H is not F.

Symbolize each figure in FOL.

B. Using the following symbolization key:

domain:

people

$K(x)$:

blank${}_{x}$ knows the combination to the safe

$S(x)$:

blank${}_{x}$ is a spy

$V(x)$:

blank${}_{x}$ is a vegetarian

$h$:

Hofthor

$i$:

Ingmar

symbolize the following sentences in FOL:

1. 1.

Neither Hofthor nor Ingmar is a vegetarian.

2. 2.

No spy knows the combination to the safe.

3. 3.

No one knows the combination to the safe unless Ingmar does.

4. 4.

Hofthor is a spy, but no vegetarian is a spy.

C. Using this symbolization key:

domain:

all animals

$A(x)$:

blank${}_{x}$ is an alligator

$M(x)$:

blank${}_{x}$ is a monkey

$R(x)$:

blank${}_{x}$ is a reptile

$Z(x)$:

blank${}_{x}$ lives at the zoo

$a$:

Amos

$b$:

Bouncer

$c$:

Cleo

symbolize each of the following sentences in FOL:

1. 1.

Amos, Bouncer, and Cleo all live at the zoo.

2. 2.

Bouncer is a reptile, but not an alligator.

3. 3.

Some reptile lives at the zoo.

4. 4.

Every alligator is a reptile.

5. 5.

Any animal that lives at the zoo is either a monkey or an alligator.

6. 6.

There are reptiles that are not alligators.

7. 7.

If any animal is an reptile, then Amos is.

8. 8.

If any animal is an alligator, then it is a reptile.

D. For each argument, write a symbolization key and symbolize the argument in FOL.

1. 1.

Willard is a logician. All logicians wear funny hats. So Willard wears a funny hat.

2. 2.

Nothing on my desk escapes my attention. There is a computer on my desk. As such, there is a computer that does not escape my attention.

3. 3.

All my dreams are black and white. Old TV shows are in black and white. Therefore, some of my dreams are old TV shows.

4. 4.

Neither Holmes nor Watson has been to Australia. A person could have seen a kangaroo only if they had been to Australia or to a zoo. Although Watson has not seen a kangaroo, Holmes has. Therefore, Holmes has been to a zoo.

5. 5.

No one expects the Spanish Inquisition. No one knows the troubles I’ve seen. Therefore, anyone who expects the Spanish Inquisition knows the troubles I’ve seen.

6. 6.

All babies are illogical. Nobody who is illogical can manage a crocodile. Berthold is a baby. Therefore, Berthold is unable to manage a crocodile.