# Section 4: Translating to QL

We now have all of the pieces of QL. Translating more complicated sentences will only be a matter of knowing the right way to combine predicates, constants, quantiﬁers, variables, and connectives. Consider these sentences:

14. Every coin in my pocket is a quarter.
15. Some coin on the table is a dime.
16. Not all the coins on the table are dimes.
17. None of the coins in my pocket are dimes.

In providing a symbolization key, we need to specify a UD. Since we are talking about coins in my pocket and on the table, the UD must at least contain all of those coins. Since we are not talking about anything besides coins, we let the UD be all coins. Since we are not talking about any speciﬁc coins, we do not need to deﬁne any constants. So we deﬁne this key:

UD: all coins
Px: $$x$$ is in my pocket.
Tx: $$x$$ is on the table.
Qx: $$x$$ is a quarter.
Dx: $$x$$ is a dime.

Sentence 14 is most naturally translated with a universal quantiﬁer. The universal quantiﬁer says something about everything in the UD, not just about the coins in my pocket. Sentence 14 means that (for any coin) if that coin is in my pocket, then it is a quarter. So we can translate it as ∀$$x$$($$Px$$ → $$Qx$$).

Since sentence 14 is about coins that are both in my pocket and that are quarters, it might be tempting to translate it using a conjunction. However, the sentence ∀$$x$$($$Px$$&$$Qx$$) would mean that everything in the UD is both in my pocket and a quarter: All the coins that exist are quarters in my pocket. This would be a crazy thing to say, and it means something very diﬀerent than sentence 14.

Sentence 15 is most naturally translated with an existential quantiﬁer. It says that there is some coin which is both on the table and which is a dime. So we can translate it as ∃$$x$$($$Tx$$&$$Dx$$).

Notice that we needed to use a conditional with the universal quantiﬁer, but we used a conjunction with the existential quantiﬁer. What would it mean to write∃$$x$$($$Tx$$ → $$Dx$$)? Probably not what you think. It means that there is some member of the UD which would satisfy the subformula; roughly speaking, there is some $$a$$ such that ($$Ta$$ → $$Da$$) is true. In SL, $$\mathcal{A}$$ → $$\mathcal{B}$$ is logically equivalent to ¬$$\mathcal{A}$$ ∨ $$\mathcal{B}$$, and this will also hold in QL. So ∃$$x$$($$Tx$$ → $$Dx$$) is true if there is some a such that (¬$$Ta$$∨$$Da$$); i.e., it 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 ∃$$x$$($$Tx$$ → $$Dx$$) is trivially 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 can do very strange things. As a general rule, do not put conditionals in the scope of existential quantiﬁers unless you are sure that you need one.

Sentence 16 can be paraphrased as, ‘It is not the case that every coin on the table is a dime.’ So we can translate it as ¬∀$$x$$($$Tx$$ → $$Dx$$). You might look at sentence 16 and paraphrase it instead as, ‘Some coin on the table is not a dime.’ You would then translate it as ∃$$x$$($$Tx$$&¬$$Dx$$). Although it is probably not obvious, these two translations are logically equivalent. (This is due to the logical equivalence between ¬∀$$x$$$$\mathcal{A}$$ and ∃$$x$$¬$$\mathcal{A}$$, along with the equivalence between ¬($$\mathcal{A}$$ → $$\mathcal{B}$$) and $$\mathcal{A}$$ & ¬$$\mathcal{B}$$.)

Sentence 17 can be paraphrased as, ‘It is not the case that there is some dime in my pocket.’ This can be translated as ¬∃$$x$$($$Px$$&$$Dx$$). It might also be paraphrased as, ‘Everything in my pocket is a non-dime,’ and then could be translated as ∀$$x$$($$Px$$ →¬$$Dx$$). Again the two translations are logically equivalent. Both are correct translations of sentence 17.

We can now translate the argument from p. 47, the one that motivated the need for quantiﬁers:

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

UD: people
Lx: $$x$$ is a logician.
Fx: $$x$$ wears a funny hat.
w: Willard

Translating, we get:

$$Lw$$
∀$$x$$($$Lx$$ → $$Fx$$)
.˙. $$Fw$$

This captures the structure that was left out of the SL translation of this argument, and this is a valid argument in QL.

## Empty predicates

A predicate need not apply to anything in the UD. A predicate that applies to nothing in the UD is called an empty predicate.

Suppose we want to symbolize these two sentences:

18. Every monkey knows sign language.
19. Some monkey knows sign language.

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

UD: animals
Mx: $$x$$ is a monkey.
Sx: $$x$$ knows sign language.

Sentence 18 can now be translated as ∀$$x$$($$Mx$$ → $$Sx$$).

Sentence 19 becomes ∃$$x$$($$Mx$$&$$Sx$$).

It is tempting to say that sentence 18 entails sentence 19; that is: if every monkey knows sign language, then it must be that some monkey knows sign language. This is a valid inference in Aristotelean logic: All $$M$$s are $$S$$, .˙. some $$M$$ is $$S$$. However, the entailment does not hold in QL. It is possible for the sentence ∀$$x$$($$Mx$$ → $$Sx$$) to be true even though the sentence ∃$$x$$($$Mx$$&$$Sx$$) is false.

How can this be? The answer comes from considering whether these sentences would be true or false if there were no monkeys.

 -A UD must have at least one member. -A predicate may apply to some, all, or no members of the UD. -A constant must pick out exactly one member of the UD. A member of the UD may be picked out by one constant, many constants, or none at all.

We have deﬁned ∀ and ∃ in such a way that ∀$$\mathcal{A}$$ is equivalent to ¬∃¬$$\mathcal{A}$$. As such, the universal quantiﬁer doesn’t involve the existence of anything— only non-existence. If sentence 18 is true, then there are no monkeys who don’t know sign language. If there were no monkeys, then ∀$$x$$($$Mx$$ → $$Sx$$) would be true and ∃$$x$$($$Mx$$&$$Sx$$) would be false.

We allow empty predicates because we want to be able to say things like, ‘I do not know if there are any monkeys, but any monkeys that there are know sign language.’ That is, we want to be able to have predicates that do not (or might not) refer to anything.

What happens if we add an empty predicate $$R$$ to the interpretation above? For example, we might deﬁne $$Rx$$ to mean ‘$$x$$ is a refrigerator.’ Now the sentence ∀$$x$$($$Rx$$ → $$Mx$$) will be true. This is counterintuitive, since we do not want to say that there are a whole bunch of refrigerator monkeys. It is important to remember, though, that ∀$$x$$($$Rx$$ → $$Mx$$) means that any member of the UD which is a refrigerator is a monkey. Since the UD is animals, there are no refrigerators in the UD and so the sentence is trivially true.

If you were actually translating the sentence ‘All refrigerators are monkeys’, then you would want to include appliances in the UD. Then the predicate $$R$$ would not be empty and the sentence ∀$$x$$($$Rx$$ → $$Mx$$) would be false.

## Picking a Universe of Discourse

The appropriate symbolization of an English language sentence in QL will depend on the symbolization key. In some ways, this is obvious: It matters whether $$Dx$$ means ‘$$x$$ is dainty’ or ‘$$x$$ is dangerous.’ The meaning of sentences in QL also depends on the UD.

Let $$Rx$$ mean ‘$$x$$ is a rose,’ let $$Tx$$ mean ‘$$x$$ has a thorn,’ and consider this sentence:

20. Every rose has a thorn.

It is tempting to say that sentence 20 should be translated as ∀$$x$$($$Rx$$ → $$Tx$$). If the UD contains all roses, that would be correct. Yet if the UD is merely things on my kitchen table, then ∀$$x$$($$Rx$$ → $$Tx$$) would only mean 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.

The universal quantiﬁer only ranges over members of the UD, so we need to include all roses in the UD in order to translate sentence 20. We have two options. First, we can restrict the UD to include all roses but only roses. Then sentence 20 becomes ∀$$xTx$$. This means that everything in the UD has a thorn; since the UD just is the set of roses, this means that every rose has a thorn. This option can save us trouble if every sentence that we want to translate using the symbolization key is about roses.

Second, we can let the UD contain things besides roses: rhododendrons, rats, riﬂes, and whatall else. Then sentence 20 must be ∀$$x$$($$Rx$$ → $$Tx$$).

If we wanted the universal quantiﬁer to mean every thing, without restriction, then we might try to specify a UD that contains everything. This would lead to problems. Does ‘everything’ include things that have only been imagined, like ﬁctional characters? On the one hand, we want to be able to symbolize arguments about Hamlet or Sherlock Holmes. So we need to have the option of including ﬁctional characters in the UD. On the other hand, we never need to talk about every thing that does not exist. That might not even make sense. There are philosophical issues here that we will not try to address. We can avoid these diﬃculties by always specifying the UD. For example, if we mean to talk about plants, people, and cities, then the UD might be ‘living things and places.’

Suppose that we want to translate sentence 20 and, with the same symbolization key, translate these sentences:

21. Esmerelda has a rose in her hair.
22. Everyone is cross with Esmerelda.

We need a UD that includes roses (so that we can symbolize sentence 20) and a UD that includes people (so we can translate sentence 21–22.) Here is a suitable key:

UD: people and plants
Px: $$x$$ is a person.
Rx: $$x$$ is a rose.
Tx: $$x$$ has a thorn.
Cxy: $$x$$ is cross with $$y$$.
Hxy: $$x$$ has $$y$$ in their hair.
e: Esmerelda

Since we do not have a predicate that means ‘... has a rose in her hair’, translating sentence 21 will require paraphrasing. The sentence says that there is a rose in Esmerelda’s hair; that is, there is something which is both a rose and is in Esmerelda’s hair. So we get: ∃$$x$$($$Rx$$&$$Hex$$).

It is tempting to translate sentence 22 as ∀$$xCxe$$. Unfortunately, this would mean that every member of the UD is cross with Esmerelda— both people and plants. It would mean, for instance, that the rose in Esmerelda’s hair is cross with her. Of course, sentence 22 does not mean that.

‘Everyone’ means every person, not every member of the UD. So we can paraphrase sentence 22 as, ‘Every person is cross with Esmerelda.’ We know how to translate sentences like this: ∀$$x$$($$Px$$ → $$Cxe$$)

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

## Translating pronouns

When translating to QL, it is important to understand the structure of the sentences you want to translate. What matters is the ﬁnal translation in QL, and sometimes you will be able to move from an English language sentence directly to a sentence of QL. 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 translate directly into QL.

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

UD: people
Gx: $$x$$ can play guitar.
Rx: $$x$$ is a rock star.
l: Lemmy

Now consider these sentences:

23. If Lemmy can play guitar, then he is a rock star.
24. If a person can play guitar, then he is a rock star.

Sentence 23 and sentence 24 have the same consequent (‘... he is a rock star’), but they cannot be translated in the same way. It helps to paraphrase the original sentences, replacing pronouns with explicit references.

Sentence 23 can be paraphrased as, ‘If Lemmy can play guitar, then Lemmy is a rockstar.’ This can obviously be translated as $$Gl$$ → $$Rl$$.

Sentence 24 must be paraphrased diﬀerently: ‘If a person can play guitar, then that person is a rock star.’ This sentence is not about any particular person, so we need a variable. Translating halfway, we can paraphrase the sentence as, ‘For any person $$x$$, if $$x$$ can play guitar, then $$x$$ is a rockstar.’ Now this can be translated as ∀$$x$$($$Gx$$ → $$Rx$$). This is the same as, ‘Everyone who can play guitar is a rock star.’

Consider these further sentences:

25. If anyone can play guitar, then Lemmy can.
26. If anyone can play guitar, then he or she is a rock star.

These two sentences have the same antecedent (‘If anyone can play guitar...’), but they have diﬀerent logical structures.

Sentence 25 can be paraphrased, ‘If someone can play guitar, then Lemmy can play guitar.’ The antecedent and consequent are separate sentences, so it can be symbolized with a conditional as the main logical operator: ∃$$xGx$$ → $$Gl$$.

Sentence 26 can be paraphrased, ‘For anyone, if that one can play guitar, then that one is a rock star.’ It would be a mistake to symbolize this with an existential quantiﬁer, because it is talking about everybody. The sentence is equivalent to ‘All guitar players are rock stars.’ It is best translated as∀$$x$$($$Gx$$ → $$Rx$$).

The English words ‘any’ and ‘anyone’ should typically be translated using quantiﬁers. As these two examples show, they sometimes call for an existential quantiﬁer (as in sentence 25) and sometimes for a universal quantiﬁer (as in sentence 26). If you have a hard time determining which is required, paraphrase the sentence with an English language sentence that uses words besides ‘any’ or ‘anyone.’

## Quantiﬁers and scope

In the sentence ∃$$xGx$$ → $$Gl$$, the scope of the existential quantiﬁer is the expression $$Gx$$. Would it matter if the scope of the quantiﬁer were the whole sentence? That is, does the sentence ∃$$x$$($$Gx$$ → $$Gl$$) mean something diﬀerent?

With the key given above, ∃$$xGx$$ → $$Gl$$ means that if there is some guitarist, then Lemmy is a guitarist. ∃$$x$$($$Gx$$ → $$Gl$$) would mean that there is some person such that if that person were a guitarist, then Lemmy would be a guitarist. Recall that the conditional here is a material conditional; the conditional is true if the antecedent is false. Let the constant p denote the author of this book, someone who is certainly not a guitarist. The sentence $$Gp$$ → $$Gl$$ is true because $$Gp$$ is false. Since someone (namely $$p$$) satisﬁes the sentence, then ∃$$x$$($$Gx$$ → $$Gl$$) is true. The sentence is true because there is a non-guitarist, regardless of Lemmy’s skill with the guitar.

Something strange happened when we changed the scope of the quantiﬁer, because the conditional in QL is a material conditional. In order to keep the meaning the same, we would have to change the quantiﬁer: ∃$$xGx$$ → $$Gl$$ means the same thing as ∀$$x$$($$Gx$$ → $$Gl$$), and ∃$$x$$($$Gx$$ → $$Gl$$) means the same thing as ∀$$xGx$$ → $$Gl$$.

This oddity does not arise with other connectives or if the variable is in the consequent of the conditional. For example, ∃$$xGx$$&$$Gl$$ means the same thing as ∃$$x$$($$Gx$$&$$Gl$$), and $$Gl$$ →∃$$xGx$$ means the same things as ∃$$x$$($$Gl$$ → $$Gx$$).

## Ambiguous predicates

Suppose we just want to translate this sentence:

27. Adina is a skilled surgeon.

Let the UD be people, let $$Kx$$ mean ‘$$x$$ is a skilled surgeon’, and let a mean Adina. Sentence 27 is simply $$Ka$$.

Suppose instead that we want to translate this argument:

The hospital will only hire a skilled surgeon. All surgeons are greedy.
Billy is a surgeon, but is not skilled. Therefore, 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:

UD: people
Gx: $$x$$ is greedy.
Hx: The hospital will hire $$x$$.
Rx: $$x$$ is a surgeon.
Kx: $$x$$ is skilled.
b: Billy

Now the argument can be translated in this way:

∀$$x$$[¬($$Rx$$&$$Kx$$) →¬$$Hx$$]
∀$$x$$($$Rx$$ → $$Gx$$)
$$Rb$$&¬$$Kb$$
.˙. $$Gb$$&¬$$Hb$$

Next suppose that we want to translate this argument:

Carol is a skilled surgeon and a tennis player. Therefore, 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 Tx mean ‘$$x$$ is a tennis player’) and a constant (let $$c$$ mean Carol). Then the argument becomes:

($$Rc$$&$$Kc$$)&$$Tc$$
.˙. $$Tc$$&$$Kc$$

This translation is a disaster! It takes what in English is a terrible argument and translates it as a valid argument in QL. The problem is that there is a diﬀerence between being skilled as a surgeon and skilled as a tennis player. Translating 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:

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

Like the English language argument it translates, 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 27 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 translating an argument that is just about dogs, it is ﬁne to deﬁne a predicate that means ‘$$x$$ is big.’ If the UD includes dogs and mice, however, it is probably best to make the predicate mean ‘$$x$$ is big for a dog.’

## Multiple quantiﬁers

Consider this following symbolization key and the sentences that follow it:

UD: People and dogs
Dx: $$x$$ is a dog.
Fxy: $$x$$ is a friend of $$y$$.
Oxy: $$x$$ owns $$y$$.
f: Fiﬁ
g: Gerald

28. Fiﬁ is a dog.
29. Gerald is a dog owner.
30. Someone is a dog owner.
31. All of Gerald’s friends are dog owners.
32. Every dog owner is the friend of a dog owner.

Sentence 28 is easy: $$Df$$.

Sentence 29 can be paraphrased as, ‘There is a dog that Gerald owns.’ This can be translated as ∃$$x$$($$Dx$$&$$Ogx$$).

Sentence 30 can be paraphrased as, ‘There is some $$y$$ such that $$y$$ is a dog owner.’ The subsentence ‘$$y$$ is a dog owner’ is just like sentence 29, except that it is about $$y$$ rather than being about Gerald. So we can translate sentence 30 as ∃$$y$$∃$$x$$($$Dx$$&$$Oyx$$).

Sentence 31 can be paraphrased as, ‘Every friend of Gerald is a dog owner.’ Translating part of this sentence, we get ∀$$x$$($$Fxg$$ → ‘$$x$$ is a dog owner’). Again, it is important to recognize that ‘$$x$$ is a dog owner’ is structurally just like sentence 29. Since we already have an $$x$$-quantiﬁer, we will need a diﬀerent variable for the existential quantiﬁer. Any other variable will do. Using $$z$$, sentence 31 can be translated as ∀$$x$$[$$Fxg$$ →∃$$z$$($$Dz$$ & $$Oxz$$)].

Sentence 32 can be paraphrased as ‘For any $$x$$ that is a dog owner, there is a dog owner who is $$x$$’s friend.’ Partially translated, this becomes

∀$$x$$[$$x$$ is a dog owner →∃$$y$$($$y$$ is a dog owner&$$Fxy$$)].

Completing the translation, sentence 32 becomes

∀$$x$$[∃$$z$$($$Dz$$ & $$Oxz$$) →∃$$y$$∃$$z$$($$Dz$$ & $$Oyz$$)&$$Fxy$$)].

Consider this symbolization key and these sentences:

UD: people
Lxy: $$x$$ likes $$y$$.
i: Imre.
k: Karl.

33. Imre likes everyone that Karl likes.
34. There is someone who likes everyone who likes everyone that he likes.

Sentence 33 can be partially translated as∀$$x$$(Karl likes $$x$$ → Imre likes $$x$$). This becomes ∀$$x$$(Lkx → Lix).

Sentence 34 is almost a tongue-twister. There is little hope of writing down the whole translation immediately, but we can proceed by small steps. An initial, partial translation might look like this:

∃$$x$$ everyone who likes everyone that $$x$$ likes is liked by $$x$$

The part that remains in English is a universal sentence, so we translate further:

∃$$x$$∀$$y$$($$y$$ likes everyone that $$x$$ likes → $$x$$ likes $$y$$).

The antecedent of the conditional is structurally just like sentence 33, with y and $$x$$ in place of Imre and Karl. So sentence 34 can be completely translated in this way

∃$$x$$∀$$y$$[∀$$z$$($$Lxz$$ → $$Lyz$$) → $$Lxy$$]

When symbolizing sentences with multiple quantiﬁers, it is best to proceed by small steps. Paraphrase the English sentence so that the logical structure is readily symbolized in QL. Then translate piecemeal, replacing the daunting task of translating a long sentence with the simpler task of translating shorter formulae.