# Section 6: Identity

Consider this sentence:

35. Pavel owes money to everyone else.

Let the UD be people; this will allow us to translate ‘everyone’ as a universal quantiﬁer. Let $$Oxy$$ mean ‘$$x$$ owes money to $$y$$’, and let $$p$$ mean Pavel. Now we can symbolize sentence 35 as ∀$$xOpx$$. Unfortunately, this translation has some odd consequences. It says that Pavel owes money to every member of the UD, including Pavel; it entails that Pavel owes money to himself. However, sentence 35 does not say that Pavel owes money to himself; he owes money to everyone else. This is a problem, because ∀$$xOpx$$ is the best translation we can give of this sentence into QL.

The solution is to add another symbol to QL. The symbol ‘=’ is a two-place predicate. Since it has a special logical meaning, we write it a bit diﬀerently: For two terms $$t$$1 and $$t$$2, $$t$$1 = $$t$$2 is an atomic formula.

The predicate $$x$$ = $$y$$ means ‘$$x$$ is identical to $$y$$.’ This does not mean merely that $$x$$ and $$y$$ are indistinguishable or that all of the same predicates are true of them. Rather, it means that $$x$$ and $$y$$ are the very same thing.

When we write $$x$$ ≠ $$y$$, we mean that $$x$$ and $$y$$ are not identical. There is no reason to introduce this as an additional predicate. Instead, $$x$$ ≠ $$y$$ is an abbreviation of ¬($$x$$ = $$y$$).

Now suppose we want to symbolize this sentence:

36. Pavel is Mister Checkov.

Let the constant $$c$$ mean Mister Checkov. Sentence 36 can be symbolized as $$p$$ = $$c$$. This means that the constants $$p$$ and $$c$$ both refer to the same guy.

This is all well and good, but how does it help with sentence 35? That sentence can be paraphrased as, ‘Everyone who is not Pavel is owed money by Pavel.’ This is a sentence structure we already know how to symbolize: ‘For all $$x$$, if $$x$$ is not Pavel, then $$x$$ is owed money by Pavel.’ In QL with identity, this becomes ∀$$x$$($$x$$ ≠ $$p$$ → $$Opx$$).

In addition to sentences that use the word ‘else’, identity will be helpful when symbolizing some sentences that contain the words ‘besides’ and ‘only.’ Consider these examples:

37. No one besides Pavel owes money to Hikaru.
38. Only Pavel owes Hikaru money.

We add the constant $$h$$, which means Hikaru.

Sentence 37 can be paraphrased as, ‘No one who is not Pavel owes money to Hikaru.’ This can be translated as ¬∃$$x$$($$x$$ ≠ $$p$$ & $$Oxh$$).

Sentence 38 can be paraphrased as, ‘Pavel owes Hikaru and no one besides Pavel owes Hikaru money.’ We have already translated one of the conjuncts, and the other is straightforward. Sentence 38 becomes $$Oph$$&¬∃$$x$$($$x$$≠ $$p$$ & $$Oxh$$).

## Expressions of quantity

We can also use identity to say how many things there are of a particular kind. For example, consider these sentences:

39. There is at least one apple on the table.
40. There are at least two apples on the table.
41. There are at least three apples on the table.

Let the UD be things on the table, and let Ax mean ‘$$x$$ is an apple.’

Sentence 39 does not require identity. It can be translated adequately as ∃$$xAx$$: There is some apple on the table— perhaps many, but at least one.

It might be tempting to also translate sentence 40 without identity. Yet consider the sentence ∃$$x$$∃$$y$$($$Ax$$&$$Ay$$). It means that there is some apple $$x$$ in the UD and some apple $$y$$ in the UD. Since nothing precludes $$x$$ and $$y$$ from picking out the same member of the UD, this would be true even if there were only one apple. In order to make sure that there are two diﬀerent apples, we need an identity predicate. Sentence 40 needs to say that the two apples that exist are not identical, so it can be translated as ∃$$x$$∃$$y$$($$Ax$$&$$Ay$$ & $$x$$ ≠ $$y$$).

Sentence 41 requires talking about three diﬀerent apples. It can be translated as ∃$$x$$∃$$y$$∃$$z$$($$Ax$$ & $$Ay$$ & $$Az$$ & $$x$$ ≠ $$y$$ & $$y$$ ≠ $$z$$ & $$x$$ ≠ $$z$$).

Continuing in this way, we could translate ‘There are at least $$n$$ apples on the table.’ There is a summary of how to symbolize sentences like these on p. 151.

Now consider these sentences:

42. There is at most one apple on the table.
43. There are at most two apples on the table.

Sentence 42 can be paraphrased as, ‘It is not the case that there are at least two apples on the table.’ This is just the negation of sentence 40:

¬∃$$x$$∃$$y$$($$Ax$$&$$Ay$$ & $$x$$ ≠ $$y$$)

Sentence 42 can also be approached in another way. It means that any apples that there are on the table must be the selfsame apple, so it can be translated as ∀$$x$$∀$$y$$[($$Ax$$&$$Ay$$) → $$x$$ = $$y$$]. The two translations are logically equivalent, so both are correct.

In a similar way, sentence 43 can be translated in two equivalent ways. It can be paraphrased as, ‘It is not the case that there are three or more distinct apples’, so it can be translated as the negation of sentence 41. Using universal quantiﬁers, it can also be translated as

∀$$x$$∀$$y$$∀$$z$$[($$Ax$$&$$Ay$$ &$$Az$$) → ($$x$$ = $$y$$∨$$x$$ = $$z$$∨$$y$$ = $$z$$)].

The examples above are sentences about apples, but the logical structure of the sentences translates mathematical inequalities like $$a$$ ≥ 3, $$a$$ ≤ 2, and so on. We also want to be able to translate statements of equality which say exactly how many things there are. For example:

44. There is exactly one apple on the table.
45. There are exactly two apples on the table.

Sentence 44 can be paraphrased as, ‘There is at least one apple on the table, and there is at most one apple on the table.’ This is just the conjunction of sentence 39 and sentence 42: ∃$$xAx$$&∀$$x$$∀$$y$$[($$Ax$$&$$Ay$$) → $$x$$ = $$y$$]. This is a somewhat complicated way of going about it. It is perhaps more straightforward to paraphrase sentence 44 as, ‘There is a thing which is the only apple on the table.’ Thought of in this way, the sentence can be translated ∃$$x$$[$$Ax$$&¬∃$$y$$($$Ay$$ & $$x$$ ≠$$y$$)].

Similarly, sentence 45 may be paraphrased as, ‘There are two diﬀerent apples on the table, and these are the only apples on the table.’ This can be translated as ∃$$x$$∃$$y$$[$$Ax$$&$$Ay$$ & $$x$$ ≠ $$y$$ &¬∃$$z$$($$Az$$ & $$x$$ ≠ $$y$$ & $$y$$ ≠ $$z$$)].

Finally, consider this sentence:

46. There are at most two things on the table.

It might be tempting to add a predicate so that $$Tx$$ would mean ‘$$x$$ is a thing on the table.’ However, this is unnecessary. Since the UD is the set of things on the table, all members of the UD are on the table. If we want to talk about a thing on the table, we need only use a quantiﬁer. Sentence 46 can be symbolized like sentence 43 (which said that there were at most two apples), but leaving out the predicate entirely. That is, sentence 46 can be translated as ∀$$x$$∀$$y$$∀$$z$$($$x$$ = $$y$$∨$$x$$ = $$z$$∨$$y$$ = $$z$$).

Techniques for symbolizing expressions of quantity (‘at most’, ‘at least’, and ‘exactly’) are summarized on p. 151.

## Deﬁnite descriptions

Recall that a constant of QL must refer to some member of the UD. This constraint allows us to avoid the problem of non-referring terms. Given a UD that included only actually existing creatures but a constant $$c$$ that meant ‘chimera’ (a mythical creature), sentences containing $$c$$ would become impossible to evaluate.

The most widely inﬂuential solution to this problem was introduced by Bertrand Russell in 1905. Russell asked how we should understand this sentence:

47. The present king of France is bald.

The phrase ‘the present king of France’ is supposed to pick out an individual by means of a deﬁnite description. However, there was no king of France in 1905 and there is none now. Since the description is a non-referring term, we cannot just deﬁne a constant to mean ‘the present king of France’ and translate the sentence as $$Kf$$.

Russell’s idea was that sentences that contain deﬁnite descriptions have a different logical structure than sentences that contain proper names, even though they share the same grammatical form. What do we mean when we use an unproblematic, referring description, like ‘the highest peak in Washington state’? We mean that there is such a peak, because we could not talk about it otherwise. We also mean that it is the only such peak. If there was another peak in Washington state of exactly the same height as Mount Rainier, then Mount Rainier would not be the highest peak.

According to this analysis, sentence 47 is saying three things. First, it makes an existence claim: There is some present king of France. Second, it makes a uniqueness claim: This guy is the only present king of France. Third, it makes a claim of predication: This guy is bald.

In order to symbolize deﬁnite descriptions in this way, we need the identity predicate. Without it, we could not translate the uniqueness claim which (according to Russell) is implicit in the deﬁnite description.

Let the UD be people actually living, let $$Fx$$ mean ‘$$x$$ is the present king of France’, and let $$Bx$$ mean ‘$$x$$ is bald.’ Sentence 47 can then be translated as ∃$$x$$[$$Fx$$&¬∃$$y$$($$Fy$$ & $$x$$ ≠ $$y$$)&$$Bx$$]. This says that there is some guy who is the present king of France, he is the only present king of France, and he is bald.

Understood in this way, sentence 47 is meaningful but false. It says that this guy exists, but he does not.

The problem of non-referring terms is most vexing when we try to translate negations. So consider this sentence:

48. The present king of France is not bald.

According to Russell, this sentence is ambiguous in English. It could mean either of two things:

48a. It is not the case that the present king of France is bald.
48b. The present king of France is non-bald.

Both possible meanings negate sentence 47, but they put the negation in different places.

Sentence 48a is called a wide-scope negation, because it negates the entire sentence. It can be translated as ¬∃$$xFx$$&¬∃$$y$$($$Fy$$ & $$x$$ ≠ $$y$$)&$$Bx$$. This does not say anything about the present king of France, but rather says that some sentence about the present king of France is false. Since sentence 47 if false, sentence 48a is true.

Sentence 48b says something about the present king of France. It says that he lacks the property of baldness. Like sentence 47, it makes an existence claim and a uniqueness claim; it just denies the claim of predication. This is called narrow-scope negation. It can be translated as ∃$$xFx$$&¬∃$$y$$($$Fy$$ & $$x$$ ≠ $$y$$)&¬$$Bx$$. Since there is no present king of France, this sentence is false.

Russell’s theory of deﬁnite descriptions resolves the problem of non-referring terms and also explains why it seemed so paradoxical. Before we distinguished between the wide-scope and narrow-scope negations, it seemed that sentences like 48 should be both true and false. By showing that such sentences are ambiguous, Russell showed that they are true understood one way but false understood another way.

For a more detailed discussion of Russell’s theory of deﬁnite descriptions, including objections to it, see Peter Ludlow’s entry ‘descriptions’ in The Stanford Encyclopedia of Philosophy: Summer 2005 edition, edited by Edward N. Zalta, http://plato.stanford.edu/archives/s.../descriptions/