9

Relations and Identity

We now bring quantificational logic up to full power by adding identity and relational statements, like “a=b” and “Lrj” (“Romeo loves Juliet”).

9.1 Identity translations

Our third rule for forming quantificational wffs introduces “=” (“equals”):

The result of writing a small letter and then “=” and then a small letter is a wff.

This lets us construct wffs like these:

· x=y = x equals y.

· r=l = Romeo is the lover of Juliet.

· ∼p=l = Paris isn’t the lover of Juliet.

We negate an identity wff by writing “∼” in front. Neither “r=l” nor “∼p=l” use parentheses, since these aren’t needed to avoid ambiguity.

The simplest use of “=” translates an “is” that goes between singular terms. Recall the difference between general and singular terms:

Use capital letters for general terms, which describe or put in a category:

I = an Italian

C = charming

F = drives a Ford

Use capitals for “a so and so,” adjectives, and verbs.

Use small letters for singular terms, which pick out a specific person or thing:

i = the richest Italian

t = this child

r = Romeo

Use small letters for “the so and so,” “this so and so,” and proper names.

Compare these two forms: 0208

Predication

Lr

Romeo is a lover

Identity

r=l

Romeo is the lover of Juliet

Use “=” for “is” if both sides are singular terms (represented by small letters). The “is” of identity can be replaced with “is identical to” or “is the same entity as,” and can be reversed (so if x=y then y=x).

We can translate “other than,” “besides,” and “alone” using identity:

· Someone other than (besides) Romeo is rich

= (∃x)(∼x=r • Rx)

For some x, x ≠ Romeo and x is rich

· Romeo alone is rich

= (Rr • ∼(∃x)(∼x=r • Rx))

Romeo is rich and it’s false that, for some x, x ≠ Romeo and x is rich

These translations also work if we switch conjuncts (“∼x=r” and “Rx”) or switch the order of letters in an identity (“∼r=x” works in place of “∼x=r”). We also can translate some numerical notions, for example:

· At least two are rich

= (∃x)(∃y)(∼x=y • (Rx • Ry))

For some x and some y: x≠y, x is rich, and y is rich

The pair of quantifiers “(∃x)(∃y)” (“for some x and some y”) doesn’t say whether x and y are identical; so we need “∼x=y” to say they aren’t. Henceforth we’ll often need more variable letters than just “x” to keep references straight. In general, it doesn’t matter which variable letters we use; we can translate “At least one is rich” as “(∃x)Rx” or “(∃y)Ry” or “(∃z)Rz.”

We can express “exactly one” and “exactly two” (and “exactly n,” for any specific whole number n):

· Exactly one is dark

= (∃x)(Dx • ∼(∃y)(∼y=x • Dy))

For some x, x is dark and there’s no y such that y≠x and y is dark

· Exactly two are dark

= (∃x)(∃y)(((Dx • Dy) • ∼x=y) • ∼(∃z)((∼z=x • ∼z=y) • Dz))

For some x and some y, x is dark and y is dark and x≠y and there’s no z such that z≠x and z≠y and z is dark

We also can express addition. Here’s a Loglish paraphrase of “1 + 1 = 2” and the corresponding formula: 0209

If exactly one being is F and exactly one being is G and nothing is F-and-G, then exactly two beings are F-or-G.

((((∃x)(Fx • ∼(∃y)(∼y=x • Fy)) • (∃x)(Gx • ∼(∃y)(∼y=x • Gy))) • ∼(∃x)(Fx • Gx)) ⊃ (∃x)(∃y)(((Fx ∨ Gx) • (Fy ∨ Gy)) • (∼x=y • ∼(∃z)((∼z=x • ∼z=y) • (Fz ∨ Gz)))))

We could prove our “1 + 1 = 2” formula by assuming its denial and deriving a contradiction. While this would be tedious, it’s interesting that it could be done. In principle, we could prove “2 + 2 = 4” and “5 + 7 = 12” – and the additions on your income tax form. Some mean teachers assign such homework problems.

9.1a Exercise: LogiCola H (IM & IT)

Translate these English sentences into wffs.

Jim is the goalie and is a student.

(j=g • Sj)

1. Aristotle is a logician.

2. Aristotle is the greatest logician.

3. Aristotle isn’t Plato.

4. Someone besides Aristotle is a logician.

5. There are at least two logicians.

6. Aristotle alone is a logician.

7. All logicians other than Aristotle are evil.

8. No one besides Aristotle is evil.

9. The philosopher is Aristotle.

10.There’s exactly one logician.

11.There’s exactly one evil logician.

12.Everyone besides Aristotle and Plato is evil.

13.If the thief is intelligent, then you aren’t the thief.

14.Carol is my only sister.

15.Alice runs but isn’t the fastest runner.

16.There’s at most one king.

17.The king is bald.

18.There’s exactly one king and he is bald.

9.2 Identity proofs

We need two new inference rules for identity. This self-identity rule holds regardless of what constant replaces “a”:

Self-identity SI

a=a

0210 This is an axiom – a basic unproved assertion that can be used to prove other things. This rule says that we may assert a self-identity as a “derived line” anywhere in a proof, regardless of earlier lines. Adding “a=a” can be useful if we already have “∼a=a” (since then we get a contradiction) or already have a line like “(a=a ⊃ Gb)” (since then we can apply an I-rule). Our self-identity line will mention this previous line; it might say “∴ b=b {self-identity, to contradict 3}.”

This substitute-equals rule is based on interchangeability of identicals: if a=b, then whatever is true of a is true of b, and vice versa. This rule holds regardless of what constants replace “a” and “b” and what wffs replace “Fa” and “Fb” – provided that the two wffs are alike except that the constants are interchanged in one or more occurrences:

Substitute Equals SE

a=b, Fa → Fb

Here’s an easy identity proof:

· I weigh 180 pounds.

My mind doesn’t weigh 180 pounds.

∴ I’m not identical to my mind.

Line 4 follows by substituting equals; if i and m are identical, then whatever is true of one is true of the other.

Here’s an easy invalid argument and its refutation:

· The bankrobber wears size-twelve shoes.

You wear size-twelve shoes.

∴ You’re the bankrobber.

· 1 Wb Invalid

· 2 Wu

· [ ∴ u=b

· 3 asm: ∼u=b

b, u

Wb, Wu, ∼u=b

Since we can’t infer anything here (we can’t do much with “∼u=b”), we set up a possible world to refute the argument. This world contains two distinct persons, the bankrobber and you, each wearing size-twelve shoes. Since the premises are all true and conclusion false in this world, our argument is invalid.

Our next example involves pluralism and monism:

· Pluralism (there’s more than one being): (∃x)(∃y)∼x=y For some x and some y: x≠y

· Monism (there’s exactly one being): (∃x)(y)y=x For some x, every y is identical to x

Here’s a proof that pluralism entails the falsity of monism: 0211

· There’s more than one being.

∴ It’s false that there’s exactly one being.

Lines 1 and 2 have back-to-back quantifiers. We can drop only quantifiers that are initial and hence outermost; so we drop quantifiers one at a time, starting from the outside. After dropping quantifiers, we substitute equals to get line 8: our “b=c” line lets us take “a=c” and replace “c” with “b,” getting “a=b.”

We didn’t bother to derive “c=c” from “(y)y=c” in line 5. From now on, it’ll often be too tedious to drop universal quantifiers using every old constant. So we’ll just derive instances likely to be useful for our proof or refutation.

Our substitute-equals rule seems to hold universally in arguments about matter or math. But it can fail with mental phenomena. Consider this argument (where “Bx” stands for “Jones believes that x is on the penny”):¹

¹ We could make the same point using relational logic (“Bjl, l=r ∴ Bjr” – where “Bxy” means “x believes that y is on the penny”) or Chapter 13’s belief logic (“j:Pl, l=r ∴ j:Pr”). Our belief logic explicitly restricts the use of the substitute-equals rule with belief formulas (§13.2).

· Jones believes that Lincoln is on the penny.

Lincoln is the first Republican president.

∴ Jones believes that the first Republican president is on the penny.

· Bl

l=r

∴ Br

If Jones is unaware that Lincoln was the first Republican president, the premises could be true while the conclusion is false. So the argument is invalid. But yet we can derive the conclusion from the premises using our substitute-equals rule. So something is wrong here.

To avoid this problem, we’ll disallow translating into quantificational logic any predicates or relations that violate the substitute-equals rule. So we won’t let “Bx” stand for “Jones believes that x is on the penny.” Statements about beliefs and other mental phenomena often violate this rule; so we have to be careful translating such statements into quantificational logic.

So the mental seems to follow different logical patterns from the physical. Does this refute the materialist project of reducing the mental to the physical? Philosophers dispute this question. 0212

9.2a Exercise: LogiCola IDC

Say whether each is valid (and give a proof) or invalid (and give a refutation).

a=b

∴ b=a

1. Fa

∴ ∼(∃x)(Fx • ∼x=a)

2. (a=b ⊃ ∼(∃x)Fx)

∴ (Fa ⊃ ∼Fb)

3. a=b

b=c

∴ a=c

4. ∼a=b

c=b

∴ ∼a=c

5. ∼a=b

∼c=b

∴ a=c

6. a=b

∴ (Fa ≡ Fb)

7. a=b

(x)(Fx ⊃ Gx)

∼Ga

∴ ∼Fb

8. Fa

∴ (x)(x=a ⊃ Fx)

9. ∴ (∃x)(y)y=x

10.∴ (∃x)(∃y)∼y=x

9.2b Exercise: LogiCola IDC

First appraise intuitively. Then translate into logic and say whether valid (and give a proof) or invalid (and give a refutation). You’ll have to figure out what letters to use; be careful about deciding between small and capital letters.

1. Keith is my only nephew.

My only nephew knows more about BASIC than I do.

Keith is a ten-year-old.

∴ Some ten-year-olds know more about BASIC than I do.

2. Some are logicians.

Some aren’t logicians.

∴ There’s more than one being.

3. This chemical process is publicly observable.

This pain isn’t publicly observable.

∴ This pain isn’t identical to this chemical process. [This attacks the identity theory of the mind, which identifies mental events with chemical processes.]

4. The person who left a lighter is the murderer. The person who left a lighter is a smoker.

No smokers are backpackers.

∴ The murderer isn’t a backpacker.

5. The murderer isn’t a backpacker.

You aren’t a backpacker.

∴ You’re the murderer. 0213

6. If Speedy Jones looks back to the quarterback just before the hike, then Speedy Jones is the primary receiver.

The primary receiver is the receiver you should try to cover.

∴ If Speedy Jones looks back to the quarterback just before the hike, then Speedy Jones is the receiver you should try to cover.

7. Judy isn’t the world’s best cook.

The world’s best cook lives in Detroit.

∴ Judy doesn’t live in Detroit.

8. Patricia lives in North Dakota.

Blondie lives in North Dakota.

∴ At least two people live in North Dakota.

9. Your grade is the average of your tests.

The average of your tests is B.

∴ Your grade is B.

10.Either you knew where the money was, or the thief knew where it was.

You didn’t know where the money was.

∴ You aren’t the thief.

11.The man of Suzy’s dreams is either rich or handsome. You aren’t rich.

∴ If you’re handsome, then you’re the man of Suzy’s dreams.

12.If someone confesses, then someone goes to jail.

I confess.

I don’t go to jail.

∴ Someone besides me goes to jail.

13.David stole money.

The nastiest person at the party stole money.

David isn’t the nastiest person at the party.

∴ At least two people stole money. [See problem 4 of §2.3b.]

14.No one besides Carol and the detective had a key.

Someone who had a key stole money.

∴ Either Carol or the detective stole money.

15.Exactly one person lives in North Dakota.

Paul lives in North Dakota.

Paul is a farmer.

∴ Everyone who lives in North Dakota is a farmer.

16.The wildcard team with the best record goes to the playoffs. Cleveland isn’t the wildcard team with the best record.

∴ Cleveland doesn’t go to the playoffs.

17.If the thief is intelligent, then you aren’t the thief.

∴ You aren’t intelligent. 0214

You aren’t intelligent.

∴ If the thief is intelligent, then you aren’t the thief.

9.3 Easier relations

Our final rule for forming quantificational wffs introduces relations:

The result of writing a capital letter and then two or more small letters is a wff.

Here are two examples:

· Lrj = Romeo loves Juliet

Gxyz = x gave y to z

Translating relational sentences into logic can be difficult. We have to study examples and catch patterns; paraphrasing into Loglish is helpful too. We’ll start with easier translations and put off multiple-quantifier relations until the next section.

Here are further examples without quantifiers:

· Juliet loves Romeo = Ljr

· Juliet loves herself = Ljj

· Juliet loves Romeo but not Paris = (Ljr • ∼Ljp)

Here are some easy examples with quantifiers:

· Everyone loves him/herself = (x)Lxx

· Someone loves him/herself = (∃x)Lxx

· No one loves him/herself = ∼(∃x)Lxx

Normally put quantifiers before relations:

Someone (everyone, no one) loves Romeo

=

For some (all, no) x, x loves Romeo.

Romeo loves someone (everyone, no one)

=

For some (all, no) x, Romeo loves x.

In the second box, English puts the quantifier last – but logic puts it first. Here are fuller translations: 0215

Someone loves Romeo

= (∃x)Lxr

For some x, x loves Romeo

Everyone loves Romeo

= (x)Lxr

For all x, x loves Romeo

No one loves Romeo

= ∼(∃x)Lxr

It’s false that, for some x, x loves Romeo

Romeo loves someone

= (∃x)Lrx

For some x, Romeo loves x

Romeo loves everyone

= (x)Lrx

For all x, Romeo loves x

Romeo loves no one

= ∼(∃x)Lrx

It’s false that, for some x, Romeo loves x

These examples are more complicated:

· Some Montague loves Juliet

· = (∃x)(Mx • Lxj)

· For some x, x is a Montague and x loves Juliet

· All Montagues love Juliet

· = (x)(Mx ⊃ Lxj)

· For all x, if x is a Montague then x loves Juliet

· Romeo loves some Capulet

· = (∃x)(Cx • Lrx)

· For some x, x is a Capulet and Romeo loves x

· Romeo loves all Capulets

· = (x)(Cx ⊃ Lrx)

· For all x, if x is a Capulet then Romeo loves x

Here are further examples:

· Some Montague besides Romeo loves Juliet

· = (∃x)((Mx • ∼x=r) • Lxj)

· For some x, x is a Montague and x ≠ Romeo and x loves Juliet

· Romeo loves all Capulets besides Juliet

· = (x)((Cx • ∼x=j) ⊃ Lrx)

· For all x, if x is a Capulet and x ≠ Juliet then Romeo loves x

· Romeo loves all Capulets who love themselves

· = (x)((Cx • Lxx) ⊃ Lrx)

· For all x, if x is a Capulet and x loves x then Romeo loves x

Finally, these examples have two different relations:

· All who know Juliet love Juliet

· = (x)(Kxj ⊃ Lxj)

· For all x, if x knows Juliet then x loves Juliet

· All who know themselves love themselves

= (x)(Kxx ⊃ Lxx)

· For all x, if x knows x then x loves x

Try to master these before starting into the harder relational translations.

9.3a Exercise: LogiCola H (RM & RT)

Using these equivalences, translate these English sentences into wffs. 0216

Lxy = x loves y

Cxy = x caused y

Gxy = x is greater than y

Ix = x is Italian

Rx = x is Russian

Ex = x is evil

t = Tony

o = Olga

g = God

God caused nothing that is evil.

∼(∃x)(Ex • Cgx)

1. Tony loves Olga and Olga loves Tony.

2. Not every Russian loves Olga.

3. Tony loves everyone who is Russian.

4. Olga loves someone who isn’t Italian.

5. Everyone loves Olga but not everyone is loved by Olga.

6. All Italians love themselves.

7. Olga loves every Italian besides Tony.

8. Tony loves everyone who loves Olga.

9. No Russian besides Olga loves Tony.

10.Olga loves all who love themselves.

11.Tony loves no Russians who love themselves.

12.Olga is loved.

13.God caused everything besides himself.

14.Nothing caused God.

15.Everything that God caused is loved by God.

16.Nothing caused itself.

17.God loves himself.

18.If God did not cause himself, then there is something that God did not cause.

19.Nothing is greater than God.

20.God is greater than anything that he caused.

9.4 Harder relations

Now we get into multiple-quantifier translations. Here’s a simple example:

· Someone loves someone

· = (∃x)(∃y)Lxy

· For some x and for some y, x loves y

This could be true because some love themselves (“(∃x)Lxx”) or because some love another (“(∃x)(∃y)(∼x=y • Lxy)”). Here are more examples:

· Everyone loves everyone

· = (x)(y)Lxy

· For all x and for all y, x loves y

· Some Montague hates some Capulet

· = (∃x)(∃y)((Mx • Cy) • Hxy)

· For some x and for some y, x is a Montague and y is a Capulet and x hates y

· Every Montague hates every Capulet

· = (x)(y)((Mx • Cy) ⊃ Hxy)

· For all x and for all y, if x is a Montague and y is a Capulet then x hates y 0217

Study carefully this next pair – which differs only in the quantifier order:

Everyone loves someone.

For all x there’s some y, such that x loves y.

(x)(∃y)Lxy

There’s someone whom everyone loves.

There’s some y such that, for all x, x loves y.

(∃y)(x)Lxy

In the first case, we might love different people. In the second, we love the same person; perhaps we all love God. Notice the difference in the contrasting pairs:

· Everyone loves someone ≠ There’s someone whom everyone loves

· Everyone lives in some house ≠ There’s some house where everyone lives

· Everyone has some job ≠ There’s some job that everyone has

· Everyone makes some error ≠ There’s some error that everyone makes

The sentences on the right make a stronger claim: some-every entails every-some, but not the other way around.

Back-to-back quantifiers of the same type can be switched: “(x)(y)” = “(y)(x)” and “(∃x)(∃y)” = “(∃y)(∃x).” But the order matters if the quantifiers are of different types: “(∃x)(y)” is stronger than “(y)(∃x).” It doesn’t matter what variable letters we use, so long as the reference pattern is the same. So in “(x)(∃y)Lxy” we could use other variables in place of “x” and “y” – as long our wff consists in a universal and then an existential (using different variables), “L,” the variable used in the universal, and finally the variable used in the existential.

Here’s a difficult every-some translation, which we’ll do step by step:

· Every Capulet loves some Montague

· For all x, if x is a Capulet then x loves some Montague

· (x)(Cx ⊃ x loves some Montague)

· (x)(Cx ⊃ for some y, y is a Montague and x loves y)

· (x)(Cx ⊃ (∃y)(My • Lxy))

Until you master these, go by “baby steps” from English to Loglish to symbols. First go from “Every Capulet loves some Montague” to “For all x, if x is a Capulet then x loves some Montague”; so the formula after “(x)” is an IF-THEN. Later go from “x loves some Montague” to “for some y, y is a Montague and x loves y”; this part is an AND. So “every Capulet” gives “if Capulet then …,” and “some Montague” gives “some are Montague and …”

As usual, we could switch conjuncts (“My” and “Lxy”). We also could put the existential further out, so the wff starts “(x)(∃y).” But the order of the quantifiers has to follow the English – so if “every” comes before “some” then “(x)” has to come before “(∃y).” This example is difficult; study it carefully.

Here are analogous but easier every-some translations: 0218

· Every Capulet loves someone

· For all x, if x is a Capulet then x loves someone

· (x)(Cx ⊃ x loves someone)

· (x)(Cx ⊃ for some y, x loves y)

· (x)(Cx ⊃ (∃y)Lxy)

· Everyone loves some Montague

· For all x, x loves some Montague

· (x) x loves some Montague

· (x) for some y, y is a Montague and x loves y

· (x)(∃y)(My • Lxy)

The first uses IF-THEN, because “Every Capulet loves someone” goes into “For all x, if x is a Capulet then x loves someone.” The second uses AND, because “x loves some Montague” goes into “for some y, y is a Montague and x loves y.”

Here’s a difficult some-every translation:

· Some Capulet loves every Montague

· For some x, x is a Capulet and x loves every Montague

· (∃x)(Cx • x loves every Montague)

· (∃x)(Cx • for all y, if y is a Montague then x loves y)

· (∃x)(Cx • (y)(My ⊃ Lxy))

“Some Capulet loves every Montague” becomes “For some x, x is a Capulet and x loves every Montague”; “(∃x)” is followed by AND. Then “x loves every Montague” becomes “for all y, if y is a Montague then x loves y”; “(x)” is followed by IF-THEN. As before, we could switch conjuncts “Cx” and “(y)(My ⊃ Lxy).” And we could start the wff with “(∃x)(y)”; here “(∃x)” has to come before “(y),” since “some” comes before “every” in the English.

Here are analogous but easier some-every translations:

· Some Capulet loves everyone

· For some x, x is a Capulet and x loves everyone

· (∃x)(Cx • x loves everyone)

· (∃x)(Cx • for all y, x loves y)

· (∃x)(Cx • (y)Lxy)

· Someone loves every Montague

· For some x, x loves every Montague

· (∃x) x loves every Montague

· (∃x) for all y, if y is a Montague then x loves y

· (∃x)(y)(My ⊃ Lxy)

The first uses AND, because “Some Capulet loves everyone” becomes “For some x, x is a Capulet and x loves everyone.” The second uses IF-THEN, because “x loves every Montague” becomes “for all y, if y is a Montague then x loves y.” 0219

Since these translations are difficult, you might want to reread a couple of times from the beginning of this section to here, until you get it.

Here are some miscellaneous translations:

· There’s an unloved lover

· = (∃x)(∼(∃y)Lyx • (∃y)Lxy)

· For some x, x is unloved (no one loves x) and x is a lover (x loves someone)

· Everyone loves a lover

· = (x)((∃y)Lxy ⊃ (y)Lyx)

· For all x, if x is a lover (x loves someone) then everyone loves x

· Romeo loves all and only those who don’t love themselves

· = (x)(Lrx ≡ ∼Lxx)

· For all x, Romeo loves x if and only if x doesn’t love x

· All who know any person love that person

· = (x)(y)(Kxy ⊃ Lxy)

· For all x and all y, if x knows y then x loves y

Relations have properties like reflexivity, symmetry, and transitivity:

· “Having the same age as” is reflexive

· = (x)Axx

· Everything has the same age as itself

· “Being taller than” is irreflexive

· = ∼(∃x)Txx

· Nothing is taller than itself

· “Being a relative of” is symmetrical

· = (x)(y)(Rxy ⊃ Ryx)

· In all cases, if x is a relative of y, then y is a relative of x

· “Being a father of” is asymmetrical

· = (x)(y)(Fxy ⊃ ∼Fyx)

· In all cases, if x is a father of y then y isn’t a father of x

· “Being taller than” is transitive

· = (x)(y)(z)((Txy • Tyz) ⊃ Txz)

· In all cases, if x is taller than y and y is taller than z, then x is taller than z

· “Being a father of” is intransitive

· = (x)(y)(z)((Fxy • Fyz) ⊃ ∼Fxz)

· In all cases, if x is a father of y and y is a father of z, then x isn’t a father of z

Love fits none of these six categories. Love is neither reflexive nor irreflexive: sometimes people love themselves and sometimes they don’t. Love is neither symmetrical nor asymmetrical: if x loves y, then sometimes y loves x in return and sometimes not. And love is neither transitive nor intransitive: if x loves y and y loves z, then sometimes x loves z and sometimes not.

9.4a Exercise: LogiCola H (RM & RT)

Using these equivalences, translate these English sentences into wffs. 0220

Lxy = x loves y

Cxy = x caused y

Gxy = x is greater than y

Ix = x is Italian

Rx = x is Russian

Ex = x is evil

t = Tony

o = Olga

Every Russian loves everyone.

(x)(Rx ⊃ (y)Lxy)

or

(x)(y)(Rx ⊃ Lxy)

1. Everyone loves every Russian.

2. Some Russians love someone.

3. Someone loves some Russians.

4. Some Russians love every Italian.

5. Every Russian loves some Italian.

6. There is some Italian that every Russian loves.

7. Everyone loves everyone else.

8. Every Italian loves every other Italian.

9. Some Italians love no one.

10.No Italians love everyone.

11.No one loves all Italians.

12.Someone loves no Italians.

13.No Russians love all Italians.

14.If everyone loves Olga, then there is some Russian that everyone loves.

15.If Tony loves everyone, then there is some Italian who loves everyone.

16.It is not always true that if a first thing caused a second, then the first is greater than the second.

17.In all cases, if a first thing is greater than a second, then the second isn’t greater than the first.

18.Everything is greater than something.

19.There’s something than which nothing is greater.

20.Everything is caused by something.

21.There’s something that caused everything.

22.Something evil caused all evil things.

23.In all cases, if a first thing caused a second and the second caused a third, then the first caused the third.

24.There’s a first cause (there’s some x that caused something but nothing caused x).

25.Anyone who caused anything loves that thing.

9.5 Relational proofs

In relational proofs, as before, we’ll reverse squiggles, then drop existentials, and lastly drop universals. Drop only initial (outermost) quantifiers. So with back-to-back quantifiers “(x)(y)” (in line 3 below), drop “(x)” first and then “(y)”: 0221

· Paris loves Juliet.

Juliet doesn’t love Paris.

∴ It’s not always true that if a first person loves a second then the second loves the first.

Our older proof strategy would have us drop each initial universal quantifier twice, once using “p” and once using “j.” But now this would be tedious; so henceforth we’ll derive only what will be useful for our proof or refutation.

Here’s another relational proof:

· There’s someone that everyone loves.

∴ Everyone loves someone.

This is valid intuitively: if there’s one specific person (God, for example) that everyone loves, then everyone loves at least one person.

For quantificational arguments without relations and identity:

1. there are mechanical strategies (like that sketched in §8.2) that always give a proof or refutation in a finite number of lines; and

2. a refutation at most needs 2n entities (where n is the number of distinct predicates in the argument).

Neither holds for relational arguments. Against 1, no possible mechanical strategy will always give a proof or refutation of a relational argument. This result is called Church’s theorem, after Alonzo Church. So working out relational arguments sometimes requires ingenuity and not just mechanical methods; the problem with our proof strategy is that it can lead into endless loops.¹ Against 2, refuting invalid relational arguments sometimes requires a possible world with an infinite number of entities. 0222

¹ The companion LogiCola computer program follows mechanical rules (algorithms) to construct proofs. Left to itself, it would go into an endless loop for some invalid relational arguments. But LogiCola is told beforehand which arguments go into an endless loop and which refutations to then give, so it can stop the loop at a reasonable point.

Instructions lead into an endless loop if they command the same sequence of actions over and over, endlessly. I’ve written computer programs with endless loops by mistake. I put an endless loop into the Index for fun:

Endless loop; see loop, endless

Loop, endless; see endless loop

Our quantificational proof strategy can lead into such a loop. If you see this coming, quit the strategy and improvise your own refutation.

Wffs that begin with a universal/existential quantifier combination, like “(x)(∃y),” often lead into endless loops. Here’s an example:¹

¹ This example is like arguing “Everyone lives in some house, so there must be some (one) house that everyone lives in.” Some great minds have committed this quantifier-shift fallacy. Aristotle argued, “Every agent acts for an end, so there must be some (one) end for which every agent acts.” St Thomas Aquinas argued, “If everything at some time fails to exist, then there must be some (one) time at which everything fails to exist.” And John Locke argued, “Everything is caused by something, so there must be some (one) thing that caused everything.”

· Everyone loves someone.

· ∴ There’s someone that everyone loves.

· (x)(∃y)Lxy Invalid

· ∴ (∃y)(x)Lxy

The premise by itself leads into an endless loop:

· Everyone loves someone.

· ∴ a loves someone.

· ∴ a loves b.

· ∴ b loves someone.

· ∴ b loves c.

· ∴ c loves someone.

· ∴ c loves d.

· … and so on endlessly …

· (x)(∃y)Lxy

· ∴ (∃y)Lay

· ∴ Lab

· ∴ (∃y)Lby

· ∴ Lbc

· ∴ (∃y)Lcy

· ∴ Lcd

· … and so on endlessly …

This argument is invalid, but we have to improvise to get the refutation; we can’t wait until our proof strategy ends (since it never will) and then use the simple formulas to construct a refutation. Instead, we have to think out the refutation by ourselves. While there’s no strategy that always works, I suggest that you:

· break out of the loop before introducing your third constant (often it suffices to use two beings, a and b; don’t multiply entities unnecessarily),

· begin your refutation with values you already have (maybe you already have “Lab” and “Laa”), and

· add other wffs to make premises true and conclusion false (maybe try adding “Lba” or “∼Lba,” and then “Lbb” or “∼Lbb,” until your refutation works).

Fiddle with the values until you find a refutation that works. 0223

Consider our example again:

· Everyone loves someone.

· ∴ There’s someone that everyone loves.

· (x)(∃y)Lxy Invalid

· ∴(∃y)(x)Lxy

If we stop the attempted proof before introducing our third constant, we may get either of these as the beginning of our refutation:

We need to add more formulas to make the premise true and conclusion false. With the first box, we need EVERYONE to love someone. Since a doesn’t love b, we need to have a love a. So we add this:

We also need b to love someone (so we need Lbb or Lba) – but without there being some one person that everyone loves (which excludes Lba, so we have to add Lbb and ∼Lba). So with ingenuity we construct this possible world, with beings a and b, that makes the premise true and conclusion false:

In this egoistic world, all love themselves but not others. This makes “Everyone loves someone” true but “There’s someone that everyone loves” false (since not everyone loves a and not everyone loves b). This refutation works too:

In this altruistic world, all love others but not themselves. This makes “Everyone loves someone” true but “There’s someone that everyone loves” false (since not everyone loves a and not everyone loves b).

Refuting relational arguments sometimes requires a universe with an infinite number of entities. Here’s an example:

· In all cases, if x is greater than y and y is greater than z then x is greater than z.

· In all cases, if x is greater than y then y isn’t greater than x.

· b is greater than a.

· ∴ There’s something than which nothing is greater.

· (x)(y)(z)((Gxy • Gyz) ⊃ Gxz) Invalid

· (x)(y)(Gxy ⊃ ∼Gyx)

· Gba

· ∴ (∃x)∼(∃y)Gyx 0224

We can imagine a world with an infinity of beings – in which each being is surpassed in greatness by another. Let’s take the natural numbers (0, 1, 2, …) as the universe of discourse. Let “a” refer to 0 and “b” refer to 1 and “Gxy” mean “x > y.” On this interpretation, the premises are all true. But the conclusion, which says “There’s a number than which no number is greater,” is false. This shows that the form is invalid.

So relational arguments raise problems about infinity (endless loops and infinite worlds) that other kinds of argument we’ve studied don’t raise.

9.5a Exercise: LogiCola I (RC & BC)

Say whether each is valid (and give a proof) or invalid (and give a refutation).

(∃x)(∃y)Lxy

∴ (∃y)(∃x)Lxy

1. (x)Lxa

∴ (x)Lax

2. (∃x)(y)Lxy

∴ (∃x)Lxa

3. (x)(y)(Lxy ⊃ x=y)

∴ (x)Lxx

4. (x)(∃y)Lxy

∴ Laa

5. (x)(y)Lxy

∴ (x)(y)((Fx • Gy) ⊃ Lxy)

6. (x)(y)(Uxy ⊃ Lxy)

(x)(∃y)Uxy

∴ (x)(∃y)Lxy

7. (x)Lxx

∴ (∃x)(y)Lxy

8. (x)Gaxb

∴ (∃x)(∃y)Gxcy

9. (x)(y)Lxy

∴ (∃x)Lax

10.Lab

Lbc

∴ (∃x)(Lax • Lxc)

11.(x)Lxx

∴ (x)(y)(Lxy ⊃ x=y)

12.(∃x)Lxa

∼Laa

∴ (∃x)(∼a=x • Lxa)

13.(x)(y)(z)((Lxy • Lyz) ⊃ Lxz)

(x)(y)(Kxy ⊃ Lyx)

∴ (x)Lxx

14.(x)Lxa

(x)(Lax ⊃ x=b)

∴ (x)Lxb

15.(x)(y)(Lxy ⊃ (Fx • ∼Fy))

∴ (x)(y)(Lxy ⊃ ∼Lyx) 0225

9.5b Exercise: LogiCola I (RC & BC)

First appraise intuitively. Then translate into logic and say whether valid (and give a proof) or invalid (and give a refutation).

1. Juliet loves everyone.

∴ Someone loves you. [Use Lxy, j, and u.]

2. Nothing caused itself.

∴ There’s nothing that caused everything. [Use Cxy.]

3. Alice is older than Betty.

∴ Betty isn’t older than Alice. [Use Oxy, a, and b. What implicit premise would make this valid?]

4. There’s something that everything depends on.

∴ Everything depends on something. [Dxy]

5. Everything depends on something.

∴ There’s something that everything depends on. [Dxy]

6. Paris loves all females.

No females love Paris.

Juliet is female.

∴ Paris loves someone who doesn’t love him. [Lxy, p, Fx, j]

7. In all cases, if a first thing caused a second, then the first exists before the second.

Nothing exists before it exists.

∴ Nothing caused itself. [Use Cxy and Bxy (for “x exists before y exists”).]

8. Everyone hates my enemy.

My enemy hates no one besides me.

∴ My enemy is me. [Hxy, e, m]

9. Not everyone loves everyone.

∴ Not everyone loves you. [Lxy, u]

10.There’s someone that everyone loves.

∴ Some love themselves.

11.Andy shaves all and only those who don’t shave themselves.

∴ It is raining. [Sxy, a, R]

12.No one hates themselves.

I hate all logicians.

∴ I am not a logician. [Hxy, i, Lx]

13.Juliet loves everyone besides herself.

Juliet is Italian.

Romeo is my logic teacher.

My logic teacher isn’t Italian.

∴ Juliet loves Romeo. [j, Lxy, Ix, r, m] 0226

14.Romeo loves either Lisa or Colleen.

Romeo doesn’t love anyone who isn’t Italian.

Colleen isn’t Italian.

∴ Romeo loves Lisa. [Lxy, r, l, c]

15.Everyone loves all lovers.

Romeo loves Juliet.

∴ I love you. [Use Lxy, r, j, i, and u. This one is difficult.]

16.Everyone loves someone.

∴ Some love themselves.

17.Nothing caused itself.

This chemical brain process caused this pain.

∴ This chemical brain process isn’t identical to this pain. [Cxy, b, p]

18.For every positive contingent truth, something explains why it’s true.

The existence of the world is a positive contingent truth.

If something explains the existence of the world, then some necessary being explains the existence of the world.

∴ Some necessary being explains the existence of the world. [Use Cx, Exy, e, and Nx. This argument for the existence of God is from Richard Taylor.]

19.That girl is Miss Novak.

∴ If you don’t like Miss Novak, then you don’t like that girl. [Use t, m, u, and Lxy; from the movie, The Little Shop around the Corner: “If you don’t like Miss Novak, I can tell you right now that you won’t like that girl. Why? Because it is Miss Novak.”]

20.Everyone who is wholly good prevents every evil that he can prevent.

Everyone who is omnipotent can prevent every evil.

If someone prevents every evil, then there’s no evil.

There’s evil.

∴ Either God isn’t omnipotent, or God isn’t wholly good. [Use Gx, Ex, Cxy (for “x can prevent y”), Pxy (for “x prevents y”), Ox, and g; from J. L. Mackie.]

21.Your friend is wholly good.

Your knee pain is evil.

Your friend can prevent your knee pain.

Your friend doesn’t prevent your knee pain (since he could prevent it only by amputating your leg – which would bring about a worse situation).

∴ “Everyone who is wholly good prevents every evil that he can prevent” is false. [Use f, Gx, k, Ex, Cxy, and Pxy. Alvin Plantinga thus attacked premise 1 of the previous argument; he proposed instead roughly this: “Everyone who is wholly good prevents every evil that he knows about if he can do so without thereby eliminating a greater good or bringing about a greater evil.”] 0227

22.For everything contingent, there’s some time at which it fails to exist.

∴ If everything is contingent, then there’s some time at which everything fails to exist. [Use Cx for “x is contingent”; Ext for “x exists at time t”; t for a time variable; and t′, t″, t‴, … for time constants. This is a critical step in St Thomas Aquinas’s third argument for the existence of God.]

23.If everything is contingent, then there’s some time at which everything fails to exist.

If there’s some time at which everything fails to exist, then there’s nothing in existence now.

There’s something in existence now.

Everything that isn’t contingent is necessary.

∴ There’s a necessary being. [Besides the letters for the previous argument, use Nx for “x is necessary” and n for “now.” This continues Aquinas’s argument; here premise 1 is from the previous argument.]

24.[Gottlob Frege tried to systematize mathematics. One of his axioms said that every sentence with a free variable¹ determines a set; so “x is blue” determines a set containing all and only blue things. While this seems sensible, Bertrand Russell showed that this entails that “x doesn’t contain x” determines a set y containing all and only those things that don’t contain themselves – and this leads to the self-contradiction “y contains y if and only if y doesn’t contain y.” The foundations of mathematics haven’t been the same since “Russell’s paradox.”]

¹ An instance of a variable is “free” in a wff if it doesn’t occur as part of a wff that begins with a quantifier using that variable; each instance of “x” is free in “Fx” but not in “(x)Fx.”

If every sentence with a free variable determines a set, then there’s a set y such that, for all x, y contains x if and only if x doesn’t contain x.

∴ Not every sentence with a free variable determines a set. [Use D for “Every sentence with a free variable determines a set,” Sx for “x is a set,” and Cyx for “y contains x.” See §16.4.]

25.All dogs are animals.

∴ All heads of dogs are heads of animals. [Use Dx, Ax, and Hxy (for “x is a head of y”). Translate “x is a head of a dog” as “for some y, y is a dog and x is a head of y.” Augustus De Morgan in the 19th century claimed that this was a valid argument that traditional logic couldn’t validate.]

9.6 Definite descriptions

Definite descriptions, phrases of the form “the so and so,” are used to pick out a definite (single) person or thing. Consider how we’ve been symbolizing these two English sentences:

· Socrates is bald = Bs

· The king of France is bald = Bk 0228

The first sentence has a proper name (“Socrates”) while the second has a definite description (“the king of France”); both seem to ascribe a property (baldness) to a particular object or entity. Bertrand Russell argued that this object-property analysis is misleading. Definite descriptions (like “the king of France”) should instead be analyzed using a complex of predicates and quantifiers:

· The king of France is bald

· = (∃x)((Kx • ∼(∃y)(∼y=x • Ky)) • Bx)

· There’s exactly one king of France, and he’s bald

· For some x, x is king of France and there’s no y such that: y≠x and y is king of France and x is bald

Russell saw his analysis as having two advantages.

First, “The king of France is bald” might be false for any of three reasons:

1. There’s no king of France;

2. there’s more than one king of France; or

3. there’s exactly one king of France, and he has hair on his head.

In fact, “The king of France is bald” is false for reason 1: France has no king. This fits Russell’s analysis. By contrast, the object-property analysis suggests that if “The king of France is bald” is false, then “The king of France isn’t bald” would have to be true – and so the king of France would have to have hair! So Russell’s analysis expresses better the logical complexity of definite descriptions.

Second, the object-property analysis of definite descriptions can easily lead us into metaphysical errors, like positing existing things that aren’t real. The philosopher Alexius Meinong argued roughly as follows (and Russell at first accepted this argument):

· “The round square does not exist” is a true statement about the round square.

· If there’s a true statement about something, then that something has to exist.

· ∴ The round square exists.

· But the round square isn’t a real thing.

· ∴ Some things that exist aren’t real things.

Russell later saw the belief in existing non-real things as foolish. Appealing to his theory of descriptions, he criticized Meinong’s argument as resting on a naïve object-property analysis of this statement:

“The round square does not exist.”

If this were a true statement about the round square, as Meinong’s first premise asserts, then the round square would have to exist – which the statement denies. Instead, the statement just denies that there’s exactly one being that’s both round and square. So Russell’s analysis keeps us from having to accept existing things that aren’t real. 0229

9.7 Copi proofs

We earlier discussed traditional Copi proofs for propositional logic and basic quantificational logic (§§7.5 and 8.6). Copi proofs can also be used for identity and relations. Copi uses our SI (Self-identity) and SE (Substitute Equals) rules, and adds a SS (Switch Sides) replacement rule: “x=y = y=x” (where you can use any variable or constant for “x” and for “y”).

9.7a and 9.7b Exercise: LogiCola IDO

Do Copi proofs for problems in §9.2a (just the valid ones, namely 3, 4, 6, 7, and 8) and §9.2b (just the valid ones, namely 1, 2, 3, 4, 6, 9, 10, 12, 13, 14, 15, and 18). These are identity arguments.

9.7c and 9.7d Exercise: LogiCola IRO and IBO

Do Copi proofs for problems in §9.5a (just the valid ones, namely 2, 5, 6, 8, 9, 10, 12, 14, and 15) and §9.5b (just the valid ones, namely 1, 2, 4, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 23, 24, and 25). These are relational arguments.