8

Basic Quantificational Logic

Quantificational logic, which builds on propositional logic, studies arguments whose validity depends on notions like “all,” “no,” and “some.” This system is stronger than syllogistic logic (Chapter 2), since it can express complex ideas like “If some are A, then all that are B or C are then D but not E.” This chapter covers the basics and the next adds relations and identity.

8.1 Easier translations

To help us evaluate arguments, we’ll construct a quantificational language. This will include propositional logic’s vocabulary, wffs, inference rules, and proofs. It adds two new vocabulary items: small letters and “∃.” Here are sample wffs:

Ir = Romeo is Italian.

Ix = x is Italian.

(x)Ix = For all x, x is Italian (all are Italian).

(∃x)Ix = For some x, x is Italian (some are Italian).

Learn to express “All are Italian” as “For all x, x is Italian.” This uses Loglish, a mix of logic and English. Loglish helps us to translate from English to logic.

“Romeo is Italian” is “Ir”; the capital letter goes first. “I” is for the general category “Italian” and “r” is for the specific individual “Romeo”:

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.

Letters here have various uses. Capitals can represent statements, general 0183 terms, or relations (which we take in the next chapter):

A capital letter alone (not followed by small letters) represents a statement.

S = it’s snowing.

A capital letter followed by a single small letter represents a general term.

Ir = Romeo is Italian.

A capital letter followed by two or more small letters represents a relation.

Lrj = Romeo loves Juliet.

Small letters can be constants or variables:

A small letter from “a” to “w” is a constant (it refers to a specific person or thing).

Ir = Romeo is Italian.

A small letter from “x” to “z” is a variable (its reference isn’t directly specified).

Ix = x is Italian.

“Ix” (“x is Italian”) is incomplete, and so not true or false, since we haven’t said whom we’re talking about. Quantifiers can complete the claim. A quantifier is a sequence of the form “(x)” or “(∃x)” – where any variable may replace “x”:

“(x)” is a universal quantifier. It says that the next formula is true for all values of x.

(x)Ix = For all x, x is Italian (all are Italian).

“(∃x)” is an existential quantifier. It says that the next formula is true for at least one value of x.

(∃x)Ix = For some x, x is Italian (some are Italian).

Quantifiers express “all” and “some” by saying in how many cases the following formula is true.

As before, grammatical formulas are wffs (well-formed formulas). Wffs now are strings we can construct using the propositional rules plus two new rules:

1. The result of writing a capital letter and then a small letter is a wff.

2. The result of writing a quantifier and then a wff is a wff.

These rules let us build wffs that we’ve already mentioned: “Ir,” “Ix,” “(x)Ix,” and “(∃x)Ix.” Don’t use additional parentheses; these forms are incorrect: “(Ir),” “(Ix),” “(x)(Ix),” “(∃x)(Ix),” “((x)Ix),” “((∃x)Ix).” Use a pair of parentheses for each quantifier and each instance of “•,” “∨,” “⊃,” and “≡”; use no other parentheses. Here are some further wffs: 0184

· ∼(x)Ix = Not all are Italian

· It’s false that, for all x, x is Italian

· ∼(∃x)Ix = No one is Italian

· It’s false that, for some x, x is Italian

· (Ix ⊃ Lx) = If x is Italian then x is a lover

· (Ix • Lx) = x is Italian and x is a lover

Translating from English to wffs can be difficult. We’ll begin with sentences that translate into wffs starting with a quantifier, or with “∼” and then a quantifier. This rule tells where to put what quantifier:

If the English begins with “all” or “every,” then begin the wff with “(x).”

If the English begins with “not all” or “not every,” then begin the wff with “∼(x).”

If the English begins with “some,” then begin the wff with “(∃x).”

If the English begins with “no,” then begin the wff with “∼(∃x).”

· All are Italian = (x)Ix

· Not all are Italian = ∼(x)Ix

· Some are Italian = (∃x)Ix

· No one is Italian = ∼(∃x)Ix

Here are harder examples:

· All are rich or Italian

· = (x)(Rx ∨ Ix)

· Not everyone is non-Italian

· = ∼(x)∼Ix

· Some aren’t rich

· = (∃x)∼Rx

· No one is rich and non-Italian

· = ∼(∃x)(Rx • ∼Ix)

When the English begins with “all,” “not all,” “some,” or “no,” put the quantifier outside all parentheses. So “All are rich or Italian” is “(x)(Rx ∨ Ix).” Don’t translate it as “((x)Rx ∨ Ix),” which means “Either everyone is rich, or x is Italian.”

If the English sentence uses a word like “or,” “and,” or “if-then,” then use the corresponding logical symbol. Otherwise, follow these rules:

With “all … is …,” use “⊃” for the middle connective.

Otherwise use “•” for the connective.

· All Italians are lovers

· = (x)(Ix ⊃ Lx)

· For all x, if x is Italian then x is a lover 0185

· Some Italians are lovers

· = (∃x)(Ix • Lx)

For some x, x is Italian and x is a lover

· No Italians are lovers

· = ∼(∃x)(Ix • Lx)

· It’s false that, for some x, x is Italian and x is a lover

With “All Italians …,” think “For all x, if x is Italian then ….” With “Some Italians …,” think “For some x, x is Italian and ….” This example is harder:

· All rich Italians are lovers

· = (x)((Rx • Ix) ⊃ Lx)

· For all x, if x is rich and Italian, then x is a lover

Here use “⊃” as the middle connective (“If rich Italian, then lover”) and “•” in the other place (“If rich and Italian, then lover”). Here are further examples:

· Not all Italians are lovers

· = ∼(x)(Ix ⊃ Lx)

· It’s false that, for all x, if x is Italian then x is a lover

· All are rich Italians

· = (x)(Rx • Ix)

· For all x, x is rich and Italian

Sometimes we must rephrase to make “is” (or “are”) the main verb:

· All dogs hate cats

· = All dogs are cat-haters

· = (x)(Dx ⊃ Hx)

· For all x, if x is a dog then x is a cat-hater

In case of doubt, say the formula in Loglish and see if it means the same as the English sentence. Our translation rules are rough and don’t always work.

The universe of discourse is the set of entities that words like “all” “some,” and “no” range over in a given context. Restricting the universe of discourse to one kind of entity (such as persons or statements) can simplify how we translate some arguments. We’ll often restrict the universe of discourse to persons. We did this implicitly when we translated “All are Italian” as “(x)Ix” instead of “(x)(Px ⊃ Ix)” (“All persons are Italian”).

Since quantificational translations are so difficult, LogiCola gives you the option to start off by having Loglish hints for these problems.

8.1a Exercise: LogiCola H (EM & ET)

Translate these English sentences into wffs.

Not all logicians run.

∼(x)(Lx ⊃ Rx)

1. x isn’t a cat.

2. Something is a cat.

3. Something isn’t a cat. 0186

4. It’s false that there is something that isn’t a cat.

5. Everything is a cat.

6. If x is a dog, then x is an animal.

7. All dogs are animals.

8. No one is evil.

9. Some logicians are evil.

10.No logician is evil.

11.All black cats are unlucky.

12.Some dogs are large and hungry.

13.Not all hungry dogs bark.

14.Some animals aren’t barking dogs.

15.Some animals are non-barking dogs.

16.All dogs who bark are frightening.

17.Not all non-dogs are cats.

18.Some cats who aren’t black are unlucky.

19.Some cats don’t purr.

20.Not every cat purrs.

21.Not all animals are dogs or cats.

22.All who are either dogs or cats are animals.

23.All who are both dogs and cats are animals.

24.All dogs and cats are animals.

25.Everyone is a crazy logician.

8.2 Easier proofs

We need quantifier inference rules. The reverse-squiggle rules hold regardless of what variable replaces “x” and what pair of contradictory wffs replaces “Fx” / “∼Fx”; here “→” means that we can infer whole lines from left to right:

Reverse squiggle RS

∼(x)Fx → (∃x)∼Fx

∼(∃x)Fx → (x)∼Fx

“Not everyone is funny” entails “Someone isn’t funny.” And “It’s false that someone is funny” (“No one is funny”) entails “Everyone is non-funny.” Our rules cover reversing squiggles on longer formulas, if the whole formula begins with “∼” and then a quantifier. Here are two examples:

·  ∼(∃x)∼Gx

–––––––––

∴(x)∼∼Gx

·  ∼(x)(Lx • ∼Mx)

––––––––––––––––

∴(∃x)∼(Lx • ∼Mx)

In the first example, we also could conclude “(x)Gx” (dropping “∼∼”). This next example is illegal in our system, since it fits poorly into our proof strategy, even though it’s logically correct:

Don’t do this:

·  (Ir ⊃ ∼(x)Gx)

––––––––––––––

∴ (Ir ⊃ (∃x)∼Gx)

0187 Reverse squiggles whenever you have a wff that begins with “∼” and then a quantifier; this moves a quantifier to the beginning of the formula, so we can drop it later.

Drop quantifiers using the next two rules (which hold regardless of what variable replaces “x” and what wffs replace “Fx” / “Fa” – provided that the two wffs are identical except that wherever the variable occurs freely¹ in the former the same constant occurs in the latter). Here’s the drop-existential rule:

¹ An instance of a variable occurs freely if it’s not part of a wff that begins with a quantifier using that variable; just the first instance of “x” in “(Fx • (x)Gx)” occurs freely. So we’d go from “(∃x)(Fx • (x)Gx)” to “(Fa • (x)Gx).”

Drop existential DE (∃x)Fx → Fa, use a new constant

Suppose someone robbed the bank; we can give this person an arbitrary name that we make up (like “Al”). Likewise, when we drop an existential, we’ll name this “someone” with a new constant – one that hasn’t yet occurred in earlier lines of the proof.¹ In proofs, we’ll use the next unused constant in alphabetical order – starting with “a,” then “b,” and so on. So if we drop two existentials, then we introduce two new constants:

¹ If more than one person robbed the bank; then our name (or constant) will refer to a random one of the robbers. Using a new name is consistent with the robber being mentioned earlier in the argument; different names (like “Al” and “Smith”) might refer to the same individual. DE should be used only when there’s at least one not-blocked-off assumption; otherwise, the symbolic version of “Someone is a thief, so Gensler is a thief” would be a two-line proof.

·  (∃x)Mx (∃x)Fx

–––––––

∴ Ma ∴ Fb

Someone is male, someone is female; let’s call the male “a” and the female “b.” It’s OK to use “a” in the first inference, since it occurs in no earlier line. But the second inference must use “b,” since “a” has now already occurred.

We can drop existentials from complicated formulas if the quantifier begins the wff and we replace the variable with the same new constant throughout. So this first inference is fine:

·  (∃x)(Fx • Gx)

–––––––––––

∴ (Fa • Ga)

This next example is wrong (because it drops the quantifier using two different constants):

·  (∃x)(Fx • Gx)

–––––––––––

∴ (Fa • Gb)

This next example is also wrong (since the formula doesn’t begin with a quantifier – instead it begins with a left-hand parenthesis):

·  ((∃x)Fx ⊃ P)

–––––––––––

∴ (Fa ⊃ P)

Drop only initial quantifiers. Here’s the drop-universal rule:

Drop universal DU

(x)Fx → Fa,

use any constant

0188 If everyone is funny, then Al is funny, Bob is funny, and so on. From “(x)Fx” we can derive “Fa,” “Fb,” and so on – using any constant. However, it’s bad strategy to use a new constant unless we really have to; normally use old constants when dropping universals.¹ As before, the quantifier must begin the wff and we must replace the variable with the same constant throughout. So this next inference is fine:

¹ Dropping a universal quantifier with a new letter assumes that something exists (or that our restricted universe of discourse is nonempty). Some systems (see §13.7) disallow this.

· (∃x)(Fx

–––––––––––

? Gx) ? (Fa ? Ga)

This next example is wrong (because it drops the quantifier using two different constants):

This next example is also wrong (since the formula doesn’t begin with a quantifier – instead it begins with a left-hand parenthesis – drop only initial quantifiers):

“((x)Fx ⊃ (x)Gx)” is an if-then and follows the if-then rules: if we have the first part “(x)Fx” true, we can get the second true; if we have the second part “(x)Gx” false, we can get the first false; if we get stuck, we make an assumption.

Here’s an example of a proof:

· All logicians are funny.

Someone is a logician.

∴ Someone is funny.

For now, use the quantificational rules in this order:

· First reverse squiggles. We did this to get “(x)∼Fx” in line 4.

· Then drop initial existentials, using a new constant each time. We did this to get “La” in line 5.

· Lastly, drop each initial universal once for each old constant. We did this to get “(La ⊃ Fa)” in line 6 and “∼Fa” in line 8.

We starred lines 2, 3, and 6; starred lines largely can be ignored in deriving further lines. Star any wff on which you reverse squiggles or drop an existential:

Here the new line has the same information. Don’t star when dropping a0189 universal; we can never exhaust an “all” by deriving instances, and we may have to derive further things from it later.

Here’s a simpler quantificational proof:

Reverse squiggles to get “(∃x)∼Fx” in line 3. Drop an existential to get “∼Fa” in line 4. Then drop a universal to get “(Fa • Ga)” in line 5. Switching lines 4 and 5 would be wrong: if we drop the universal first using “a,” then we can’t drop the existential later using “a” (since then “a” would be old).

In doing proofs, first assume the conclusion’s opposite; then use quantificational rules plus S- and I-rules to derive all you can. If you find a contradiction, apply RAA. If you’re stuck and need to break a NOT-BOTH, OR, or IF-THEN, then make another assumption. If you get no contradiction and yet can’t do anything further, then try to refute the argument. Here’s a fuller statement of our strategy’s quantificational steps:

1. FIRST REVERSE SQUIGGLES: For each unstarred, not-blocked-off line that begins with “∼” and then a quantifier, derive a line using the reverse-squiggle rules. Star the original line.

2. THEN DROP EXISTENTIALS: For each unstarred, not-blocked-off line that begins with an existential quantifier, derive an instance using the next available new constant (but don’t drop an existential if you already have a not-blocked-off instance in previous lines – so don’t drop “(∃x)Fx” if you already have “Fc”). Star the original line.

3. LASTLY DROP UNIVERSALS: For each not-blocked-off line that begins with a universal quantifier, derive instances using each old constant. Don’t star the original line; you may have to use it again. (Drop a universal using a new constant only if you’ve done everything else possible, making further assumptions if needed, and still have no old constants.)

Drop existentials before universals. Introduce a new constant each time you drop an existential, and use the same old constants when you drop a universal. And drop only initial quantifiers.

8.2a Exercise: LogiCola IEV

Prove each of these arguments to be valid (all are valid). 0190

· ∼ (∃x)Fx

· ∴ (x)∼(Fx • Gx)

1. (x)Fx

∴ (x)(Gx ∨ Fx)

2. ∼(∃x)(Fx • ∼Gx)

∴ (x)(Fx ⊃ Gx)

3. ∼(∃x)(Fx • Gx)

(∃x)Fx

∴ (∃x)∼Gx

4. (x)((Fx ∨ Gx) ⊃ Hx)

∴ (x)(∼Hx ⊃ ∼Fx)

5. (x)(Fx ⊃ Gx)

(∃x)Fx

∴ (∃x)(Fx • Gx)

6. (x)(Fx ∨ Gx)

∼(x)Fx

∴ (∃x)Gx

7. (x)∼(Fx ∨ Gx)

∴ (x)∼Fx

8. (x)(Fx ⊃ Gx)

(x)(Fx ⊃ ∼Gx)

∴ (x)∼Fx

9. (x)(Fx ⊃ Gx)

(x)(∼Fx ⊃ Hx)

∴ (x)(Gx ∨ Hx)

10.(x)(Fx ≡ Gx)

(∃x)∼Gx

∴ (∃x)∼Fx

8.2b Exercise: LogiCola IEV

First appraise intuitively. Then translate into logic (using the letters given) and prove to be valid (all are valid).

1. All who deliberate about alternatives believe in free will (at least implicitly).

All deliberate about alternatives.

∴ All believe in free will. [Use Dx and Bx; from William James.]

2. Everyone makes mistakes.

∴ Every logic teacher makes mistakes. [Use Mx and Lx.]

3. No feeling of pain is publicly observable.

All chemical processes are publicly observable.

∴ No feeling of pain is a chemical process. [Use Fx, Ox, and Cx. This attacks a form of materialism that identifies mental events with material events. We also could test this argument using syllogistic logic (Chapter 2).]

4. All (in the electoral college) who do their jobs are useless.

All (in the electoral college) who don’t do their jobs are dangerous.

∴ All (in the electoral college) are useless or dangerous. [Use Jx for “x does their job,” Ux for “x is useless,” and Dx for “x is dangerous.” Use the universe of discourse of electoral college members: take “(x)” to mean “for every electoral college member x” and don’t translate “in the electoral college.”] 0191

5. All that’s known is experienced through the senses.

Nothing that’s experienced through the senses is known.

∴ Nothing is known. [Use Kx and Ex. Empiricism (premise 1) plus skepticism about the senses (premise 2) yields general skepticism.]

6. No pure water is burnable.

Some Cuyahoga River water is burnable.

∴ Some Cuyahoga River water isn’t pure water. [Use Px, Bx, and Cx. The Cuyahoga is a river in Cleveland that used to catch fire.]

7. Everyone who isn’t with me is against me.

∴ Everyone who isn’t against me is with me. [Use Wx and Ax. These claims from the Gospels are sometimes thought to be incompatible.]

8. All basic laws depend on God’s will.

∴ All basic laws about morality depend on God’s will. [Bx, Dx, Mx]

9. Some lies in unusual circumstances aren’t wrong.

∴ Not all lies are wrong. [Lx, Ux, Wx]

10.Nothing based on sense experience is certain.

Some logical inferences are certain.

All certain things are truths of reason.

∴ Some truths of reason are certain and aren’t based on sense experience. [Bx, Cx, Lx, Rx]

11.No truth by itself motivates us to action.

Every categorical imperative would by itself motivate us to action.

Every categorical imperative would be a truth.

∴ There are no categorical imperatives. [Use Tx, Mx, and Cx. Immanuel Kant claimed that commonsense morality accepts categorical imperatives (objectively true moral judgments that command us to act and that we must follow if we are to be rational); but some thinkers argue against the idea.]

12.Every genuine truth claim is either experimentally testable or true by definition.

No moral judgments are experimentally testable.

No moral judgments are true by definition.

∴ No moral judgments are genuine truth claims. [Use Gx, Ex, Dx, and Mx. This is logical positivism’s argument against moral truths.]

13.Everyone who can think clearly would do well in logic.

Everyone who would do well in logic ought to study logic.

Everyone who can’t think clearly ought to study logic.

∴ Everyone ought to study logic. [Tx, Wx, Ox]

8.3 Easier refutations

Applying our proof strategy to an invalid argument leads to a refutation: 0192

· Someone is short.

· Someone is tall.

· ∴ Someone is both short and tall.

· * 1 (∃x)Sx Invalid

· * 2 (∃x)Tx

· * [∴ (∃x)(Sx • Tx)

· * 3 asm: ∼(∃x)(Sx • Tx)

· * 4 ∴ (x)∼(Sx • Tx) {from 3} * 5 ∴ Sa {from 1}

· * 6 ∴ Tb {from 2}

· * 7 ∴ ∼(Sa • Ta) {from 4}

· * 8 ∴ ∼(Sb • Tb) {from 4}

· 9 ∴ ∼Ta {from 5 and 7}

· 10 ∴ ∼Sb {from 6 and 8}

a, b

Sa, ∼Ta

Tb, ∼Sb

Reverse a squiggle (line 4). Drop two existentials, using a new constant each time (lines 5 and 6). Drop the universal twice, using “a” and “b” (lines 7 and 8). Getting no contradiction, we gather simple wffs for a refutation (here a “simple wff” is one containing only capital letters, zero or more constants, and zero or one squiggles). We get a little possible world with two people, a and b, where a is short and not tall, but b is tall and not short. The argument is invalid, since this possible world makes the premises all true (someone is short and someone is tall) but the conclusion false (no one is both short and tall).

If we try to prove an invalid argument, we’ll instead be led to a refutation – a little possible world with various individuals (like a and b) and simple truths about them (like Sa and ∼Sb) that make the premises all true and conclusion false. In evaluating premises and conclusion, use these rules to evaluate each formula or subformula that starts with a quantifier:

An existential wff is true if and only if at least one case is true.

A universal wff is true if and only if all cases are true.

Premise “(∃x)Sx” is true because at least one case (“Sa”) is true, and premise “(∃x)Tx” is true because at least one case (“Tb”) is true.¹ But conclusion “(∃x)(Sx • Tx)” is false because both cases are false:

¹ SOME is like OR: something holds in this case OR that case OR that case … – so a single true case makes a SOME true. ALL is like AND: something holds in this case AND that case AND that case … – so a single false case makes an ALL false.

(Sa • Ta) = (1 • 0) = 0

(Sb • Tb) = (0 • 1) = 0

Always check that your refutation works. If you don’t get premises all 1 and conclusion 0, then you did something wrong; look at what you did with the wff that came out wrong (a premise that’s 0 or ?, or a conclusion that’s 1 or ?).

These two rules are crucial for working out proofs and refutations: 0193

· For each initial existential quantifier, introduce a new constant.

· For each initial universal quantifier, derive an instance for each old constant.

If you have two existentials, don’t drop both using the same constant – and don’t drop just one existential. And if you have two constants, then drop any universals using both constants; if in our example we dropped the universal in “(x)∼(Sx • Tx)” using “a” but not “b,” then our refutation would fail:

a, b

Sa, ∼Ta, Tb

a is short and not tall, b is tall

Since “Sb” is unknown, our conclusion “(∃x)(Sx • Tx)” would also be unknown (because the second case with “b” is unknown):

(Sa • Ta) = (1 • 0) = 0

(Sb • Tb) = (? • 1) = ?

The “Someone is both short and tall” conclusion is unknown, since our world doesn’t exclude b being short (besides being tall). We avoid such problems if we drop each initial universal quantifier using each old constant; here we’d go from “(x)∼(Sx • Tx)” to “∼(Sb • Tb),” which would lead to “∼Sb.”

As we refute arguments, we’ll often have to evaluate premises or conclusions that don’t start with quantifiers, such as these wffs:

Identify any subformulas that start with quantifiers (as highlighted here). Evaluate each subformula to be 1 or 0, and then apply “∼” to reverse the result. On our short-tall refutation, “(x)Sx” = 0 and so “∼(x)Sx” = 1. Likewise, “(x)(Sx ∨ Tx)” = 1, and so “∼(x)(Sx ∨ Tx)” = 0; and “(∃x)(Sx • Tx)” = 0, and so “∼(∃x)(Sx • Tx)” = 1. In evaluating a wff that starts with a squiggle and then a quantifier, evaluate the wff without the squiggle and then give the original wff the opposite value. Divide and conquer!

Possible worlds for refutations must contain at least one entity. We seldom need more than two entities.

8.3a Exercise: LogiCola IEI

Prove each of these arguments to be invalid (all are invalid). 0194

a, b

∼Fa, ∼Ga, Gb

1. (∃x)Fx

∴ (x)Fx

2. (∃x)Fx

(∃x)Gx

∴ (∃x)(Fx • Gx)

3. (∃x)(Fx ∨ Gx)

∼(x)Fx

∴ (∃x)Gx

4. (∃x)Fx

∴ (∃x)∼Fx

5. ∼(∃x)(Fx • Gx)

(x)∼Fx

∴ (x)Gx

6. (x)(Fx ⊃ Gx)

∼(x)Gx

∴ (x)∼(Fx • Gx)

7. (x)((Fx • Gx) ⊃ Hx)

(∃x)Fx

(∃x)Gx

∴ (∃x)Hx

8. (∃x)(Fx ∨ ∼Gx)

(x)(∼Gx ⊃ Hx)

(∃x)(Fx ⊃ Hx)

∴ (∃x)Hx

9. (∃x)∼(Fx ∨ Gx)

(∃x)Hx

∼(∃x)Fx

∴ ∼(x)(Hx ⊃ Gx)

10.(∃x)∼Fx

(∃x)∼Gx

∴ (∃x)(Fx ≡ Gx)

8.3b Exercise: LogiCola IEC

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

1. Some butlers are guilty.

∴ All butlers are guilty. [Use Bx and Gx.]

2. No material thing is infinite.

Not everything is material.

∴ Something is infinite. [Use Mx and Ix.]

3. Some smoke.

Not all have clean lungs.

∴ Some who smoke don’t have clean lungs. [Use Sx and Cx.]

4. Some Marxists plot violent revolution.

Some faculty members are Marxists.

∴ Some faculty members plot violent revolution. [Mx, Px, Fx] 0195

5. All valid arguments that have “ought” in the conclusion also have “ought” in the premises.

All arguments that seek to deduce an “ought” from an “is” have “ought” in the conclusion but don’t have “ought” in the premises.

∴ No argument that seeks to deduce an “ought” from an “is” is valid. [Use Vx for “x is valid,” Cx for “x has ‘ought’ in the conclusion,” Px for “x has ‘ought’ in the premises,” Dx for “x seeks to deduce an ‘ought’ from an ‘is,’” and the universe of discourse of arguments. This one is difficult to translate.]

6. Every kick returner who is successful is fast.

∴ Every kick returner who is fast is successful. [Kx, Sx, Fx]

7. All exceptionless duties are based on the categorical imperative.

All non-exceptionless duties are based on the categorical imperative.

∴ All duties are based on the categorical imperative. [Use Ex, Bx, and the universe of discourse of duties; from Kant, who based all duties on his supreme moral principle, called “the categorical imperative.”]

8. All who aren’t crazy agree with me.

∴ No one who is crazy agrees with me. [Cx, Ax]

9. Everything can be conceived.

Everything that can be conceived is mental.

∴ Everything is mental. [Use Cx and Mx; from George Berkeley, who attacked materialism by arguing that everything is mental and that matter doesn’t exist apart from mental sensations; so a chair is just a collection of experiences. Bertrand Russell thought premise 2 was confused.]

10.All sound arguments are valid.

∴ All invalid arguments are unsound. [Use Sx and Vx and the universe of discourse of arguments.]

11.All trespassers are eaten.

∴ Some trespassers are eaten. [Use Tx and Ex. The premise is from a sign on the Appalachian Trail in northern Virginia. Traditional logic (§2.8) takes “all A is B” to entail “some A is B”; modern logic takes “all A is B” to mean “whatever is A also is B” – which can be true even if there are no A’s.]

12.Some necessary being exists.

All necessary beings are perfect beings.

∴ Some perfect being exists. [Use Nx and Px. Kant claimed that the cosmological argument for God’s existence at most proves premise 1; it doesn’t prove the existence of a perfect God unless we add premise 2. But premise 2, by the next argument, presupposes the central claim of the ontological argument – that some perfect being is a necessary being. So, Kant claimed, the cosmological argument presupposes the ontological argument.]

13.All necessary beings are perfect beings.

∴ Some perfect being is a necessary being. [Use Nx and Px. Kant followed traditional logic (see problem 11) in taking “all A is B” to entail “some A is B.”] 0196

14.No one who isn’t a logical positivist holds the verifiability criterion of meaning.

∴ All who hold the verifiability criterion of meaning are logical positivists. [Use Lx and Hx. The verifiability criterion of meaning says that every genuine truth claim is either experimentally testable or true by definition.]

15.No pure water is burnable.

Some Cuyahoga River water isn’t burnable.

∴ Some Cuyahoga River water is pure water. [Use Px, Bx, and Cx.]

8.4 Harder translations

We’ll now start using statement letters (like “S” for “It’s snowing”) and individual constants (like “r” for “Romeo”); here’s an example:

If it’s snowing, then Romeo is cold = (S ⊃ Cr)

Here “S,” since it’s a capital letter not followed by a small letter, represents a whole statement. And “r,” since it’s a small letter between “a” and “w,” is a constant that stands for a specific person or thing.

We’ll also start using multiple and non-initial quantifiers. From now on, use this expanded rule about what quantifier to use and where to put it:

Where the English has “all” or “every,” put this in the wff: “(x).”

Where the English has “not all” or “not every,” put this in the wff: “∼(x).”

Where the English has “some,” put this in the wff: “(∃x).”

Where the English has “no,” put this in the wff: “∼(∃x).”

If all are Italian, then Romeo is Italian = ((x)Ix ⊃ Ir)

Since “if” translates as “(,” likewise “if all” translates as “((x).” As you translate, mimic the English word order:

· all not = (x)∼

not all = ∼(x)

· all either = (x)(

either all = ((x)

· if all either = ((x)(

if either all = (((x)

Use a separate quantifier for each “all,” “some,” and “no”:

· If all are Italian, then all are lovers

= ((x)Ix ⊃ (x)Lx)

· If not everyone is Italian, then some aren’t lovers

= (∼(x)Ix ⊃ (∃x)∼Lx)

· If no Italians are lovers, then some Italians are not lovers

= (∼(∃x)(Ix • Lx) ⊃ (∃x)(Ix • ∼Lx)) 0197

“Any” differs in subtle ways from “all” (which translates into a “(x)” that mirrors where “all” occurs in the English sentence). “Any” has two different but equivalent translation rules; here’s the easier rule, with examples:

To translate “any,” first rephrase the sentence so it means the same thing but doesn’t use “any”; then translate the second sentence.

“Not any …” = “No ….”

“If any …” = “If some ….”

“Any …” = “All ….”

· Not anyone is rich = No one is rich

= ∼(∃x)Rx

· Not any Italian is a lover = No Italian is a lover

= ∼(∃x)(Ix • Lx)

· If anyone is just, there will be peace = If someone is just, there will be peace

= ((∃x)Jx ⊃ P)

Our second rule usually gives a formula that’s different but equivalent:

To translate “any,” put a “(x)” at the beginning of the wff, regardless of where the “any” occurs in the sentence.

· Not anyone is rich = For all x, x isn’t rich

= (x)∼Rx

· Not any Italian is a lover = For all x, x isn’t both Italian and a lover

= (x)∼(Ix • Lx) ⇐ Note “•” here!

· If anyone is just, there will be peace = For all x, if x is just there will be peace

= (x)(Jx ⊃ P)

“Any” at the beginning of a sentence usually just means “all.” So “Any Italian is a lover” means “All Italians are lovers.”

8.4a Exercise: LogiCola H (HM & HT)

Translate these English sentences into wffs. Recall that our translation rules are rough guides and sometimes don’t work; so read your formula carefully to make sure it reflects what the English means.

If everyone is evil, then Gensler is evil.

((x)Ex ⊃ Eg)

1. Gensler is either crazy or evil.

2. If Gensler is a logician, then some logicians are evil.

3. If everyone is a logician, then everyone is evil.

4. If all logicians are evil, then some logicians are evil.

5. If someone is evil, it will rain.

6. If everyone is evil, it will rain.

7. If anyone is evil, it will rain. 0198

8. If Gensler is a logician, then someone is a logician.

9. If no one is evil, then no one is an evil logician.

10.If all are evil, then all logicians are evil.

11.If some are logicians, then some are evil.

12.All crazy logicians are evil.

13.Everyone who isn’t a logician is evil.

14.Not everyone is evil.

15.Not anyone is evil.

16.If Gensler is a logician, then he’s evil.

17.If anyone is a logician, then Gensler is a logician.

18.If someone is a logician, then he or she is evil.

19.Everyone is an evil logician.

20.Not any logician is evil.

8.5 Harder proofs

Now we come to proofs using formulas with multiple or non-initial quantifiers. Such proofs, while needing no new inference rules, are often tricky and require multiple assumptions. As before, drop only initial quantifiers:

Both of these are wrong:

· ((x)Fx ⊃ (x)Gx)

–––––––––––––

∴ (Fa ⊃ (x)Gx)

· ((x)Fx ⊃ (x)Gx)

–––––––––––––

∴ (Fa ⊃ Ga)

“((x)Fx ⊃ (x)Gx)” is an if-then; to infer with it, we need the first part true or the second part false – as in these examples:

Both of these are right:

· ((x)Fx ⊃ (x)Gx)

(x)Fx

–––––––––––––

∴ (x)Gx

· ((x)Fx ⊃ (x)Gx) ∼(x)Gx

–––––––––––––

∴ ∼(x)Fx

· If we get stuck, we may need to assume one side or its negation.

Here’s a proof using a formula with multiple quantifiers: 0199

· If some are enslaved, then all have their freedom threatened.

∴ If this person is enslaved, then I have my freedom threatened.

After the assumption, we apply an S-rule to get lines 3 and 4. Then we’re stuck, since we can’t drop the non-initial quantifiers in 1. So we make a second assumption in line 5, get a contradiction, and derive 8. We soon get a second contradiction to complete the proof.

Here’s an invalid argument:

· If all are enslaved, then all have their freedom threatened.

∴ If this person is enslaved, then I have my freedom threatened.

· 1 ((x)Sx ⊃ (x)Tx) Invalid [ ∴ (St ⊃ Ti)

· * 2 asm: ∼(St ⊃ Ti)

· 3 ∴ St {from 2}

· 4 ∴ ∼Ti {from 2}

· ** 5 asm: ∼(x)Sx {break 1}

· ** 6 ∴ (∃x)∼Sx {from 5}

· 7 ∴ ∼Sa {from 6}

t, i, a

St, ∼Ti, ∼Sa

In evaluating the premise, first identity and evaluate subformulas that start with quantifiers (these are highlighted here), and then plug in 1 or 0 for these:

So the premise is true. Since the conclusion is false, the argument is invalid.

As we refute invalid arguments, we’ll often have complex premises or conclusions to evaluate, such as these wffs:

Identity any subformulas that start with quantifiers (as highlighted 0200 here). Evaluate each such subformula to be 1 or 0, replace it with 1 or 0, and figure out whether the whole formula is 1 or 0. Divide and conquer!

8.5a Exercise: LogiCola I (HC & MC)

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

· (x)(Mx ∨ Fx)

∴ ((x)Mx ∨ (x)Fx)

(This is like arguing that, since everyone is male or female, thus either everyone is male or everyone is female.)

· 1 (x)(Mx ∨ Fx) Invalid

· [ ∴ ((x)Mx ∨ (x)Fx)

· * 2 asm: ∼((x)Mx ∨ (x)Fx)

· * 3 ∴ ∼(x)Mx {from 2}

· * 4 ∴ ∼(x)Fx {from 2}

· * 5 ∴ (∃x)∼Mx {from 3}

· * 6 ∴ (∃x)∼Fx {from 4}

· 7 ∴ ∼Ma {from 5}

· 8 ∴ ∼Fb {from 6}

· * 9 ∴ (Ma ∨ Fa) {from 1}

· * 10 ∴ (Mb ∨ Fb) {from 1}

· 11 ∴ Fa {from 7 and 9}

· 12 ∴ Mb {from 8 and 10}

a, b

Fa, ∼Ma

Mb, ∼Fb

1. (x)(Fx ∨ Gx)

∼Fa

∴ (∃x)Gx

2. (x)(Ex ⊃ R)

∴ ((∃x)Ex ⊃ R)

3. ((x)Ex ⊃ R)

∴ (x)(Ex ⊃ R)

4. ((∃x)Fx ∨ (∃x)Gx)

∴ (∃x)(Fx ∨ Gx)

5. ((∃x)Fx ⊃ (∃x)Gx)

∴ (x)(Fx ⊃ Gx)

6. (x)((Fx ∨ Gx) ⊃ Hx)

Fm

∴ Hm

7. Fj

(∃x)Gx

(x)((Fx • Gx) ⊃ Hx)

∴ (∃x)Hx

8. ((∃x)Fx ⊃ (x)Gx)

∼Gp

∴ ∼Fp

9. (∃x)(Fx ∨ Gx)

∴ ((x)∼Gx ⊃ (∃x)Fx)

10.∼(∃x)(Fx • Gx)

∼Fd

∴ Gd

11.(x)(Ex ⊃ R)

∴ ((x)Ex ⊃ R)

12.(x)(Fx • Gx)

∴ ((x)Fx • (x)Gx)

13.(R ⊃ (x)Ex)

∴ (x)(R ⊃ Ex)

14.((x)Fx ∨ (x)Gx)

∴ (x)(Fx ∨ Gx)

15.((∃x)Ex ⊃ R)

∴ (x)(Ex ⊃ R)

8.5b Exercise: LogiCola I (HC & MC)

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

1. Everything has a cause.

If the world has a cause, then there is a God.

∴ There is a God. [Use Cx for “x has a cause,” w for “the world,” and G for “There is a God” (which we needn’t here break down into “(∃x)Gx” – “For some x, x is a God”). A student of mine suggested this argument; but the next example shows that premise 1 can as easily lead to the opposite conclusion.] 0201

2. Everything has a cause.

If there is a God, then something doesn’t have a cause (namely, God).

∴ There is no God. [Use Cx and G. The next example qualifies “Everything has a cause” to avoid the problem; some prefer an argument based on “Every contingent being or set of such beings has a cause.”]

3. Everything that began to exist has a cause.

The world began to exist.

If the world has a cause, then there is a God.

∴ There is a God. [Use Bx, Cx, w, and G. This “Kalam argument” is from William Craig and James Moreland; they defend premise 2 by appealing to the Big Bang theory, the law of entropy, and the impossibility of an actual infinite.]

4. If everyone litters, then the world will be dirty.

∴ If you litter, then the world will be dirty. [Lx, D, u]

5. Anything enjoyable is either immoral or fattening.

∴ If nothing is immoral, then everything that isn’t fattening isn’t enjoyable. [Ex, Ix, Fx]

6. Anything that can be explained either can be explained as caused by scientific laws or can be explained as resulting from a free choice of a rational being.

The totality of basic scientific laws can’t be explained as caused by scientific laws (since this would be circular).

∴ Either the totality of basic scientific laws can’t be explained or else it can be explained as resulting from a free choice of a rational being (God). [Use Ex for “x can be explained,” Sx for “x can be explained as caused by scientific laws,” Fx for “x can be explained as resulting from a free choice of a rational being,” and t for “the totality of scientific laws.” This one is from R. G. Swinburne.]

7. If someone knows the future, then no one has free will.

∴ No one who knows the future has free will. [Kx, Fx]

8. If everyone teaches philosophy, then everyone will starve.

∴ Everyone who teaches philosophy will starve. [Tx, Sx]

9. No proposition based on sense experience is logically necessary.

∴ Either no mathematical proposition is based on sense experience, or no mathematical proposition is logically necessary. [Use Sx, Nx, and Mx, and the universe of propositions; from the logical positivist A. J. Ayer.]

10.Any basic social rule that people would agree to if they were free and rational but ignorant of their place in society (whether rich or poor, white or black, male or female) is a principle of justice.

The equal-liberty principle and difference principle are basic social rules that people would agree to if they were free and rational but ignorant of their place in society.

∴ The equal-liberty principle and difference principle are principles of justice. [Use Ax, Px, e, and d; from John Rawls. Equal-liberty says that everyone is entitled to the greatest liberty compatible with an equal liberty for all others; difference says that wealth is to be distributed equally, except for inequalities that provide incentives that ultimately benefit everyone and are equally open to all.] 0202

11.If there are no necessary beings, then there are no contingent beings.

∴ All contingent beings are necessary beings. [Use Nx and Cx. Aquinas accepted the premise but not the conclusion.]

12.Anything not disproved that’s of practical value to one’s life to believe ought to be believed.

Free will isn’t disproved.

∴ If free will is of practical value to one’s life to believe, then it ought to be believed. [Use Dx, Vx, Ox, f (for “free will”), and the universe of discourse of beliefs; from William James.]

13.If the world had no temporal beginning, then some series of moments before the present moment is a completed infinite series.

There’s no completed infinite series.

∴ The world had a temporal beginning. [Use Tx for “x had a temporal beginning,” w for “the world,” Mx for “x is a series of moments before the present moment,” and Ix for “x is a completed infinite series.” This one and the next are from Immanuel Kant, who thought our intuitive metaphysical principles lead to conflicting conclusions and thus can’t be trusted.]

14.Everything that had a temporal beginning was caused to exist by something previously in existence.

If the world was caused to exist by something previously in existence, then there was time before the world began.

If the world had a temporal beginning, then there was no time before the world began.

∴ The world didn’t have a temporal beginning. [Use Tx for “x had a temporal beginning,” Cx for “x was caused to exist by something previously in existence,” w for “the world,” and B for “There was time before the world began.”]

15.If emotivism is true, then “X is good” means “Hurrah for X!” and all moral judgments are exclamations.

All exclamations are inherently emotional.

“This dishonest income tax exemption is wrong” is a moral judgment.

“This dishonest income tax exemption is wrong” isn’t inherently emotional.

∴ Emotivism isn’t true. [T, H, Mx, Ex, Ix, t]

16.If everything is material, then all prime numbers are composed of physical particles.

Seven is a prime number.

Seven isn’t composed of physical particles.

∴ Not everything is material. [Mx, Px, Cx, s]

17.If everyone lies, the results will be disastrous.

∴ If anyone lies, the results will be disastrous. [Lx, D] 0203

18.Everyone makes moral judgments.

Moral judgments logically presuppose beliefs about God.

If moral judgments logically presuppose beliefs about God, then everyone who makes moral judgments believes (at least implicitly) that there is a God.

∴ Everyone believes (at least implicitly) that there is a God. [Use Mx for “x makes moral judgments,” L for “Moral judgments logically presuppose beliefs about God,” and Bx for “x believes (at least implicitly) that there is a God.” This is from the Jesuit theologian Karl Rahner.]

19.“x = x” is a basic law.

“x = x” is true in itself, and not true because someone made it true.

If “x = x” depends on God’s will, then “x = x” is true because someone made it true.

∴ Some basic laws don’t depend on God’s will. [Use e (for “x = x”), Bx, Tx, Mx, and Dx.]

20.Nothing that isn’t caused can be integrated into the unity of our experience. Everything that we could experientially know can be integrated into the unity of our experience.

∴ Everything that we could experientially know is caused. [Use Cx, Ix, and Ex; from Immanuel Kant. The conclusion is limited to objects of possible experience – since it says “Everything that we could experientially know is caused”; Kant thought the unqualified “Everything is caused” leads to contradictions (see # 1 and 2).]

21.If everyone deliberates about alternatives, then everyone believes (implicitly) in free will.

∴ All who deliberate about alternatives believe (implicitly) in free will. [Dx, Bx]

22.All who are consistent and think that abortion is normally permissible will consent to the idea of their having been aborted in normal circumstances.

You don’t consent to the idea of your having been aborted in normal circumstances.

∴ If you’re consistent, then you won’t think that abortion is normally permissible. [Use Cx, Px, Ix, and u. See my article in January 1986 Philosophical Studies or the synthesis chapter of my Ethics: A Contemporary Introduction, 3rd ed. (New York: Routledge, 2018).]

8.6 Copi proofs

We earlier discussed the traditional Copi proof method for propositional logic (§7.5). This method can also be used for quantificational logic.

For each quantifier (universal and existential), Copi has instantiation rules (to drop a quantifier) and generalization rules (to add a quantifier). Existential instantiation (EI) is the same as our drop-existential rule (DE). It holds regardless of what variable replaces “x,” what constant replaces “a,” and what wffs replace “Fx” / “Fa” – provided that the two wffs are identical except that wherever the 0204 variable occurs freely¹ in the former the same constant occurs in the latter:

¹ An instance of a variable occurs freely in a formula if it’s not part of a wff that begins with a quantifier using that variable; just the first instance of “x” in “(Fx • (x)Gx)” occurs freely.

EI Existential instantiation

(∃x)Fx → Fa,

use a new constant

Here the constant must not have occurred in any previous step of the proof or in the original conclusion. As before, “→” in all these rules means that we can infer whole lines from left to right. Existential generalization works the opposite way, and isn’t subject to the restriction that the constant has to be new:

EG Existential generalization

Fa → (∃x)Fx

Another form of the rule uses a variable in place of “a”:

EG Existential generalization

Fy → (∃x)Fx

This form holds regardless of what variables replace “x” and “y” and what wffs replace “Fx” / “Fy” – provided that the two wffs are identical except that wherever the variable that replaces “x” occurs freely in the former the variable that replaces “y” occurs freely in the latter.

Universal instantiation (UI) is like our drop-universal rule (DE), except that it also has two forms. The first form holds regardless of what variable replaces “x,” what constant replaces “a,” and what wffs replace “Fx” / “Fa” – provided that the two wffs are identical except that wherever the variable occurs freely in the former the same constant occurs in the latter:

UI Universal instantiation

(x)Fx → Fa

Here the constant needn’t be new, and so it could have occurred earlier in the proof or in the original conclusion. And again a second form uses a variable in place of “a.” This form holds regardless of what variables replace “x” and “y” and what wffs replace “Fx” / “Fy” – provided that the two wffs are identical except that wherever the variable that replaces “x” occurs freely in the former the variable that replaces “y” occurs freely in the latter:

UI Universal instantiation

(x)Fx → Fy

0205 Universal generalization works the opposite way:

UG Universal generalization

Fy → (x)Fx,

“y” can’t occur in an assumption

Here the variable that replaces “y” can’t occur in an assumption.

Copi has a replacement rule much like our reverse-squiggle rules (any variable can uniformly replace “x” and any wff can uniformly replace “P”):

QN Quantifier negation

(x)P = ∼(∃x)∼P

(∃x)P = ∼(x)∼ P

This lets us switch, for example, one instance of “(x)Fx” and “∼(∃x)∼Fx” anywhere in a wff. I’ll take the Copi proof system for quantificational logic to include these five rules: EI, EG, UI, UG, and QN.

To see how these work, I’ll now prove the valid arguments in this chapter’s explanation sections. Here’s a Copi proof for the first example in §8.2:

Conclusion: (∃x)Fx

1. (x)(Lx ⊃ Fx)

2. (∃x)Lx

3. La {EI 2}

4. (La ⊃ Fa) {UI 1}

5. Fa {MP 3+4}

6. (∃x)Fx {EG 5}

And here’s a Copi proof for the second example in §8.2:

Conclusion: (x)Fx

1. (x)(Fx • Gx)

2. (Fy • Gy) {UI 1}

3. Fy {SM 2}

4. (x)Fx {UG 3}

And here’s a Copi proof for the valid example in §8.5:

Conclusion: (St ⊃ Ti)

1. ((∃x)Sx ⊃ (x)Tx)

2. St {Assume} *

3. (∃x)Sx {EG 2} *

4. (x)Tx {MP 1+3} *

5. Ti {UI 4} *

6. (St ⊃ Ti) {CP 2+5} 0206

If you’ve mastered Copi propositional proofs, the quantificational proofs won’t be too difficult. If you’re really confused on how to start, try assuming the opposite of the conclusion and deriving a contradiction. Again, use Copi proofs only on valid arguments; if you try the Copi procedure on an invalid argument, you won’t derive the conclusion and you won’t reach a natural “stopping point” that gives you a refutation of the argument’s validity.

8.5a and 8.5b Exercise: LogiCola IEO

Do Copi proofs for problems in §§8.2a and 8.2b (all are valid). These are easier problems.

8.5c and 8.5d Exercise: LogiCola IHO and IMO

Do Copi proofs for problems in §8.5a (just the valid ones, namely 1, 2, 4, 6, 8, 9, 11, 12, 13, 14, and 15) and §8.5b (just the valid ones, namely 1, 2, 3, 5, 6, 7, 10, 12, 13, 14, 15, 16, 18, 19, 20, and 22). These are harder problems.