In either case the derivation continues as follows (with {x, y} meaning line x if we began the first way, and line y if we began the second way):

because N = min (T, N).

6.3 Exercises

SECTION 6.1

1 Prove that the rule RSIMP is derivable in L3A:

RSIMP (Right Conjunct Simplification). From P Q infer Q.

2 Show that the inference P → Q

(P → ¬P) → Q Q,

which is captured by the derived rule MCD in L3A, is valid in L3.

3Produce derivations to show that the following formulas are theorems of L3A:

a.(P Q) → (¬P → Q)

b.(P Q) → (Q P)

c.¬(P → Q) → ¬Q

d.¬(P → Q) → (Q → P)

e.¬(P → Q) → (P → ¬Q)

f.P → (¬Q → ¬(P → Q))

4Prove the Modified Deduction Theorem for L3A in two parts:

a.Show that if P → (P → Q) is a theorem then Q is derivable from P (this part is pretty easy).

b.Show that if Q is derivable from P then P → (P → Q) is a theorem. You can show this by proving that if we have a derivation

2R1

3R2

. . . . . .

n Rn n+1 Q

in L3A, where R1, . . . , Rn are the intermediate formulas in the derivation, then every one of the following formulas is a theorem of L3A:

P → (P → P)

P → (P → R1)

P → (P → R2)

. . .

P → (P → Rn) P → (P → Q)

Specifically, show that

i.P → (P → P) is a theorem.

ii.If any of the Ri—or Q—are axioms of L3A, then P → (P → Ri) (or P → (P → Q)) is a theorem.

iii.If any of the Ri—or Q—follows from previous formulas Rj and Rk by the rule MP, then P → (P → Ri) (or P → (P → Q)) follows from P → (P → Rj) and P → (P → Rk) (although not, of course, by a single application of MP—you will find L3D11 useful in this part).

SECTION 6.2

5a. Show that the rules L3P5.2.2, L35.3.2, L35.4.2, and L35.7.2 are derivable from other rules in the system L3PA, excluding all of the .2 versions of the L3P5 rules.

b.Show that the rule L3P5.5.2 is derivable from other axioms in the system L3PA, excluding all of the .2 versions of the L3P5 axioms.

c.Show that the rule L3P5.6.2 is derivable from other axioms in the system L3PA, excluding all of the .2 versions of the L3P5 axioms.

d.Show that the rule L3P5.8.2 is derivable from other axioms in the system L3PA, excluding all of the .2 versions of the L3P5 axioms.

e.Show that the rule L3P5.9.2 is derivable from other axioms in the system

L3PA, excluding all of the .2 versions of the L3P5 axioms.

6Show that L3P6.3.1 and L3P6.3.2 are derivable in L3PA given L3P6.1.1 and L3P6.1.2, and vice versa.

7Show that L3P7.1 is derivable from the other axioms of L3PA.

8Justify the truth-values for v3 in each row of the truth-value table for the MP rule in L3PA.

9Derive the following theorems in L3PA:

a.[(n → P) → (¬P → n), T]

b.[(P → f) → (t → (Q → (P → R))), T]

c.[(n → P) → (n → (Q → P)), T]

d.[(t → P) → ((n → Q) → (n → (P → Q))), T]

e.[(t → (P → Q))→ (P → Q), T]

f.[(t → (P → Q)) → ((t → P) → (t → Q)), T]

10Derive graded versions of the following rules for L3PA. In each case, make sure that you have assigned the strongest possible graded value to the inferred formula.

a.LSIMP

b.RSIMP

c.SUB

Note: The graded rule should look like this:

SUB (Substitution). From [P → Q, T], [Q → P, T], and a graded formula [R, v] such that R contains P as a subformula, infer any graded formula [R*, x] in which R* is the result of replacing one or more occurrences of

P in R with Q and x is . . .(fill in the blank).

d.MT

e.DN

f.TRAN

g.GCON

h.GHS

i.GMP

j.DS

k.DC

11Prove that if we can derive a graded formula [Q, v] from some (possible empty) set of graded assumptions in system L3PA, then for any value v less than v, we can also derive [Q, v ] from those same graded assumptions.

12a. Derive [(¬P → n), T] from [P, N] in L3PA.

b.Derive [Q → P, N] from [P, N] in L3PA.

c.Derive [t → (P ¬P), N] in L3PA.

13Given the (weak or strong) completeness of L3A for L3, we can prove that if a formula P (without truth-value constants) is a quasi-tautology of L3 then [P, N] is a theorem of L3PA as follows:

If P is a quasi-tautology of L3 then the formula ¬P → P is always true (given that P always has the value T or N) and is therefore a tautology. Because L3A is complete, ¬P → P is a theorem of L3A and so [¬P → P, T] is a theorem of L3PA. [(¬P → P) → P, N] is also a theorem:

and so we can derive [P, N] from these two theorems by Modus Ponens.

Given the strong completeness of L3A for L3, prove that if a set of formulasquasi-entails P (where none of the formulas contain truth-value constants) then [P, N] is derivable in L3PA from the graded set of formulas G in which each member of has the grade N.

x-variant v of a variable assignment v is an assignment such that v (y) = v(y) for every variable y other than x.

The truth-conditions for formulas on three-valued interpretations are defined in terms of satisfaction and dissatisfaction by variable assignments. Here are the truth-conditions for the first-order generalization of L3, which we will call L3 :1

1.An atomic formula Pt1 . . . tn is

satisfied by a variable assignment v on an interpretation I if <I*(t1), . . . , I*(tn)> ext(P), where I*(ti) is I(ti) if ti is a constant and is v(ti) if ti is a variable,

dissatisfied if <I*(t1), . . . ,I*(tn)> cxt(P), and

neither satisfied nor dissatisfied if <I*(t1), . . . ,I*(tn)> fge(P).

When a formula is neither satisfied nor dissatisfied by a variable assignment, we will say that it is undetermined by that assignment. Clauses 2–6 reflect the Lukasiewicz truth-tables for propositional logic:

2.A formula ¬P is

satisfied by a variable assignment v on an interpretation I if P is dissatisfied by v on I,

dissatisfied if P is satisfied by v on I, and

undetermined otherwise.

3.A formula P Q is

satisfied by a variable assignment v on an interpretation I if both P and Q are satisfied by v on I,

dissatisfied if either P or Q is dissatisfied by v on I, and

undetermined otherwise.

4.A formula P Q is

satisfied by a variable assignment v on an interpretation I if either P or Q is satisfied by v on I,

dissatisfied if both P and Q are dissatisfied by v on I, and

undetermined otherwise.

5.A formula P → Q is

satisfied by a variable assignment v on an interpretation I if P is dissatisfied by v on I, or Q is satisfied by v on I, or P and Q are both undetermined by v on I,

dissatisfied if P is satisfied by v on I and Q is dissatisfied by v on I, and

undetermined otherwise.

6.A formula P ↔ Q is

satisfied by a variable assignment v on an interpretation I if either P and Q are both satisfied by v on I, or P and Q are both dissatisfied by v on I, or P and Q are both undetermined by v on I,

1As in Chapter 6, we will omit the subscript L on the Lukasiewicz connectives but include them on the connectives for the other first-order systems.

ext(N) = {<h1, h2>: h1 D, h2 D, and h1 – h2 ≥ 3 } fge(N) = {<h1, h2>: h1 D, h2 D, and 1 < h1 – h2 < 3 } cxt(N) = {<h1, h2>: h1 D, h2 D, and h1 – h2 ≤ 1 }

I(a) = 6 I(b) = 5 6 I(c) = 4 11 I(d) = 4 10 I(e) = 4 9

We’ll call this interpretation VH, for vague heights. Intuitively, T means is tall; we have somewhat arbitrarily—but not unreasonably—chosen the cutoff points for being tall and being not tall. V means very tiny and N means is noticeably taller than—again, the cutoffs are somewhat arbitrary but seem okay. Note that we’ve decided that none of the heights in the domain are very tiny. The formula Ta has the value T on interpretation VH because I(a) is a member of ext(T); Tc, Td, and Te have the value F since I(c), I(d), and I(e) are members of cxt(T), and Tb has the value N because I(b) is a member of fge(T). Va, Vb, Vc, and Vd all have the value F while Ve has the value N. Naa has the value F, while Nab, Nac, Nad, and Nae all have the value T. Ncd and Nde have the value F, and Nce has the value N.

The truth-values of compound formulas without quantifiers are as expected for a Lukasiewicz system; for example, Ta Tb has the value T on VH, Tb Tc has the value N, and Tb Tc has the value F. Ta → Ta, Tb → Tb, Tc → Tc, Td → Td, and Te → Te all have the value T. Ta → Tb and Tb → Tc both have the value N, while Tc → Td and Td → Te both have the value T, as do Tb → Ta, Tc → Tb, Td → Tc, and Te → Td.

The universally quantified formula ( x)Tx has the value F on interpretation VH because at least one x-variant of each variable assignment will dissatisfy Tx, namely, any x-variant that assigns a member of cxt(T) to x. ( x)Vx has the value F for a similar reason. The existentially quantified formula ( x)Tx has the value T because at least one x-variant of each variable assignment does satisfy Tx: any x-variant that assigns a member of ext(T) to x will do so. But the formula ( x)Vx has the value N because no x-variant of any variable assignment satisfies Vx (so ( x)Vx isn’t true), but not all x-variants dissatisfy Vx: those that assign a member of fge(V) to x neither satisfy nor dissatisfy Vx (so ( x)Vx isn’t false either). ( x)¬Vx has the value T because the counterextension of V is nonempty; x-variants that assign members of cxt(V ) to x will satisfy ¬Vx.

( x)( y)Nyx has the value F on interpretation VH, precisely because it’s false that for every height in the domain you can find one that’s noticeably taller. Consider any variable assignment v: the x-variant v of v that assigns 6 7 to x dissatisfies the formula ( y)Nyx (as do all x-variants that assign any height greater than 6 4 to x). v dissatisfies this formula because every y-variant v of v dissatisfies Nyx: for every height in the domain that we can assign to y, <v (y), v (x)> ( = <v (y), 6 7 >) is a member of cxt(N) because v (y) – 6 7 is less than 1 .

The formula ( x)(¬Vx → ( y)Nxy), which we may read as every height that isn’t very tiny is noticeably taller than some height, has the value N on VH. We’ll first show that for any variable assignment v,

i.the x-variants of v that assign members of cxt(V) to x either satisfy or neither satisfy nor dissatisfy the formula ¬Vx → ( y)Nxy, and

ii.the x-variants of v that assign members of fge(V) to x cannot dissatisfy ¬Vx → ( y)Nxy.

For (i), consider the x-variants v that assign members of cxt(V) to x. These x-variants will all satisfy ¬Vx. Some of them, those that assign heights ≥ 4 10 , also satisfy ( y)Nxy—because each height ≥ 4 10 is noticeably taller than at least one height in the domain. But among these heights those that lie strictly between 4 9 and 4 10 neither satisfy nor dissatisfy ( y)Nxy—for each of these heights v (x) there is a y-variant v of v such that <v (x), v (y)> fge(N), but there is no y-variant v such that <v (x), v (y)> ext(N). So some x-variants that assign members of cxt(V) to x satisfy the conditional ¬Vx → ( y)Nxy while others neither satisfy nor dissatisfy the conditional. For (ii), consider the x-variants that assign members of fge(V) to x: none of these can dissatisfy ¬Vx → ( y)Nxy because none of them satisfy ¬Vx.

Finally, there are no x-variants that assign members of ext(V) to x, because ext(V) is empty, so between (i) and (ii) we have considered all of the x-variants of v, with the result that some x-variants of v satisfy the conditional ¬Vx → ( y)Nxy, some x-variants leave the conditional undetermined, and no x-variants dissatisfy it. Thus the universally quantified formula ( x)(¬Vx → ( y)Nxy) has the value N.

The formula ( x)(Tx → Tx) has the value T on interpretation VH, as well as on all other interpretations forL3 . Every variable assignment satisfies the universal quantification because every x-variant will satisfy Tx→ Tx—the identical antecedent and consequent are either both satisfied or both dissatisfied or both undetermined on any variable assignment. On the other hand, the tall version of the Law of Excluded Middle, ( x)(Tx ¬Tx), has the value N on interpretation VH. This is because the formula Tx ¬Tx is undetermined by variable assignments that assign a value

the value T, but Tx ¬T is not dissatisfied by any—so ( x)(Tx ¬Tx) doesn’t have the value F. No variable assignment can dissatisfy Tx ¬Tx because no variable assignment can dissatisfy both disjuncts.

The negation ¬( x)(Tx ¬Tx) of the Law of Excluded Middle is equivalent to ( x)(¬Tx ¬¬Tx) in L3 . This is the formula at issue in Max Black’s Problem of the Fringe: we would like to affirm that a predicate is vague by stating that there is at least one object in its fringe, and it would seem that we can express this with the formula ( x)(¬Tx ¬¬Tx): there is at least one object that is neither tall nor not tall. But by the Principle of Double Negation, which holds in L3 as it does in classical logic, the formula ( x)(¬Tx ¬¬Tx) is equivalent to ( x)(¬Tx Tx)—and that formula seems to be a contradiction. So in claiming that the predicate tall is vague we would seem to be committed to the truth of a contradiction—that’s the

problem. In Chapter 3 we noted that the fringe formula is indeed a contradiction of classical first-order logic.

Neither ( x)(¬Tx ¬¬Tx) nor ( x)(¬Tx Tx) is a contradiction (i.e., a formula that always has the value F) in L3 ; both formulas have the value N on interpretation VH. We’ll just consider the formula ( x)(¬Tx Tx). This formula is undetermined on every variable assignment v: no x-variant v of v can satisfy ¬Tx Tx, so ( x) (¬Tx Tx) isn’t satisfied by v; but ( x)(¬Tx Tx) is not dissatisfied by any v, either, because not every x-variant v of v dissatisfies ¬Tx ¬¬Tx (any assignment that assigns a member of fge(T) to x, for example, neither satisfies nor dissatisfies ¬Tx Tx). The formula ( x)(¬Tx Tx) therefore has the value N on interpretation VH.

But the fact that ( x)(¬Tx ¬¬Tx) isn’t a contradiction in L3 doesn’t address Black’s problem, because this formula was supposed to affirm the existence of borderline cases and consequently to assert that the predicate T is vague. The formula has the value N on VH, so it can’t assert the vagueness of the predicate T there, and more importantly, there is no interpretation on which the formula is true. (Extending the terminology of Chapter 5, the formula is quasi-contradictory: it can only have one of the values T, N). So it is impossible for this formula truly to assert the vagueness of the predicate T.

Does this mean that we have no way within L3 to assert the vagueness of predicates? Not at all. Recall that Bochvar’s external connectives are all definable in L3, and so they are also definable in L3 . In particular, external negation is definable here, where ¬BEP is satisfied by a variable assignment if P is either dissatisfied or undetermined and ¬BEP is dissatisfied otherwise. We can use the formula ( x) (¬BETx ¬BE¬Tx), in which the first two negations are Bochvar’s external negation and the third is Lukasiewicz’s negation, to assert that the predicate tall is vague. If we regard Lukasiewicz’s negation as expressing the English prefix un, and take short to mean the same as untall, then this formula asserts: At least one height is neither tall nor short. Since Tx and ¬Tx are both undetermined on any variable assignment that assigns a value in fge(T) to x, both ¬BETx and ¬BE¬Tx will be satisfied by these assignments and the formula ( x)(¬BETx ¬BE¬Tx) is therefore true on interpretation VH.

The Sorites paradox also has a solution in L3 . The argument

Ts1

Es2s1 Es3s2 Es4s3

. . .

Es193s192

( x) ( y) ((Tx Eyx) → Ty)

Ts193

is valid in L3 . The proof is exactly like the proof showing that the argument is valid in classical first-order logic; the key point is that if the premise( x)( y)((Tx Eyx) → Ty)

has the value T in L3 then Ty must be satisfied by every variable assignment that satisfies Tx Ey. But the validity of the Sorites argument is not enough to generate the paradox: the paradox depends on the argument’s premises’ actually being true, for it is only when the premises are in fact true that the conclusion must be true as well. So let’s determine the truth-values of the premises and conclusion of the argument on the following interpretation, which we will call VH*:

D: set of heights between 4 7 and 6 7 by 1/8 increments, inclusive ext(T) = {<h>: h D and h ≥ 5 11 }

cxt(T) = {<h>: h D and h ≤ 5 3 }

fge(T) = {<h>: h D and 5 3 < h < 5 11 }

. . .

I(s193) = 4 7

The formula Ts1 has the value T on this interpretation, since I(s1) is in ext(T). All of the premises Esi+1si in this argument have the value T, because each pair <I(si+1), I(si)> is a member of ext(E). The conclusion of the argument, Ts193, has the value F since I(s193) is in cxt(T). But the premise ( x) ( y) ((Tx Eyx) → Ty) is not true on VH*; it has the value N because it is undetermined by every variable assignment. Consider any variable assignment v. We will show that

i.some x-variants v of v satisfy ( y) ((Tx Eyx) → Ty), and

ii.some x-variants v of v neither satisfy nor dissatisfy ( y) ((Tx Eyx) → Ty), but

iii.no x-variants v of v dissatisfy ( y) ((Tx Eyx) → Ty).

For (i), consider the x-variant v such that v (x) = 6 7 (some other x-variants will work as well). v satisfies ( y) ((Tx Eyx) → Ty) because every y-variant v of v

satisfies (Tx Eyx) → Ty. Why? First consider the variant with v (y) = 6 67/ (and, of

8

course, v (x) is still 6 7 ). v satisfies Tx, Ey, and Ty—and so it satisfies the conditional (Tx Eyx) → Ty. Now consider any other y-variant v of v —these are variants that

assign values other than 6 67/ to y. Such a y-variant must dissatisfy Eyx since no

8

height in the domain other than 6 67/ is 1/ less than 6 7 , and so it will also

8 8

dissatisfy Tx Eyx and therefore satisfy (Tx Eyx) → Ty.

To show (ii), consider the x-variant v such that v (x) = 5 11 . The y-variant v

of v that assigns 5 107/ to y satisfies both Tx and Eyx but fails to satisfy or dissatisfy

8

Ty and so it fails to satisfy or dissatisfy (Tx Eyx) → Ty. Thus v doesn’t satisfy ( y) ((Tx Eyx) → Ty). All other y-variants of v satisfy (Tx Eyx) → Ty because they dissatisfy Eyx – so v doesn’t dissatisfy ( y) ((Tx Eyx) → Ty) either: v neither satisfies nor dissatisfies ( y) ((Tx Eyx) → Ty).

Finally, to show (iii), we note that for no value of x can we dissatisfy ( y) ((Tx Eyx) → Ty). To dissatisfy the formula there would have to be a value of y such that Tx and Eyx are satisfied but Ty is dissatisfied. However, there is no pair of heights h1 and h2 such that h1 is in the extension of T, h2 is 1/8 less than h1, and h2 is in the counterextension of T.

On interpretation VH*, therefore, the Principle of Charity premise is not true. Despite the Sorites’ semantic validity inL3 weare not forced to accept its conclusion on this interpretation. Moreover, in denying the truth of the Principle of Charity we have not run afoul of an ancillary problem that we noted in Chapter 1. There we said that when we deny the Principle of Charity we must accept its negation, which says that there are two heights, 1/8 apart, such that it is true that one is tall and false that the other is—a claim that is certainly not true! We may now add that this problem arises only within the framework of classical logic, where denying the truth of a claim commits us to its falsity and hence to the truth of its negation. But when the Principle of Charity has the value N in L3 , the principle’s negation will also have the value N. So this solution to the Sorites paradox in L3 doesn’t endorse the ludicrous claim that 1/8 can take us from a height that is tall to a height of which it is false that it is tall.2

Is VH* a reasonable interpretation for the Sorites argument? Well, yes. Even if we haven’t got the cutoff points for the extension and counterextension of tall exactly right, the conditions that diffuse the paradox here will undoubtedly be met by any reasonable interpretation of tall. The paradox is dissolved as long as the premise ( x)( y)((Tx Eyx) → Ty) can turn out to be neither true nor false, and this will be the case as long as there is at least one pair of heights x and y such that (Tx Eyx) → Ty is neither satisfied nor dissatisfied but no pair of heights x and y for which the formula is dissatisfied. As long as we have a nonempty fringe along with a nonempty extension (or a nonempty counterextension) for the predicate T there will be a pair of heights x and y such that (Tx Eyx) → Ty is neither satisfied nor dissatisfied— we naturally assume that the interpretation of E remains the same—and that seems right for vague predicates. And as long as there are no two consecutive heights such that one is in the extension of the predicate T while the other is in its counterextension, a situation also guaranteed by a nonempty fringe, the formula (Tx Eyx) → Ty can’t be dissatisfied by any values of x and y.

7.2 Quantifiers Based on the Other Three-Valued Systems

Because conjunction and disjunction are defined in KS3 in exactly the same way as in L3, the quantifier clauses for the two systems are the same. The only difference

a height of which it is neither true nor false that it is tall. We will return to this very legitimate concern in Chapter 10.

The universally quantified formula ( BIx)Tx has the value N on interpretation VH. This is because at least one value that can be assigned to x will cause Tx to be neither satisfied nor dissatisfied, and that is sufficient for concluding that the universal quantification has neither the value T nor the value F in BI3 . This contrasts with L3 (and KS3 ), where ( x)Tx has the value F because there is at least one value that can be assigned to x that dissatisfies Tx. For a similar reason the existentially quantified formula ( BIx)Tx has the value N on interpretation VH in BI3 , whereas in the other two systems the existential quantification ( x)Tx has the value T. The Bochvarian internal Law of Excluded Middle ( BIx)(Tx BI ¬BITx) also has the value N. In fact, any (closed) quantified formula that contains Tx will have the value N in BI3 on interpretation VH since at least one value of x fails to satisfy or dissatisfy Tx and this failure is contagious. So the Principle of Charity premise of the Sorites argument, along with Black’s fringe formula, also have the value N in BI3 on interpretations VH* and VH (while the Sorites argument remains valid).

In Bochvar’s external system a conjunction is true if both conjuncts are true and is false otherwise, and a disjunction is true if at least one conjunct is true and is false otherwise. So a universally quantified formula should be true if every variable assignment satisfies the formula following the quantifier and should be false otherwise, while an existentially quantified formula should be true if there is at least one variable assignment that satisfies the formula following the quantifier and should be false otherwise. The satisfaction clause for the universal quantifier in BE3 is thus

7.A formula ( BEx)P is

satisfied by a variable assignment v on I if P is satisfied by every x-variant of v on I,

dissatisfied otherwise.

The satisfaction clause for the BE3 existential quantifier is left as an exercise. Clauses for formulas formed with the propositional connectives in BE3 will reflect their truth-conditions in BE3. For example, the clause for negated formulas is

2.A formula ¬BEP is

satisfied by a variable assignment v on I if P is not satisfied by v on I, and

dissatisfied otherwise.

The formula ( BEx)Tx has the value F on interpretation VH since any variable assignment that assigns to x a height less than 5 11 will fail to satisfy Tx. The Law of Excluded Middle formula ( BEx)(Tx BE ¬BETx) has the value T on VH, since every variable assignment will satisfy Tx BE ¬BETx. Indeed, the Law of Excluded Middle is true on every interpretation in BE3 , since ¬BETx must be satisfied by any variable assignment that fails to satisfy Tx. Black’s fringe formula ( BEx) (¬BETx BE ¬BE¬BETx) has the value F in BE3 on interpretation VH and indeed on every interpretation because the external negation clause guarantees that it is