LING 501, Fall 2004: Laws of logic and the definitions of the connectives

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1 LING 501, Fall 2004: Laws of logic and the definitions of the connectives Modified October 24, adding a test yourself section at the end. Modified October 21, correcting errors noted in class on October 20. Modified October 18, adding a new introduction, and making various other changes. Modified October 17, changing title, adding material on logical laws and their relation to the definition of connectives. Laws of logic Logic is the theory of entailment over a set of objects, for example the members of the set PL of well-formed formulas of the propositional logic PL. The entailment relation can be defined in terms of the form of those objects, i.e. syntactically, or in terms of their meaning, i.e. semantically. Syntactic entailment for propositional logic is generally represented in the form (A), called a syntactic sequent, where P = {p 1,, p n } PL is a set of premises, - (called the syntactic turnstile) is the syntactic entailment relation, and q is the conclusion. Semantic entailment is represented in the form (B), called a semantic sequent, where each p i P is an equivalence class of members of PL, = (called the semantic turnstile) is the semantic entailment relation, and q is an equivalence class of members of PL. Since the syntactic and semantic forms of entailment for PL are equivalent, I will use the = symbol for both relations. 1 (A) (B) P - q P = q Certain correct sequent schemata have achieved a special status as laws of logic, including the following. 2 Simplification For all p, q S: [p & q] = p and [p & q] = q Addition For all p, q S: p = [p q] Generalization For all p, q, r S: If p = r and q = r, then [p q] = r Modus ponens For all p, q S: p, [p q] = q Contradiction For all p, q S: p, ~p = Each of these laws is expressed using a particular truth functor: Simplification uses conjunction, Addition and Generalization disjunction, Modus ponens, conditionality, and Contradiction negation. These laws can help identify the truth functors in a natural language. For example, we 1 If the conclusion is omitted, the resulting sequent P = is equivalent to for any q: P = q. Similarly if the premises are omitted, the resulting sequent = q is equivalent to = q. 2 I will generally omit the brackets around the premises.

2 identify and as [p & q] in the sentence Ann was sick yesterday and Bea stayed home yesterday because of the correctness of the sequents Ann was sick yesterday and Bea stayed home yesterday = Ann was sick yesterday, and Ann was sick yesterday and Bea stayed home yesterday = Bea stayed home yesterday. Similarly, we identify if then as [p q] in the sentence If Ann was sick yesterday, then Bea stayed home yesterday, because of the correctness of the sequent Ann was sick yesterday, If Ann was sick yesterday, then Bea stayed home yesterday = Bea stayed home yesterday. Note that Addition allows any proposition at all, including a seemingly irrelevant one, to be disjoined to another in a correct sequent, for example Ann was sick yesterday = Ann was sick yesterday or Bea s cat eats alfalfa sprouts. Similarly, Contradiction allows any proposition at all, including irrelevant and false ones, to follow from a proposition together with its negation, for example Ann was sick yesterday, Ann was not sick yesterday = Mahatma Gandhi was Jawaharlal Nehru. Entailment structure as a model of consistency of belief Let S be a set of strings in PL, and = an entailment relation satisfying the axioms of entailment. A valuation V of S is an assignment of truth or falsity to each atomic proposition in PL, and assignment of truth or falsity to complex propositions in accordance with the structure of those propositions as in the Semantics of propositional logic handout. Each such valuation partitions S into two subsets: Φ, which contains all and only all the false propositions, and Θ, which contains all and only all the true ones. Consistency of belief requires for all valuations, that Θ be closed under entailment, namely that for all sequents p 1, p n = q, if p 1, p n Θ, then q Θ. 3 The valuation of complex propositions in PL can be determined recursively from bottom up, using the equivalences in 1 through 4 below, provided also that Φ is closed under = -1, the dual of =. 4 Just as = is truth preserving, so = -1 is falsity preserving; that is for all sequents p 1, p n = -1 q, if p 1, p n Φ, then q Φ. As the notation suggests, = -1 is the inverse of =: if p = -1 q, then q = p; and if p 1, p n = -1 q, then for every r S, if p 1 = r, and p n = r, then q = r. 1. ~ p Θ if and only if p Φ. 2. [p & q] Θ if and only if p Θ and q Θ. 3. [p q] Φ if and only if p Φ and q Φ. 4. [p q] Φ if and only if p Θ and q Φ. Axiomatization of entailment In this section, I state certain axioms for the entailment relation that do not depend on the specific syntax (e.g. conditions governing the occurrence of specific symbols) or semantics (e.g. truth preservation) of the domain of entailment. Using this method, I define the truth functors not by their truth tables, but by their role in entailment, and also rename them logical operators. I then determine which of their truth theoretic properties can be proved. 2 3 Saying that Θ is closed under entailment is the same as saying that entailment is truth preserving. 4 Demonstration of how the truth tables can be constructed from the requirements of truth and falsity preservation will not be covered in this course. See me if you're interested.

3 3 An entailment structure <S, => consists of a set S and an entailment relation = over S which satisfies the following axiom schemata. Projection For all p 1,, p n, S, and for any k = 1,, n: p 1, p n = p k Permutation For all p 1,, p n, q (n > 1): if p 1, p n = q, then for any permutation π of <1,, n>: p π (1), p π(n) = q Simplification For all p 1,, p n, q S (n > 0), if for some i = 1,, n: p 1, p i, p i, p n = q, then p 1, p i, p n = q Dilution (or Thinning, or Monotonicity) For all p 1,, p n, p n+1, q S (n 0): if p 1, p n = q, then p 1, p n, p n+1 = q Cut For all p 1,, p n, q, s 1,, s m, r S (m, n 0): if p 1, p n = q, and q, s 1, s m = r, then p 1, p n, s 1, s m = r From Projection, we see that = is reflexive, by setting n = 1. From Cut, we see that = is transitive, by setting n = 1 and s = 0. The antisymmetry of = follows if we consider every logically equivalent expression to be the same member of S. The Permutation and Simplification axioms are consequences of considering the premises of an entailment relation to be a subset of S, and so can be omitted. If the Dilution axiom is dropped, other versions of entailment can be defined, including defeasible reasoning. Using the laws of logic to define the logical operators Suppose we understand how a relation with the properties of entailment holds over a set S, but its syntax does not wear the logical operators on its sleeve, so to speak. Then we can use the relation to determine which members of S are the negations of others, which the conjunctions, which the disjunctions and so forth. The method was discovered by Gerhard Gentzen about 70 years ago, and was worked out in full detail by Arnold Koslow, A Structuralist Theory of Logic, Cambridge University Press, Each definition makes use of a logical law involving the connective. Negation 1. for all s, t S: s, ~s = t (law of contradiction: "de contradictione, quodlibet") 2. ~s is the weakest member of S to satisfy 1. That is, for all K such that for all s, t S, if s, K = t, then K = ~s. Conjunction 1. For all s, t S: s & t = s; (law of simplification) 2. For all s, t S: s & t = t; ( )

4 3. s & t is the weakest member of S to satisfy 1 and 2. That is, for all K such that for all s, t S, if K = s, and K = t, then K = s & t. Disjunction 1. For all s, t, u S, if s = u and t = u, then s t = u; (law of generalization) 2. s t is the weakest member of S to satisfy 1. That is, for all K such that if for all s, t, u S, if s = u and t = u then K = u, then K = s t. Conditional 1. For all s, t S, s, s t = t; (modus ponens) 2. s t is the weakest member of S to satisfy 1. That is, for all K such that for all s, t S, if s, K = t, then K = s t. The superset relation is an entailment relation The definition of the logical operators in terms of logical laws allows us to identify them with respect to other entailment relations. For example, let S be a set, and S the powerset of S. Then = S is an entailment relation defined as follows. For every set Γ S and s S, Γ = S s if and only if s is a superset of the intersection of the members of Γ. For example, let S 3 = {a, b, c}. Then S3 = {{a, b, c}, {a, b}, {a, c}, {b, c}, {a}, {b}, {c}, }. Using the general definitions of the connectives given above, we establish the following specific definitions, where s, t S3. Negation: ~ s is the complement of s with respect to S 3. For example, ~ {a, b} = c; ~ {a} = {b, c}; ~ {a, b, c} = ; ~ = {a, b, c}. Conjunction: s & t is the intersection of s with t. For example, {a, b} & {a, c} = {a}; {b} & {c} = ; {a, b, c} & {a, c} = {a, c}. Disjunction: s t is the union of s with t. For example, {a, b} {a, c} = {a, b, c}; {b} {c} = {b, c}; {a, b, c} {a, c} = {a, b, c}. Conditional: s t is the union of t with the complement of s with respect to S 3. For example, {a, b} {a, c} = {a, c}; {b} {c} = {a, c}; {a, b, c} {a, c} = {a, c}. The Hasse diagram for = S 3 is shown in Figure 1. 4

5 5 {a, b, c} {a, b} {a, c} {b, c} {a} {b} {c} Figure 1 The subset relation is also an entailment relation Let S be a set, and S the powerset of S. Then = S is another entailment relation defined as follows. For every set Γ S and s S, Γ = S s if and only if s is a subset of the union of the members of Γ. For example, let S 3 = {a, b, c} as in the preceding section. Then S3 = {{a, b, c}, {a, b}, {a, c}, {b, c}, {a}, {b}, {c}, } as in the preceding section. Using the general definitions of the connectives, we establish the following specific definitions, where s, t S3. Negation: ~ s is the complement of s with respect to S 3. For example, ~ {a, b} = c; ~ {a} = {b, c}; ~ {a, b, c} = ; ~ = {a, b, c}. Conjunction: s & t is the union of s with t. For example, {a, b} & {a, c} = {a, b, c}; {b} & {c} = {b, c}; {a, b, c} & {a, c} = {a, b, c}. Disjunction: s t is the intersection of s with t. For example, {a, b} {a, c} = {a}; {b} {c} = ; {a, b, c} {a, c} = {a, c}. Conditional: s t is the intersection of t with the complement of s with respect to S 3. For example, {a, b} {a, c} = {c}; {b} {c} = {c}; {a, b, c} {a, c} =. The entailment relation = S 3 is the dual of = S 3. In the dual of an entailment relation such as = S 3, conjunction and disjunction change places, but negation is unaffected. The Hasse diagram for = S 3 is shown in Figure 2.

6 6 {a} {b} {c} {a, b} {a, c} {b, c} {a, b, c} Figure 2 Classical entailment relations Any entailment relation on a set S for which the laws of double negation and excluded middle hold, and for which the connectives are everywhere defined, is a classical entailment relation. For example, logical consequence in propositional logic is a classical entailment relation, as are = S 3 and = S 3. Double negation For all p S: ~~p p Excluded middle For all p S: = [p ~ p] A nonclassical entailment relation Let S 3 be the set of three atomic individuals {a, b, c}, and let a+b, a+c, b+c, and a+b+c be mereological sums of those individuals. Form the set S 7 = {a, b, c, a+b, a+c, b+c, a+b+c} of all these individuals. Finally, let = S 7 be the mereological whole part entailment relation defined as follows. 5 For every set Γ S 7 and i S 7, Γ = S 7 i if and only if i is part of the mereological sum of the members of Γ. For this entailment relation, every member i of S 7 except a+b+c has a negation, namely that individual j which is disjoint from i. However, a+b+c has no negation, since there is no individual in S 7 which is disjoint from a+b+c. Also observe that a+b+c is also not the negation of any individual in S 7. That is, negation is a function from S 6 to S 6, where S 6 = S 7 {a+b+c}, namely {a, b, c, a+b, a+c, b+c}, but not from S 7 to S 7. 5 This consequence relation is the basis of the calculus of individuals originally defined in Henry Leonard and Nelson Goodman, The calculus of individuals and its uses. Journal of Symbolic Logic 5: 45-55, 1940.

7 7 Similarly, disjunction is the mereological product (overlap) of those individuals which have a part in common, for example a+b a+c = (a+b) (a+c) = a, but is undefined for those which do not, for example a+b and c. On the other hand conjunction is defined for every pair of individuals, namely their mereological sum. For example a&c = a+c, and a+b&b+c = a+b+c. Note that English and expresses mereological sum, e.g. Veronica and Archie denotes the mereological sum of Veronica and Archie. Entailment relations such as = S 7 are nonclassical, since certain classical laws such as the law of Double negation and the law of the Excluded middle fail. For example, ~ ~ a+b+c is undefined, not a+b+c, as required by the law of Double negation. Similarly, the disjunction of a with the negation of a = a b+c, which is undefined, not a tautology. There are no tautologies in = S 7. The Hasse diagram for = S 7 is given in Figure 3. Notice that it has a bottom but no top. a b c a+b a+c b+c a+b+c Figure 3

8 8 Test yourself section. Click here for answers. Let L = a*b*, and let L n be the largest sublanguage of L containing strings of length n or less. Let = s be the substring entailment relation defined as follows. s = s t if and only if s is a substring of t; s 1,... s k = s t if and only if the maximal substring that s 1,... s k have in common is a substring of t. 1. Draw the Hasse diagrams using = s for the following languages: a. L 1 b. L 2 c. L 3 2. What is the conjunction of a with b in each of the systems in 1? 3. In which of the systems in 1 is the conjunction of ab with bb defined? What is it for each system in which it is defined? 4. Is the conjunction of every pair of members of every L n defined? Why or why not? 5. In which of the systems in 1 is the disjunction of a with b defined? What is it for each system in which it is defined? 6. Is the disjunction of every pair of members of L defined? Why or why not? 7. What is the negation of a in each of the systems in 1? 8. In L 2 and L 3 the double negation of a is distinct from a. What is it in each case? Let M = (a b)* and let and let M n be the largest sublanguage of M containing strings of length n or less. Let = s be the substring entailment relation as defined above. 9. Draw the Hasse diagrams using = s for the following languages: a. M 1 b. M What is the conjunction of a with b in each of the systems in 9? 11. What is the conjunction of ab with bb in M 2? 12. Why is the conjunction of ab with ba in M 2 not defined? 13. Why is the disjunction of a with b in M 2 not defined? 14. What is the negation of a in each of the systems in 9?