A Mathematical Toolkit


 Candice Flowers
 1 years ago
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1 A Mathematical Toolkit Philosophy Naive Set Theory Cantor, in his 1895 work Contributions to the Founding of the Theory of Transfinite Numbers, gives the following definition of a set: By a set [Menge] we are to understand any collection into a whole M of definite and separate objects m of our intuition or our thought. These objects are called the elements of M. This definition wavers between the circular and the confused, but it s about the best that can be done for the notion of set. So I won t try to do better, and I ll just take set to be a primitive, along with the relation of set membership. Sets can be specified either extensionally or intensionally. An extensional specification gives the members of a set via an explicit list: {1, 3, 7} An intensional specification gives the members of a set via a property, with the set then being the set of all and only objects having that property: {x : x is prime} Extensional specifications can be reduced to intensional ones by using disjunctive properties of the form x = 1 x = 3 x = 7. NOTE: In a more formal development, we would make precise the range of properties, by allowing property specifications of the form of arbitrary sentences in a single free variable from a firstorder language whose only nonlogical symbol is the binary predicate. For our purposes, we can take set theory to be governed by the following two basic principles: 1. Axiom of Extensionality: If sets x and y have exactly the same members, then they are the same set. x y( z(z x z y) x = y) 1
2 2. Axiom of (Unrestricted) Comprehension: Given any property P, there is a set of all and only things having property P. x(x {y : Φ(y)} Φ(x)) In fact the axiom of comprehension is inconsistent (consider the property of nonselfmembership (x x)), and a more sophisticated set theory such as ZFC will have a more complex set of axioms. We will just avoid the inconsistencies and remain in naive set theory, though. We can now define a number of auxiliary notions: 1. Subset: x y if every member of x is a member of y. x y(x y z(z x z y)) 2. Empty set: is the set with no members. = {x : x x} 3. Union: x y is the set containing all members of x and all members of y. x y(x y = {z : z x z y}) 4. Intersection: x y is the set containing all elements that are in both of x and y. x y(x y = {z : z x z y}) 5. Set subtraction: x y is the set of everything in x but not in y. x y(x y = {z : z x z y}) 6. Indexed union: Given an index set I and a collection of sets X i indexed by I, i I X i is the set of everything that is a member of any X i. i I X i = {z : i I z X i } 7. Indexed intersection: Given an index set I and a collection of sets X i indexed by I, i I X i is the set of everything that is a member of every X i. i I X i = {z : i I z X i } 8. Power set: (x) is the set of all subsets of x. x( (x) = {y : y x}) 2
3 2 Relations and Functions Because of the axiom of extensionality, the order of the elements of a set is irrelevant to the identity of the set. Thus {1, 3} = {3, 1}. However, for some purposes we want order to matter, so we want the notion of an ordered pair. As it happens, we can construct ordered pairs out of sets, by defining: < x, y >= {x, {x, y}} From the set {x, {x, y}} we can recover information about the order of the elements, so this will serve as an ordered pair. However, in practice we will never care about the method of defining ordered pairs in set theory, so this definition will never be mentioned again. Ordered pairs can then be used to represent binary relations. A relation such as loves can be represented as a collection of ordered pairs, each of which represents the fact that the first member of the ordered pair loves the second member of the ordered pair. More generally, given any two sets X and Y, we define the cartesian product of X and Y X Y as the set of all ordered pairs whose first element comes from X and whose second element comes from Y. X Y = {< x, y >: x X y Y} A binary relation from X to Y can then be thought of as a subset of X Y. When a binary relation is simply relating one group to itself, we can speak of a binary relation on X, which is a subset of X X. One kind of relation of special interest is that of functions. Functions are binary relations which never relate one object (in the first position of an ordered pair) to more than one object (in the second position of an ordered pair). Thus if a function contains the ordered pair < 1, 3 >, it cannot also contain the ordered pair < 1, 2 >, although it can contain the ordered pair < 2, 3 >. Formally: f is a function if and only if x y z((< x, y > f < x, z > f ) y = z) The domain of a function is the set of all objects appearing in the first position of ordered pairs in the function; the range of the function is the set of all objects appearing in the second position of ordered pairs in the function. We can say that f is a function from A to B f : A B if the domain of f is A and the range of f is a subset of B. If B is identical to the range of f, then we say that f is surjective, or onto. If f never maps two elements of the domain to the same element of the range, then we say that f is injective, or into. If f is both surjective and injective, then we say that it is bijective, or a onetoone mapping from A to B. 3
4 3 Recursively Generated Sets Suppose that we start with the set X 0 = {3, 7}, and we have the function f (x) = x + 3. If we expand X 0 by applying f to all of its members, we get the set X 1 = {3, 6, 7, 10}. If we expand again by applying f to all members of X 1, we get the set X 2 = {3, 6, 7, 9, 10, 13}. Suppose we perform this expansion infinitely many times. The result will be the set X{3n + 3, 3n + 7 : n N}. This is an example of a recursively generated set. More generally, a recursively generated set is a set that results from combining some base set X 0 and some collection f 1,..., f n of generating functions. We can think of the recursively generated set in two ways: 1. The generating functions can be used to produce a sequence of expanding sets: X 1 = X 0 {y : x X 0 i f i (x) = y} X 2 = X 1 {y : x X 1 i f i (x) = y}... X n+1 = X n {y : x X n i f i (x) = y} We then define the recursively generated set X to be the union of this sequence: X = i N X i 2. We can say that a set Z is closed under a function f if the application of f to members of Z produces nothing that was not already in Z: Z is closed under f if and only if {y : x Z y = f (x)} Z Given a base set X 0 and a collection of functions f i, there will then be many different sets containing X 0 and closed under all of the f i. The recursively generated set (on X 0 and the f i s) is then the smallest such set. We can obtain that smallest set by intersecting all of the closed sets (more precisely, we take one set containing X 0 and closed under the f i s, and take its power set. We then take every set in the power set containing X 0 and closed under the f i s, and form the indexed intersection of that collection. The result is the recursively generated set.) For our purposes, there will be two particularly important types of recursively generated sets: 1. Formal languages form recursively generated sets. For example, in a propositional language, we can take the base set to be the set of sentence letters, and then take the generating functions to be the syntactic operations of combining using sentential connectives. The recursively generated set is then the set of all sentences in the language. 4
5 2. Proof systems form recursively generated set. In an axiomatic system, we can take the base set to be the set of axioms, and take the generating functions to be the extension of proofs via application of a proof rule. The recursively generated set is then the set of all proofs in the proof system. 4 Orderings An ordering relation, such as the relations < and on the natural numbers (or on the integers, or on the real numbers), or such as the relations taller than and older than on people, is a special case of the general category of binary relations. Syntactically, a binary relation is a symbol which combines with two terms to form a complete sentence. Thus taller than combines with the terms Aristotle and Socrates to form the sentence Aristotle is taller than Socrates. Semantically, a binary relation can be thought of as a collection of ordered pairs, where each ordered pair represents two objects between which the relation holds (and the direction in which it holds between them). Thus the relation taller than would contain ordered pairs such as <Yao Ming, George Bush> and <George Bush, Danny DeVito> (but not <George Bush, Yao Mind> or <Danny DeVito, George Bush>). We can now identify some interesting properties of binary relations: 1. A relation is reflexive if every object bears the relation to itself. The identity relation is thus reflexive, as is the relation at least as tall as. 2. A relation is irreflexive if no object bears the relation to itself. The relation > on the natural numbers is thus irreflexive, as are the relation taller than and not identical to. Notice that most relations are neither reflexive nor irreflexive. Given a set X, if I just take a random assortment of ordered pairs drawn from X (that is, ordered pairs both of whose members are members of X), I ll probably get some ordered pairs of the form < a, a >, but not all ordered pairs of this form. A relation such as admires, which doesn t have any particular internal logical constraint on its structure, will typically be neither reflexive nor irreflexive. (Some people admire themselves, but not all do.) 3. A relation is symmetric if, whenever a bears the relation to b, b bears the relation to a. Thus R is symmetric if, whenever < a, b > R, < b, a > R. Is a sibling of is a symmetric relation, as is is distance d from. 4. A relation is asymmetric if, whenever a bears the relation to b, b does not bear the relation to a. Thus R is asymmetric if, whenever < a, b > R, 5
6 < b, a > R. Is a parent of is an asymmetric relation, as is the relation < on the natural numbers. 5. Antisymmetry is a variation on asymmetry that is useful in some cases. A relation is antisymmetric if, when a bears the relation to b, b does not also bear the relation to a unless a and b are identical. The difference between asymmetric and antisymmetric relations is the difference between, for example, < and. < is asymmetric we can never have both a < b and b < a., on the other hand, is antisymmetric. We can have both a b and b a, but only in the special case of a = b. As with reflexivity and irreflexivity, most relations are not symmetric, asymmetric, or antisymmetric. A random assortment of ordered pairs will include some mirror images, but not all. Admires, for example, is not symmetric, asymmetric, or antisymmetric. There must be some special logical feature to a relation to guarantee the internal structure of symmetry, asymmetry, or antisymmetry. 6. A relation is transitive if, whenever a bears the relation to b and b bears the relation to c, a bears the relation to c. Thus R is transitive if, whenever < a, b >, < b, c > R, also < a, c > R. Identity is a transitive relation, as is < on the natural numbers and taller than and is an ancestor of on people. 7. A relation is intransitive if, whenever a bears the relation to b and b bears the relation to c, a does not bear the relation to c. Thus R is transitive if, whenever < a, b >, < b, c > R, < a, c > R. The successor relation on the natural numbers is irreflexive, as is the relation is a parent of on people. Again, most relations are neither transitive nor intransitive. Absent some special logical feature of the relation, neither transitivity nor intransitivity will apply. Notice that the three features reflexivity, symmetry, and transitivity all share the following feature: If R is not reflexive/symmetric/transitive, then it can be expanded so that it is. That is, there is some relation R such that R is reflexive/symmetric/transitive and R R. We can thus talk about the reflexive closure, symmetric closure, or transitive closure of R: The reflexive closure of R is the smallest relation R such that R is reflexive and R R. The symmetric closure of R is the smallest relation R such that R is symmetric and R R. 6
7 The transitive closure of R is the smallest relation R such that R is transitive and R R. (Are these definitions wellformed? Is there a unique smallest reflexive relation R containing R, for example? Yes take the intersection of all reflexive relations containing R. (If you are particularly fastidious, when R is a relation on A, take the intersection of all reflexive subsets of the power set of A A containing R.)) An ordering relation is any relation R which is either: or: Irreflexive, asymmetric, and transitive Reflexive, antisymmetric, and transitive. The choice is between a strict ordering like < and a loose ordering like. The former is irreflexive and asymmetric; the latter is reflexive and antisymmetric. Given a strict ordering, we can always create a loose ordering by adding in the identity relation. Given a loose ordering, we can always create a strict ordering by removing the identity relation. The difference between the two thus tends to be rather superficial. An ordering relation in this sense needn t be linear it is a partial order. The subset relation, for example, is a partial order, and generates orders such as: {1, 2, 3, 4} {1, 2, 3} {1, 2, 4} {1, 3, 4} {2, 3, 4} {1, 2} {1, 3} {1, 4} {2, 3} {2, 4} {3, 4} {1} {2} {3} {4} If we want a linear ordering, we add one further condition to the relation: 8. A relation is connected if, given any two objects a and b, either a bears the relation to b or b bears the relation to a (or a and b are identical). Thus either < a, b > R, or < b, a > R (or < a, b >=< a, a >=< b, b >). Suppose we have some set X on which we have some sort of successor relation some way, given a member x of X, to find the next thing in X. For example, 7
8 our set might be the natural numbers, with a successor relation of adding 1. Or our set might be the even natural numbers, with a successor relation of adding two. Or our set might be the collection of subsets of {1, 2, 3, 4}, with a successor relation of adding a single element. This last example shows that a member x of X needn t have a unique successor {1, 2}, {1, 3}, and {1, 4} are all successors of {1} in this sense. The paradigm case of a set on which we have a successor relation is a recursively generated set. If X is a recursively generated set formed from a base of X via generating functions f 1,..., f n, then each generating function creates successors. The successors of x will be f 1 (x),..., f n (x). Say that x < s y if y is a successor of x. Then we can, when circumstances are correct, use the successor relation to form an ordering of X. Let < be the transitive closure of < s. Then < will be an ordering relation so long as irreflexivity is preserved. When will irreflexivity be preserved? When repeated application of the successor relation can never bring one back to one s starting point. In the natural numbers, for example, no number of applications of adding one to 3 can bring one back to 3. Suppose, on the other hand, that sentence φ is a successor of sentence ψ if φ can be obtained by combining ψ with some axiom(s) via modus ponens. Then we can get back to our starting point. Thus consider the following line of deduction (this gets a bit messy; skip over it if you just want to take my word for it): 1. p q 2. (p q) (p (p q)) 3. p (p q) 4. (p (p q)) ((p p) (p q)) 5. (p p) (p q) 6. (p ((q p) p)) ((p (q p)) (p p)) 7. p ((q p) p) 8. (p (q p)) (p p) 9. p (q p) 10. p p 11. p q 8
9 This derivation shows that we can travel from p q to p (p q) via repeated application of the successor relation, and then from p (p q) to p q by more applications of the successor relation. Thus the successor relation can take us from p q via a long journey back to p q. Taking the transitive closure will thus yield a relation that is not irreflexive. OK, so when will we never get back to where we started? The following condition is stronger than necessary (that is, it s a sufficient condition, but not necessary), but it will do for our purposes. A recursively generated set X is freely generated just in case all of the generating functions are injective and no two generating functions for X can ever produce the same outputs. That is, there are no f i and f j and x, y X such that f i (x) = f j (y). The toy language from Handout I is freely generated. Applying the rule flank with q can never produce the same output as applying the rule prefix with rr. If a set X is freely generated, then the transitive closure of < s on it is an ordering (a partial ordering, if there is more than one generating function). This gives us a quick way to generate orderings out of freely generated recursively defined sets. The paradigm case of recursively defined sets, for our purposes, will be the formal language. In all the cases we will consider, the syntax of the formal language will be freely generated. Then, roughly speaking, we will have φ < ψ just in case φ is a syntactic constituent of ψ. 5 Proof By Induction Proof by induction is, in its simplest form, a technique for showing that all natural numbers have a certain property. After discussing the application of induction to the natural numbers, we will consider its extension to certain other collections. One such collection that will be of particular interest to us is the set of sentences in a formal language such as PL. Suppose you have a line of dominoes, and you want to be sure that they are all knocked down. It suffices to know two things: 1. The first domino is knocked down. 2. The dominoes are spaced such that, if any domino falls, it knocks down the next domino in line. Given these two conditions, we can immediately see that the second domino is knocked down, since the first domino is knocked down (by the first condition), and when it falls it knocks down the second domino (by the second condition). And we can then see that the third domino is knocked down, since the second 9
10 condition guarantees that the second domino knocks it down when it falls. It then follows that the fourth domino is knocked down, and so on. Even if there are an infinite number of dominoes, we have enough information to see that they all fall. Of course, it would take infinitely many steps of reasoning, along the above lines, to see that they all fall proof by induction is then, in effect, a technique for compressing those infinitely many steps into just one. Suppose we know two things about a certain property P: 1. The natural number 0 has property P. 2. Given any natural number n, if n has property P, then n + 1 also has property P. Then proof by induction allows us to conclude that every natural number has property P. Carrying out a proof by induction thus requires that we establish these two features of property P. The claim that 0 has P is called the base clause of the proof by induction. Establishing the base clause typically (but not always) is a matter of simple calculation. The claim that if n has property P, then n + 1 also has property P is called the inductive clause of the proof by induction. Establishing the inductive clause typically (but not always) proceeds by way of conditional proof: we assume that n has property P, and then use this assumption to conclude that n + 1 has property P. We then conditionalize to conclude without assumption that if n has P, then n + 1 also has P. Here are some examples of proofs by induction: 1. Claim: 3 2n n+2 is divisible by 7 for all values of n in the natural numbers. Proof: As with all proofs by induction, the proof has two parts: establishing the base clause and establishing the inductive clause. Base Clause: When n = 0, 3 2n n+2 = = = 7, and 7 is divisible by 7 as desired. Inductive Clause: Suppose that 3 2k k+2 is divisible by 7. We must now show that the claim holds for k+1 as well. Thus we are interested in 3 2(k+1)+1 +2 (k+1)+2. We have 3 2(k+1)+1 +2 (k+1)+2 = 3 2k+3 +2 k+3. We then factor out 3 2 from the first term and 2 1 from the second term to yield 9 3 2k k+2. We factor out in this way because we are trying to isolate the k case within the k + 1 case so that we can use the inductive hypothesis. We now rewrite this as 7 3 2k k k+2. Factoring out the 2, we have 7 3 2k+1 + 2(3 2k k+2 ). By the inductive hypothesis, 3 2k k+2 is divisible by 7, so twice it is as well. And, of course, 10
11 7 3 2k+1 is also divisible by 7. The sum of two numbers divisible by 7 is again divisible by 7, so 3 2(k+1) (k+1)+2 is divisible by 7. This establishes the inductive clause. Given the base clause and the inductive clause, the claim follows by induction. 2. Claim: Σ n i=0 2i = 2 n+1 1. Proof: By induction: Base Clause: When n = 0, the claim is that 2 0 = = = 2 1 = 1. Inductive Clause: Suppose that Σ k i=0 2i = 2 k+1 1. We then want to show that Σ k+1 i=0 2i = 2 k = 2 k+2 1. We first note that Σ k+1 i=0 2i = (Σ k i=0 2i ) + 2 k+1, by stripping off the last member of the summation. We remove the last member so that we can isolate the k case within the k + 1 case and use the inductive hypothesis. By that hypothesis, we can replace Σ k i=0 2i with 2 k+1 1, yielding Σ k+1 i=0 2i = 2 k k+1 = 2 2 k+1 1 = 2 k+2 1, as desired. This establishes the inductive clause. Given the base clause and the inductive clause, the claim follows by induction. 3. Claim: Given a chess board of side length 2 n squares for any natural number n, with one square removed from the area of the board, the board can be completely covered with Lshaped threesquare tiles without any overlapping tiles or any tiles extending outside the boundary of the board. Proof: By induction. Base Clause: When n = 0, the side of the board is of length 1. The board thus consists of a single square. When a square is removed, a null board results. Trivially, such a board can be successfully Ltiled. bf Inductive Clause: Suppose that a board of side 2 k, with a square removed, can be Ltiled. Now consider a board of side 2 k+1. This board can be split into four equal pieces, each a board of side 2 k. One of these four pieces will have the removed square, and then by the inductive hypothesis can be Ltiled. Now consider the other three boards. Remove the three squares where these three boards meet in the center of the 2 k+1 board. These three squares form the shape of a single Ltile. The result of this removal is three boards, each of side 2 k, each with a single square removed. By the inductive hypothesis, each can then be Ltiled. Since each of the described parts of the 2 k+1 board can thus be Ltiled, the entire board can also be. This establishes the inductive clause. 11
12 Given the base clause and the inductive clause, the claim follows by induction. Next we ll consider two small generalizations of proof by induction. First, the base case can be shifted. In the above examples, we were proving that every natural number had a certain property. But we could just as easily prove, for example, that every natural number greater than 11 has a certain property. Thus consider the claim n! > 5 n. For n = 0 through n = 11, this claim is false. But for n = 12, it is true, as a quick calculation will show. Given its truth at n = 12, we can proceed to prove the inductive clause: Suppose k! > 5 k, for some k > 11. Then (k + 1)! = (k + 1) k! > 5 k! > 5 5 k = 5 k+1. This establishes the inductive clause. In this manner, we can move the base clause arbitrarily (to numbers either larger or smaller than 0). We just have to compute the base clause at the appropriate point, and then be sure, when considering the inductive clause, to keep in mind what the relevant range of values for that clause is. Second, we can make use of strong induction. In normal proofs by induction, of the sort considered above, the inductive clause states that if k has property P, then k + 1 also has property P. In strong induction, we make the stronger assumption that all natural numbers less than or equal to k (and greater than or equal to our base case, whatever that is) have property P, and then seek to extend P to k + 1. Consider an example: Claim: Define the sequence {s i } as follows: s 1 = 1, s 2 = 2, s 3 = 3, and s k = s k 1 + s k 2 + s k 3 for k > 3. Then s n 2 n for all n. Proof: We check base cases of n=1, 2, and 3 by simple calculation. Now, given k, assume s j 2 j for all j k. We want to show that s k+1 2 k+1. By definition, s k+1 = s k + s k 1 + s k 2. By the inductive hypothesis, s k 2 k, s k 1 2 k 1, and s k 2 2 k 2. Thus s k+1 2 k + 2 k k 2. But we saw earlier that the sum of the first n powers of 2 is 2 n+1 1, so we have s k+1 2 k+1 1 < 2 k+1, as desired. Notice that we had to assume the inductive hypothesis for k, k 1, and k 2. This required strong induction. Strong induction isn t really a new principle it s just a clever application of regular induction. Suppose we want to do strong induction on property P. Let property P be the property that a number n has if every number less than or equal to it has property P. Then regular induction on P will be equivalent to strong induction on P. In the standard Peano axiomatization of the natural numbers and arithmetic, the principle of induction is taken as an axiom. We can view this as expressing 12
13 a confidence that induction is a valid proof method on the natural numbers, or as using the validity of induction as a method for defining what counts as the natural numbers. So much for induction on natural numbers. But induction is a much broader principle than this. It can be used to establish results on a wide class of structures other than the natural numbers. For example, consider the toy language L defined by the following two rules: 1. p is a sentence of L 2. If φ is a sentence of L, then qφq is a sentence of L. 3. If φ is a sentence of L, then rrφ is a sentence of L. We can then do the following proof by induction on L: Claim: Any sentence of L contains exactly one occurrence of p. Proof: By induction on the sentences of L. Base Case: The base case is the atomic sentence p. Trivially, this sentence contains exactly one occurrence of p. Inductive Case: Suppose that φ has exactly one occurrence of p. We want to show that both qφq and rrφ also contain exactly one occurrence of p. But this is immediate. The φ portion of both contains one occurrence, and neither the flanking q portions nor the prefixed rr portion contains p at all. Thus the property of containing exactly one occurrence of p perpetuates through the sentences of L. Thus the claim follows by induction. Now, why does this proof make sense? We ll look at two ways of making sense of it: one which ties it to induction on the natural numbers, and one which generalizes the use of induction. 1. We can assign to sentences of L a ranking. The ranking will indicate how many rule applications are needed to create the sentence. We thus define the ranking function recursively: (a) The rank of p is 0. (b) If the rank of φ is k, then the rank of qφq is k + 1. (c) If the rank of φ is k, then the rank of rrφ is k + 1. Thus the sentence rrrrp has a rank of 2, and the sentence qrrqpqq has a rank of 3. We can now perform a straightforward proof by induction on the natural numbers. We want to show that every natural number n has the following property: 13
14 Every sentence of L of rank n contains exactly one occurrence of p. Showing that every n has this property is equivalent to showing that every sentence of L has the property of containing exactly one occurrence of p. The proof proceeds in the usual way: The base case is n = 0. The only sentence of rank 0 is p, and (as in the proof above) we immediately verify that p contains exactly one occurrence of p. For the inductive case, assume that every sentence of rank k (or up through rank k) has exactly one occurrence of p. We now want to show the same for sentences of rank k + 1. But, by the nature of the ranking function, a sentence of rank k + 1 is formed from a sentence of rank k via one of the two formation rules flanking on both sides by q, or prefixing of rr. Thus we let φ be a sentence of rank k, and thus assume (by the inductive hypothesis) that it contains exactly one occurrence of p. We then have to show the same of qφq and rrφ, which is what is done in the proof above. The key to this approach is finding a good ranking function. There are many ways of mapping all of the sentences of L to the natural numbers, but most of them won t yield good proofs by induction. We need the ranking function to mirror some real structure of L, so that the property in question will propagate from one rank to the next, allowing the inductive clause to be proved. 2. We can do induction directly on the set of sentences of L. First we will define a partial ordering on that set. Given two sentences φ and ψ, say that φ < ψ if and only if ψ is the result of applying one of the syntactic rules to φ. Then let be the transitive closure of <. The transitive closure of a relation is the smallest transitive relation containing that relation. Transitive closures are often useful in defining ordering relations. In this case, < relates sentences that are a single formation step apart. then relates sentences that are any number of formation steps apart. is now a wellordering. This means that, given any subset of L, that subset contains minimal elements under the ordering. Now suppose that, for a given property P, we know two things about P in L: (a) The minimal sentences of L have P. (In fact, p will be the unique minimal sentence of L. (b) If every predecessor of φ has P, then φ has P. It will then follow that every sentence of L has P. 14
15 Proof: Suppose not. Then let X be the set of sentences of L that lack P. Since is a wellordering, there is some φ that is a minimal member of X. Thus every predecessor of φ is not in X. But if those predecessors are not in X, then they do have property P. Thus every predecessor of φ has property P. By the second assumption above, it follows that φ itself has property P. This contradicts the assumption that φ is in X, so in fact X must be empty. Thus every sentence of L has P. The two assumptions we made about P are, of course, exactly the generalized versions of the base clause and the inductive clause for a proof by induction. Here s the picture in general. Suppose we have a set X with a wellordering. Let X 0 be the set of minimal elements of X. Then if: (a) Every member of X 0 is P. (b) Given x X, if every predecessor of x is P, then x is P. it follows that every member of X is P. But where is the required wellordering to come from? A couple of points can be helpful here: (a) One particularly simple form of wellordering is a discrete ordering in which (a) every element has only immediate successors and predecessors (except endpoints, if any) and (b) every element is only finitely many steps away from the starting point(s). Such a wellordering lends itself naturally to the creation of a ranking function. (b) Suppose a set is recursively generated in the sense given earlier, with a base set X 0 and generating functions f 1,..., f n. Then we can say that x < y if y is the result of applying some generating function to x, and let be the transitive closure of <. will then be a wellordering of the recursively generated set. Since, in particular, logical languages are typically recursively generated sets, this method will be particularly useful to us. One final point about proofs by induction. These proofs can also be used on transfinite structures. We won t worry about transfinite induction in anything we do, but it s still something worth knowing. Proof by induction requires only that the set in question be wellordered. While the simplest wellorderings mirror the finitary structure of the natural numbers, not all wellorderings have this feature. We could, for example, do proofs by induction on ω 1 (the first uncountable ordinal). We would just need to show that, for any α ω 1, if every predecessor of α had the desired property, then α also had it. In this case, such a proof would break into two cases: 15
16 (a) α is a successor ordinal then α comes immediately after some other ordinal, so we can use the features of its immediate predecessors to show that α also has the desired features. (b) α is a limit ordinal then α has no immediate predecessor, so we will have to use the fact that all of α s predecessors have the desired feature. Proofs by induction will allow us to establish many important features of logical languages. We will give one example here. The axiomatic proof system given above identifies a set of sentences as theorems sentences that can be proved via the axioms and modus ponens. We will write φ to indicate that φ is a theorem. The semantics for PL already identify a set of sentences as logical truths sentences that are true in every model. We will write φ to indicate that φ is a logical truth. If our proof system is to be a good one, it must match our prior notion of logical consequence. This matching is typically divided into two parts: 1. The proof system must not prove anything that is not a real (semantic) consequence. Thus, if Σ φ, then Σ φ. This direction is called the soundness of the proof system. 2. The proof system must prove everything that is a real consequence. Thus, if Σ φ, then Σ φ. This direction is called the completeness of the proof system. The claim that every theorem is a logical truth is thus a special case of soundness (the case in which the set Σ of premises is empty). We will prove this special case by induction. Claim: If φ, then φ. Proof: By induction on the set of theorems. We will treat the set of theorems as a recursively generated set, with a base set of the axioms and a generating rule of modus ponens. Base Clause: Here we must show that every axiom is a logical truth. This is mostly a simple matter of inspection we look over the axioms listed above, and check a truth table for each one. However, there is the slight complication that we allow substitution instances of these axioms to be axioms as well. To show that all of the substitution instances are logical truths, we appeal to the fact that the substituted sentence must contribute some truth value to the larger sentence, and whatever truth value it contributes will be one of the possible truth values that the sentence letter substituted into could have contributed. But whatever truth value that sentence letter would have contributed, the axiom would have come out true. Thus the substitution instance of the axiom is also true in every model. 16
17 Inductive Clause: Suppose modus ponens is used to generate a theorem ψ. Then φ and φ ψ must already have been theorems, for some φ. Thus, by the inductive hypothesis, each must be a logical truth. Let M be an arbitrary model of PL. Then φ and φ ψ are both true in M. It follows immediately that ψ is also true in M. Thus ψ is a logical truth, as desired. The claim thus follows by induction. 6 The Maximal Generalization of Induction So recursively generated sets give rise to ordering relations. So what? Why is that long trek worth making? To see why, we have to notice one more feature of the ordering relations thus generated. An ordering relation is wellfounded under the following condition: Suppose < is an ordering relation on a set A. Let B be an arbitrary (nonempty) subset of A. Then B has some minimal element under the < relation. An element x B is minimal in B if there is no element y B such that y < x. Thus < is wellfounded if, for each B A, there is some x in B such that y x for all y B. Notice that if < is a partial ordering, then there may be multiple minima. We may have x, y B such that neither x nor y have anything below them in B, because x and y are incommensurable, so x y and y x. This definition of wellfoundedness is designed for strict orderings; things need to be tweaked a bit for loose orderings. An ordering relation that is wellfounded is called a wellordering. The natural numbers are wellordered by < any subset of the natural numbers has a least element. The integers, on the other hand, are not wellordered by <, since (for example) the set of all integers does not have a least element under <. Similarly, the real numbers are not wellordered by <, since the set (0, 1) does not have a least element under <. The set of subsets of {1, 2, 3, 4} is wellordered by. A couple of immediate consequences of wellorderings: 1. If < is a wellordering (on A), then < is lowerbounded. That is, there is some x A such that y x for all y A. In a partial ordering, there may be many lower bounds. 17
18 2. If < is a wellordering, then < is not dense it is not always the case that if x < y, then there is some z such that x < z and z < y. We will now prove two important results: 1. Claim: If X is a freely recursively generated set with a partial ordering < s defined as above, then < s is a wellordering. Proof: We begin by proving a lemma: Lemma: Let Z be a set with a strict ordering (partial or linear) R. Given z Z, let a zchain be a sequence z = z 0, z 1,... of members of Z such that z i+1 Rz i for all i in the length of the sequence. Call a member z of Z hereditarily finite if every zchain from it is finite in length. (An element is hereditarily finite if there is no infinite sequence of objects successively less than it.) If every z Z is hereditarily finite, then R is wellfounded. Proof: Suppose R is not wellfounded. Then there is a subset Y Z which has no least element. Pick some y 0 Y. Then since Y has no least element, there is some y 1 Y prior to y 0 in the ordering. Similarly, there is some y 2 Y prior to y 1 in the ordering. Proceeding in this manner, we generate an infinite descending chain from y, showing that it is not hereditarily finite. Thus if every element of Z is hereditarily finite, R must be wellfounded. We now show that, given any x X, x is hereditarily finite. Recall that X = i=0 X i, where the X i are formed by successive application of the generating functions to the base set X. Since X is freely generated, for any x X there will be a unique N such that X N marks the introduction of x into X. But then no xchain can be longer than N, since the ordering < s places y < s z just in case y first appears in X i, z first appears in X j, and i < j. Thus x is hereditarily finite. Given the lemma, it thus follows that < s on X is wellfounded, and hence that it is a wellordering. 2. Claim: If R is a wellordering of X, then proof by induction on X (with respect to R) is valid. That is, if X 0 is the set of Rminimal elements of X, then if: (a) Every member of X 0 is P. (b) If y is P for all y such that yrx, then x is P. Rancestors are P.) (x is P if all of its then every member of X is P. Proof: Suppose X is wellordered by R, and conditions (a) and (b) hold for some property P. Suppose, for reductio, that there is some x X which is not P. Now let P be the subset of X consisting of all nonp members of 18
19 X. Since R is a wellordering, there is some Rminimal member y of P. There are now two possibilities: (a) y has no Rpredecessors (in X it is, of course, given that y has no Rpredecessors in P ). Then y is Rminimal in X, and hence is in X 0. But by condition (a), every member of X 0 is P, so this contradicts the assumption that y P. (b) y has Rpredecessors. Since y is Rminimal in P, every Rpredecessor of y must be in X P. But then every Rpredecessor of y has the property P. But then by condition (b), y itself has the property P. This contradicts the assumption that y P. This completes the reductio, and shows that there cannot be an x X which lacks property P. So here s the picture in full generality. Proof by induction is a valid technique on all and only sets which are wellordered (we proved only the all part of this above). Given any freely recursively generated set, there is a wellordering of that set in which x comes before y just in case y is eventually generated from x. So proof by induction can be used on any recursively generated set, where the induction follows the path of generation of that set. Finally, both formal languages and proof systems on formal languages form nice recursively generated sets, where: The ordering on the whole language follows the path of syntactic generation of sentences, so that p comes before p, and p comes before q p. Thus induction can be performed by showing that each syntactic rule preserves the desired property. The ordering on the proof system follows the path of derivational generation, so that proof P 1 comes before proof P 2 just in case P 2 is an extension of proof P 1. Thus proof by induction can be performed by showing that each proof rule preserves the desired property. That s the high theory. Here s the practical upshot. Given a standard formal language L and a property P, if you can show that (a) every atomic sentence of L has property P and (b) property P is preserved by each of the syntactic rules of L (typically, preserved under construction of a complex sentence for each connective/logical operator in L), then you can conclude that every sentence of L has property P. And given a standard proof system, if you can show that (a) every starting point of the proof system (typically, the axioms) has property P, and (b) property P is preserved under each of the derivational rules of the proof system, then you can conclude that every theorem of the proof system has property P. 19
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