Structural Induction

Transcription

1 Structural Induction In this lecture we ll extend the applicability of induction to many universes, ones where we can define certain kinds of objects by induction, in addition to proving their properties by induction. This kind of technique goes under the general name of structural induction, though there are other names for it. The technique started in CS when it was realized that to build compilers for programming languages, it was really convenient to have a grammatical structure for expressions of the language, instead of just some ad hoc rules for what was a syntactically correct program. Logicians had been using the technique for 30 years before that, though they didn t think too much about it. 1

2 Recursive definitions of sets Let S be the subset of N defined by the following rules: 1. 3 S; 2. if x S and y S then x + y S; There is an extra condition: nothing gets into S unless you can prove it s there using the rules. Example: 1. 3 S (rule 1) 2. 6 = S by 1. and rule = S by 2. and rule 2. 2

3 Using induction along with recursive definitions Proposition The set S on the last slide is the set of positive multiples of 3. Proof: Let P be the set of positive multiples of 3. We show P S and S P. For the first inclusion, we show that every integer of the form 3n, for n 1, is in S. We do this by induction on n. Basis. When n = 1, the number 3 1 S by rule 1. Induction step. Assume that 3k is in S. We want 3(k + 1) S. But 3(k + 1) = 3k + 3. By inductive hypothesis 3k S, and 3 is already in S, so 3k + 3 S by rule 2. 3

4 Proof continued: S P. To show this, we rely on the extra condition, which says nothing is in S unless you can show it in a finite number of uses of rules 1 and 2. Our method is now rule induction, a special and common case of structural induction. Basis. This is the case when we apply a rule with no hypotheses. We have to show that what the rule puts into S is a positive multiple of 3. This is rule 1, which shows 3 S, and 3 = 3 1 is a positive multiple of 3. Induction step. This is the case where we use a rule with hypotheses, and in this case we get to assume our proposition about objects in the hypotheses of the rule. Thus for the hypothesis x S we get to assume x is a multiple of 3, and for y S we get to assume y is a multiple of 3. We now complete the inductive step by considering what the conclusion of the rule puts into S. This is x + y, and we have to show that this number is a multiple of 3. But our inductive assumptions are that x = 3k for some positive k, and y = 3j for some positive j. Thus x + y = 3(j + k) and j + k is positive, so x + y is also a positive multiple of 3. 4

5 Summarizing rule induction To show: every element of a set S defined by rules has a certain property P. Basis step: Show that everything introduced by rules with no hypotheses has property P. Induction step: For each rule, assume that the objects in the hypotheses of the rule have property P. Show that the object introduced in the conclusion of the rule also has property P. 5

6 Inductive (recursive) definition of propositional logic formulas These are expressions like (p (q ( r))). They are fully parenthesized, because their definition is easier that way. The way we define these formulas is similar to the way we evaluate them. We start with the variables, then work our way up to the top, using rules of syntax. This will be clearer with an example: Consider the formula (q ( r)). This can be proved to be syntactically correct if we use the following rules: 1. The symbols T and F are formulas. 2. Any propositional variable is a formula. 3. If P is a formula, then so is ( P ). 4. If P and Q are formulas, then so is (P Q). 5. Similar rules for the connectives,,. To show that the given expression is correct, we start with the variable r, a formula by rule 2. Then by rule 2, ( r) is a correct formula. Since also by rule 2, q is a correct formula, then by rule 4 (q ( r)) is correct, too. 6

7 Proving things about formulas using their recursive definitions For this part, we focus only on formulas built up using the connectives,, and. We would like to show that any such formula is logically equivalent to one where the negation sign applies only to propositional variables. Call such formulas negation-normal-form (NNF) formulas. We will do this via structural induction, but we first need a lemma, which is again proved by structural induction: Lemma If P is a NNF formula, then ( P ) is equivalent to a NNF formula. We assume the truth of this lemma for now, and proceed to the proof of our main result. This is done by structural induction. (Continued) 7

8 Continuing the proof We are trying to show that any formula built from variables, T, and F, and then using the rules for the connectives,, and, is logically equivalent to a NNF formula. The base case of the induction is given by considering those rules that have no hypotheses. So if P is T or F, then these are already in NNF. If P is a variable p, it is also already is in NNF. The inductive step of this proof corresponds to the formation rules for the connectives,, and. Suppose, then that P has been formed as (Q R). We get to assume our result about the formulas Q and R. Thus Q Q where Q is in NNF, and R R where R is in NNF. But then (Q R ) is in NNF, and P = (Q R) (Q R ), which is what we need. The case when P = (Q R) is completely similar. This leaves us with the case P = ( P ). We get to assume that P is equivalent to a NNF formula S. By our lemma, though, S is equivalent to another NNF formula S. Then P = ( P ) ( S ) S as we wanted. This completes the structural induction. 8

9 Proving the lemma Now we show that if P is a NNF formula, then ( P ) is equivalent to a NNF formula. Again we use structural induction. Basis: If P is T or F, or a variable, then P is clearly equivalent in each case to a NNF formula. Induction: We consider the formation rule for : If P is (Q R), and is in NNF, then both Q and R must be in NNF. Now P is equivalent to to Q R by DeMorgan s law, and Q is in NNF, so by inductive hypothesis Q Q, where Q is in NNF. Similarly R R where R is in NNF. Then P ( Q R) (Q R ), which is in NNF. The case where P is (Q R) is exactly similar(use the other DeMorgan law). If P is Q, and is in NNF, then Q must already be a variable, so we are done: already P is in NNF. 9

10 Defining other sets recursively Let Σ be a finite set of characters, for example Σ = {a, b}. We let σ be the set of all strings formed from characters in Σ, including the very convenient null string λ, a string with 0 length and with no characters in it. We let w be a typical string in Σ. Various subsets of Σ can be defined by structural induction, assuming that we have available the concatenation operator. This is just the operation that juxtaposes two strings w and x to give wx. We define λw = wλ = w. For example, the set Σ can be defined by three rules: 1. λ Σ ; 2. Σ Σ ; 3. If w Σ, and σ Σ, then wσ Σ. 10

11 Other recursively defined sets of strings Let Σ = {a, b}. We want to define the set S of all strings which end with the character b. We can do this via the following rules: (i) b S. (ii) If x S then ax and bx are both in S (there are two rules here). You will learn in 376 that these types of rules are abbreviated as follows: (i) S b. (ii) S as and S bs. This setup is called a grammar. We can use a grammar to define the set of propositional expressions using the variable symbols p, q, r, and the symbols T and F. Our set of characters is Σ = {p, q, r, T, F, (, ),,, }. It includes connectives and left and right parentheses. The rules can be written This abbreviates 8 rules. S p q r T F (S S) (S S) ( S). 11