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Induction: in Many Forms Induction: in many forms Matthew Hennessy Draft March 26, 2015 Induction Matthew Hennessy The natural numbers N Two rules for constructing natural numbers: (a) base rule: 0 is in N (b) inductive rule: if k is in N then so is its successor (k + 1) Every natural number can be constructed using these two rules Definition principle for N: To define a function f : N ! X : (a) base rule: describe the result of applying f to 0 (b) inductive rule: assuming f (k) has already been defined, describe the result of applying f to its successor (k + 1) Result: with (a) (b) we know function f is defined for every natural number. Induction Matthew Hennessy Examples Summation: sum : N ! N is defined by: (a) base rule: sum(0) = 0 (b) inductive rule: sum(k + 1) = sum(k) + (k + 1) Factorial: fac : N ! N is defined by: (a) base rule: fac(0) = 1 (b) inductive rule: fac(k + 1) = fac(k) × (k + 1) Induction Matthew Hennessy Proof principle for N To prove a property P(n) for every natural number n: (a) Base case: prove P(0) is true using known mathematical facts (b) Inductive case: I assume the inductive hypothesis: that P(k) is true I from this hypothesis prove that P(k + 1) follows using known mathematical facts If (a) (b) are established it follows that: P(n) is true for every natural number n Induction Matthew Hennessy Example n∗(n+1) Prove sum(n) = 2 for every natural number n n∗(n+1) Property P(n) : sum(n) = 2 Proof: We must show: (a) Base case, P(0) : sum(0) = 0 Follows by definition of sum (b) Inductive case: k∗(k+1) I Assume the inductive hypothesis: IH is sum(k) = 2 (k+1)∗(k+2) I Use IH to deduce P(k + 1) : sum(k + 1) = 2 use algebraic manipilations n∗(n+1) Result: sum(n) = 2 is true for every natural number n Induction Matthew Hennessy Inductive structures Example: binary trees BT Each node is either I a leaf: I or has two siblings Induction Matthew Hennessy Constructing binary trees (a) Base case: is a binary tree (b) Inductive case: If L and R are binary trees then so is L R Induction Matthew Hennessy Syntax for binary trees BT bTree 2 BT ::= leaf j Branch(bTree; bTree) Construction rules: (a) Base case: leaf is a binary tree (b) Inductive case: If L and R is a binary tree then so is Branch(L; R) Examples Branch(leaf; Branch(leaf; leaf)) Branch(Branch(leaf; Branch(leaf; leaf)); Branch(leaf; leaf)) Induction Matthew Hennessy Definition principle for binary trees To define a function f : BT ! X : (a) Base rule: describe the result of applying f to leaf (b) Inductive rule: assuming f (T1) and f (T2) have already been defined, describe the result of applying f to the tree Branch(T1; T2) Result: with (a) (b) we know function f is defined for every binary tree. Induction Matthew Hennessy Example definitions Number of leaves in a tree: leaves : BT ! N defined by: I base case: leaves(leaf) = 1 I Inductive case: leaves(Branch(T1; T2)) = leaves(T1) + leaves(T2) Number of branches in a tree: branches : BT ! N defined by: I base case: branches(leaf) = 0 I Inductive case: branches(Branch(bTree1; bTree2)) = branches(bTree1) + branches(bTree2) + 1 Induction Matthew Hennessy Structural induction for binary trees To prove a property P(T ) for every binary tree T (a) Base case: prove P(leaf) is true using known mathematical facts (b) Inductive case: I assume the inductive hypothesis: that P(T1) and P(T2) are both true I from this hypothesis prove that P(Branch(T1; T2; )) follows using known mathematical facts If (a) (b) are established it follows that: P(T ) is true for every binary tree T Induction Matthew Hennessy Example proof leaves(T ) = branches(T ) + 1 for every binary tree T Property P(T ) is: leaves(T ) = branches(T ) + 1 I base case: P(leaf): we must prove leaves(leaf) = branches(leaf) + 1 follows by definition I Inductive case: assume P(T1) and P(T2) are true- (IH) From (IH) prove P(Branch(T1; T2)) follows leaves(Branch(T1; T2)) = leaves(T1) + leaves(T2) = branches(T1) + 1 + branches(T2) + 1 (IH) = (branches(T1) + branches(T2) + 1) + 1 = branches(Branch(T1; T2)) + 1 Induction Matthew Hennessy Arithmetic expressions E 2 Exp ::= n j (E + E) j (E × E) Constructing arithmetic expressions: I Base cases: n is an arithmetic expression for every n 2 N I Inductive cases: If E1 and E2 are arithmetic expressions so are I E1 + E2 I E1 × E2 I an infinite number of base cases I two inductive cases Induction Matthew Hennessy Definition principle for arithmetic expressions To define a function f : Exp ! X : (a) Base rule: describe the result of applying f to n for every n in Nums (b) Inductive rule: assuming f (E1) and f (E2) have both already been defined, describe the result of I applying f to (E1 + E2) I applying f to (E1 × E2) Result: with (a) (b) we know function f is defined for every arithmetic expression. Induction Matthew Hennessy Structural induction for arithmetic expressions To prove a property P(E) for every arithmetic expression E (a) Base case: prove P(n) is true for every natural number n (b) Inductive case: I assume the inductive hypothesis: that P(E1) and P(E2) are both true I from this hypothesis prove that I P(E1 + E2) follows I P(E1 × E2) follows If (a) (b) are established it follows that: P(E) is true for every arithmetic expression E Induction Matthew Hennessy Example: normalisation of big-step semantics For every arithmetic expression E there exists some natural number k such that `big E + k P(E) is: `big E + k for some natural number k Proof by structural induction: (a) Base case: We have to show P(n) for every n in N (b) Inductive case: Assume P(E1) and P(E2) are true. We have to show I P(E1 + E2) is true I P(E1 × E2) is true Induction Matthew Hennessy Example: small-step semantics `sm E ! F implies `ch E !ch F for all expressions E; F P(E) is E ! F implies E !ch F Proof by structural induction: (a) Base case: We have to show n !ch F implies n ! F for every n in N (b) Inductive case: Assume the inductive hypotheses (IH) I E1 ! F implies E1 !ch F I E2 ! F implies E2 !ch F From (IH) we have to show I E1 + E2 ! F implies E1 + E2 !ch F I E1 × E2 ! F implies E1 × E2 !ch F Induction Matthew Hennessy More examples I Determinacy of big-step semantics: `big E + m and `big E + n implies m = n I Determinacy of small-step semantics: ∗ ∗ I E ! m and E ! n implies m = n ∗ ∗ I E !ch m and E !ch n implies m = n I Consistency: ∗ ∗ I E ! n implies E !ch n ∗ ∗ I E !ch n implies E ! n ∗ I `big E + n implies E ! n ∗ I E ! n implies `big E + n I Some proofs are not easy ∗ I Some require proof principle for Induction Matthew Hennessy Rule Induction I When there is no structure ? I When structure is infinite ? Solution: Perform induction on size of derivations Induction Matthew Hennessy Example (ax) (plus) n D m n D 0 n D(m + n) Derivations: (ax) 7 D 0 (ax) (plus) 2 D 0 7 D 7 (plus) (plus) 2 D 2 7 D 14 (plus) (plus) 2 D 4 7 D 21 Size of derivations: 2 D 4 < 7 D 21 Induction Matthew Hennessy Example proof To prove: If n D m then m = n × k for some natural number k I Let P(n; m) be: m = n × k for some natural number k. I We prove n D m implies P(n; m) by strong mathematical induction on size of derivation of n D m. I Which was the rule last used? I (ax): m must be 0 and P(n; 0) holds for every n I (plus): m must be m1 + n where n D m1 has a smaller derivation I use induction on n D m1 to finish proof Induction Matthew Hennessy Rule induction (ax) (plus) n D m n D 0 n D(m + n) To prove n D m implies P(n; m) I Axiom (ax): prove P(n; 0) for every n 2 N I Rule (plus): I assume P(n; m) - because of hypothesis n D m I from this assumption deduce P(n; m + n) follows - because of conclusion n D(m + n) Induction Matthew Hennessy What is going on? Inductively defined sets Given a set T - world of discourse I Axiom: an element of T h1;h2;:::hn I Rule: c where n > 0 I each hi an element of T hypotheses I c an element of T conclusion I Deductive system D : set of axioms and rules D(T): All elements of T which can be proved from the axioms using the rules Induction Matthew Hennessy Rule induction for deductive systems To prove P(t) for every t in D(T ) (a) prove P(a) for every axiom in D h1;h2;:::hn (b) for every rule c I assume P(hi ) for every hypothesis hi I from these assumptions show P(c) follows From (a), (b) conclude P(t) for every t in D(T ) Alternative: Strong mathematical induction on the size of derivations in the deductive system D.
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