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 ∈ BT ::= leaf | 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 ∈ Exp ::= n | (E + E) | (E × E) Constructing arithmetic expressions: I Base cases: n is an arithmetic expression for every n ∈ 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 ∈ 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.

Induction Matthew Hennessy Strong mathematical induction

To prove Prop(n) for every n ∈ N: (i) Assume the inductive hypothesis (IH) which says that Prop(k) is true for all k strictly less than some m m an arbitrary number

(ii) Show that Prop(m) follows from (IH).

Note: Apriori no base case

Induction Matthew Hennessy

Strong mathematical induction: an example

Prop(n): if n > 1 then n = p1 × ... × pl for some prime numbers pi , 1 ≤ i ≤ l

Recall: p is prime if p = a × b means a is either 1 or p itself. Proof: (i) Assume (IH): Prop(k) is true for every k

I m = 1

I m is prime

I m is not prime. So m = m1 × m2

Induction Matthew Hennessy Another example

E ∈ Exploc ::= x ∈ Vars | n ∈ Nums | (E + E) | (E × E) | let x = E in E

Prop(P) : If `big P ⇓ n, then `big P ⇓ m implies n = m Proof method:

I strong mathematical induction on the size of P

Better proof method:

I strong mathematical induction on the size of the proof of

`big P ⇓ n rule induction

Induction Matthew Hennessy

An example of rule induction

C, D ∈ Com ::= l := E | if B then C else C | C ; C | skip | while B do C

0 ∗ 0 Prop(C) : If `big hC, si ⇓ s then hC, si → hskip, s i Proof method: 0 I Induction on the size of the proof of `big hC, si ⇓ s

Case analysis on C:

I Five possibilities

I C of form while B do D most interesting

Induction Matthew Hennessy The interesting case

C is while B do D:

The proof of hC, si ⇓ s0 looks like: ...... (b-?) (b-?) (b-?) 0 hB, si ⇓ true hD, si ⇓ s1 hwhile B do D, s1i ⇓ s (b-while.t) hwhile B do D, si ⇓ s0

This contains lots of information we can use

Induction Matthew Hennessy