Cs7120 (Prasad)L9-RECUR-IND1 Recursion and Induction.

cs7120 (Prasad) L9-RECUR-IND 1

Recursion and Induction


• Define sets by inductioninduction

zero N

n N succ(n) N

• Define functions on sets by recursionrecursion

n N : plus(zero, n) = n m, n N :

plus(succ(m), n) = succ(plus(m,n))

• Prove properties about the defined functions using principle of structural inductionprinciple of structural induction.


Example0 + n = n (obvious)n + 0 = n (not so obvious!)

• Prove that the two rules for “+” are adequate to rewrite (n+0) to n.

(Induction on the structure of the first argument)

• Show that “+” is commutative, that is, (x + y) = (y + x).

• Motivation To ensure that sufficient relevant information has

been encoded for automated reasoning.


Induction Proof

• Definition of “+” 0 + m = ms(n) + m = s(n+m)

• Proof that 0 is the identity w.r.t. +

0+m = m+0 = m

• Basis: 0 + 0 = 0• Induction Hypothesis:

k >= 0:

k + 0 = k

• Induction Step: Show s(k) + 0 = s(k)

s(k) + 0= s(k + 0) (*rule 2*)

= s(k) (*ind hyp*)

• Conclusion: By principle of mathematical induction

m N: m + 0 = m


• Basis: n : n + 0 = n

n : 0 + n = n n: n + 0 = n + 0

• Induction Hypothesis: k >= 0, n : k + n = n + k

• Induction Step:

s(k)+n = n+s(k) s(k)+n = (*rule2*) s(k+n) = (*ind. hyp.*) s(n+k) = (*rule2*) s(n)+k(* STUCK!!!

our goal: n+s(k) *)

So prove the auxiliary result.

s(k)+n = k+s(n)

n

mProof proceedsrow by row


• Auxiliary result

s(i)+ m = i+s(m)Basis: s(0) + m = (*rule2*) s(0 + m)= (*rule1*) s(m)= (*rule1*) 0 + s(m)Induction step:

s(s(j)) + m=(*rule2*) s(s(j)+m)=(*ind.hyp.*) s(j+s(m))=(*rule2*) s(j)+s(m)

• Overall result

s(k) + n =(*auxiliary result*) k + s(n)=(*induction hyp.*) s(n) + k=(*auxiliary result*) n + s(k)(* End of proof of

commutativity *)


Motivation for formal proofs

• In mathematics, proving theorems enhances our understanding of the domain of discourse and our faith in the formalization.

• In automated theorem proving, these results demonstrate the adequacy of the formal description and the symbol manipulation system.

• These properties also guide the design of canonical forms for (optimal) representation of expressions and for proving equivalence.


Semantic Equivalence vs Syntactic Identity• Machines can directly test only syntactic identity. • Several distinct expressions can have the same

meaning (value) in the domain of discourse. To formally establish their equivalence, the domain is first axiomatized, by providing axioms (equations) that characterize (are satisfied by) the operations.

• In practice, an equational specification is transformed into a set of rewrite rules, to normalize expressions (into a canonical form).

(Cf. Arithmetic Expression Evaluation)


Induction Principle for Lists

• P(xs) holds for any finite list xs if:– P([]) holds, and– Whenever P(xs) holds, it implies that

for every x, P(x::xs) also holds.

• Prove:filter p (map f xs)

= map f (filter (p o f) xs)


• Basis:filter p (map f []) = filter p [] = []

map f(filter (p o f) []) = map f []= []

• Induction Step: map f (filter (p o f) (x::xs))= map f

(if ((p o f) x) then x:: (filter (p o f) xs)

else filter (p o f) xs )

case 1: (p o f) x = truecase 2: (p o f) x = false


• case 1:map f ( x:: (filter (p o f) xs) )

= f x :: map f (filter (p o f) xs)

= f x :: filter p (map f xs) (* induction hypothesis *) = filter p (f x :: map f xs) (* p (f x) holds *) = filter p (map f (x::xs))• case 2: filter p (map f (x::xs)) = filter p (f x :: map f xs)

(* p (f x) does not hold *) = filter p (map f xs) = map f ( filter (p o f) xs ) (* induction hypothesis *)


Tailoring Induction Principle

fun interval m n = if m > n then [] else m:: interval (m+1) n(* Quantity (n-m) reduces at each recursive call. *)• Basis:

P(m,n) holds for m > n• Induction step:

P(m,n) holds for m <= n, given that P(m+1,n) holds.


Induction Proof with Auxiliariesfun [] @ xs = xs | (y::ys) @ xs = y:: (ys@xs);fun rev [] = [] | rev (x::xs) = (rev xs) @ [x];

Prove : rev (rev xs) = xs• Basis: rev (rev []) = rev [] = []• Induction step: rev(rev (y::ys)) = rev ( (rev ys) @ [y] ) = (* via auxiliary result *) y :: ( rev (rev ys) ) = y :: ys (* ind. hyp. *)


Auxiliary result

rev ( zs @ [z] ) = z:: rev zsInduction Step:

rev ((u::us) @ [z])= rev ( u :: (us @ [z])) (* @ def *)

= (rev (us @ [z])) @ [u] (* rev def*) = (z :: (rev us)) @ [u] (* ind hyp *)

= z :: ((rev us) @ [u]) (* @ def *)

= z :: rev (u::us) (* rev def*)

(*Creativity required in guessing a suitable auxiliary result.*)


Weak Induction vs Strong Inductiondatatype exp = VarVar of string | OpOp of exp * exp;

• Prove that the number of occurrences of the constructors in a legal exp are related thus:

#VarVar(e) = #OpOp(e) + 1• To show this result, we need the result on

all smaller exps, not just the exps whose “node count” or “height” is one less.– Motivates Strong/Complete Induction

Principle.


McCarthy’s 91-function

fun f x =

if x > 100 then x - 10

else f(f(x+11))

fun f x =

if x > 100 then x - 10

else 91


Is f total?

fun f x =

if (x mod 2) = 0

then x div 2

else f(f(3*x+1))

View int x as (2i + 1) 2^k - 1

Cs7120 (Prasad)L9-RECUR-IND1 Recursion and Induction.

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