Structural Induction

CS 611

Structural Induction

• Useful for proving properties about languages.

• Useful for proving properties about algorithms defined on languages.

• A generalization of the induction you know and love.

Induction on the Naturals

0

n

n+1numbers. newget you how is "1" And

1. number that of trueis P

then number, a of trueis P if

and case basis theof trueis P

)1()(..2

)0( 1.

prove )(. :prove To

nPnPNn

P

nPNn

Example

Prove: every natural number is odd or even.Base: 0 is even. Induction: if n is odd or even then n+1 is odd or even.Suppose n is odd, then n+1 is evenSuppose n is even then n+1 is odd.

0

n

n+1

Induction on Aexp

constvar

AexpAexp

+

AexpAexp

-

Suppose I want to prove a property of all arithmetic expressions. Like “eval (aexp) is finite” for some evaluation function eval.

Aexp has 2 atoms and 2 constructors.

So induction on Aexp has ___ base cases and ___ inductive steps?

What is the induction theorem for arithmetic expressions?

Induction on Aexp

) - ()()(.Aexp,4.

) ()()(.Aexp,3.

var)(2.

const)(.1

:prove ),(.Aexp prove To

baPbPaPba

baPbPaPba

P

P

bPb

constvar

AexpAexp

+

AexpAexp

-

Example

Eval (aexp a)if a = var then val(a)if a = const then aif a = (b+c) then eval(b)*eval(c)if a = (b-c) then eval(b)/eval(c)

Prove: forall a:aexp. eval(a) terminates

Base 1: eval(var) terminatesBase 2: eval(const) terminatesInd1: suppose eval(b) and eval(c) terminate

prove eval (b+c) termiantesInd2: suppose eval(b) and eval(c) terminate

prove eval(b-c) terminates

Example

Eval (aexp a)if a = var then val(a)if a = const then aif a = (b+c) then eval(b)*eval(c)if a = (b-c) then eval(b)/eval(c)

Prove: forall a:aexp. eval(a) = correct_val(a)

Base 1: eval(var) = correct_val(var)Base 2: eval(const) = correct_val(const)Ind1: suppose eval(b) = correct_eval(b)

and eval(c) = correct_eval(c)prove eval(b+c) = correct_eval(b+c)

Ind2: suppose eval(b) = correct_eval(b) and eval(c) = correct_eval(c)prove eval(b-c) = correct_eval(b-c)

Example

Eval (aexp a)if a = var then val(a)if a = const then aif a = (b+c) then eval(b)*eval(c)if a = (b-c) then eval(b)/eval(c)

Prove: forall a:aexp. eval(a) is finite

Base 1: eval(var) is finiteBase 2: eval(const) is finiteInd1: suppose eval(b) is finite

and eval(c) is finiteprove eval(b+c) is finite

Ind2: suppose eval(b) is finite and eval(c) is finiteprove eval(b-c) is finite

Reminders

• Reverse engineer proof of Thm 3.1 to scratch work for Monday.

• Start on proofs of Thms 3.2, 3.3, probably for Tuesday.

• Thm 3.4 and Cor 3.2 likely for Friday

• Paper 1 on Friday

• Cancel Godel encoding part of hwk.