Winter 2004/5Pls – inductive – Catriel Beeri1 Inductive Definitions (our meta-language for specifications) Examples Syntax Semantics Proof Trees

Winter 2004/5 Pls – inductive – Catriel Beeri 1

Inductive Definitions(our meta-language for specifications)

Examples Syntax Semantics Proof Trees Rule set as a function Proof by induction Function (& relation) definition by induction Generalizing rule form


An inductive definition by

one axiom, one proper rule

Let be defined by

(r1) 4 ,

(r2) , , 2 10

Ex. 1

if &

then 2 3

S

S

x y S y x y

x y S

Examples


Questions:

1. Do we know which elements are in S?

(at least, can we enumerate them?)

2. Given c, can we answer :is c a member of S?

3. Which is the underlying universe here ?

A: any of N, Z, Q, ….

It is common to omit the universe, U, since:– It is often known from the context

– The rules “tell” us what it is

– Even if there are several candidates, the defined set is often the same in them


4. Is there just one set in N that satisfies (r1), (r2)? If not, which one is the one being defined?A: The intention is the least set that satisfies the rulesBut, “least’ is often omittedNote difference between minimal, least (in a collection)

* There is an abuse of language. It should be

Easier to use S for both: defined and a set variable

Let be the set that satisfies

(r1) 4 (r2) if , &...

leas

then

t

2 3

S

x y x

X

X XyX


Can we enumerate?

Solve membership?

Is the defined set independent of the universe?

: let S be defined by

(r1) 4 , (r2) 7 ,

, (r3)

2

Ex.2

3

S S

x y S

x y S


A new problem: goal resolution

Given c, can we find all y s.t. S(c,y)?

c – a goal (on 1st position)

y, such that S(c,y) – a solution

Can we solve when the goal is on 2nd position?


(r1) S(4,7) (r2) S(4,9)

( , ) ( ,z) (r3)

Ex.3

( 1,2 3 )

S x y S x

S x y z


Q: is the defined set independent of universe?

What about goal resolution on either column?


( , ) ( ,z) (r1) S(x, x+1) (r2)

(

Ex.

, )

4

S x y S y

S x z


Inductive definition: a set R of rules

A rule: conjunction of premises (הנחות)

implies conclusion

or: body implies head

Our style: (no commas!)

Syntax

1 ... ne S e S

e S

membershi -- a pr mip e seie S - a premise, an arbitrary formula

does include

conditio

membersnot

n

hips


Empty body (0 premises) – an axiom:

Otherwise, a proper rule

1st Example:

Here, x, y are individual variables

(r1) 4 (an axiom)S

2 10(r2) (a proper rule)

2 3

x S y S y x y

x y S

e S


SemanticsVariable assignment :

maps individual variables to constants (in U)

Applied to a rule a ground (var-free) rule instance

{ 3, 4} givesx y 3 4(r2)

18

S S true

S

S satisfies rule r if for each r-instance, if body is true in S then head is true in S (Individual vars are universally quantified)

S satisfies R, if it satisfies each r in R


• If U (the universe)– contains the constants in R,

– is closed under the operations in R,

then it satisfies R – satisfying set exists

• Intersection property :

If each set satisfies R,

so does

Hence, a (unique) least set that satisfies R exists

This set is the semantics of R – ind{R}

,iS i I

i I iS


Extensions • Define several sets simultaneously

– Can use in rules, for each defined

• Can assume certain sets are given (known)– Can use in premises

(This is not a real extension, can use these in conditions)

These enable construction of definition hierarchies.

j ie S iS

{ }jQ

ie Q

iS


Example of a proof of membership in an inductive set:

Each leaf – an axiom instance

Each internal node – inferred from its children

– a rule instance

4 420 4 44

52 20164

S SS S SS

S SS

Proof Trees


We abbreviate to :

Element a is inferred from by rule r if ...

Q: What when r is an axiom?

1,..., na a

1 ... ( )( ) , & ( ) is na S a Sr true

a S

1 is from , ..., by rule ifinferre d na S a S a S r


Proof TreeA tree in which each node is• labeled by a is the element at the node

• Associated with a rule r such that: is inferred from the children of the node by r (for a leaf --- associated rule is an axiom)(Labels of a node & its children ~ rule instance)

Proof trees contain only membership formulas

Proof tree – a composition of inferences

a S

a S


Fact: a is in ind{R} iff it has a proof tree Proof tree is a proof of membership for its root

How do we prove this?

1. An arbitrary satisfying set contains every element that has a proof tree --- induction on depth of tree

2. The set of elements that have proof trees is a satisfying set


Rule set as a function

Generalize:

For example 1:

( ) ( )R rr R

f Q f Q

( ) the elements inferred by from elements of rf Q r Q

( ) {4} { | 2 3 , , , 2 10 }Rf Q z z x y x y Q y x y

({2,3}) {4,10,15,13,12}Rf


Define: Q is closed under r/R if it is closed under :

Fact: closed under r/R = satisfies r/R

Fact:

same for

is :

monotone

( ) ( )r

r r

f

P Q f P f Q

( ) / ( )r Rf Q Q f Q Q

Rf

/r Rf f


Thm: [Knaster-Tarski, 1955]

A monotone function f on sets has the

intersection property for its closed sets:

if some set is closed under f, a least closed set exists

for any R, a least set closed under R exists (for a large enough universe)

(remember: closed under = satisfies)

ii I

if , are closed under

then S is closed under

iS i I f

f


• Prove for each axiom that P(e) holds• Prove for each rule

that if (and ) then P(e) holds (actually, prove it for each variable assignment)

proof by rule induction

Example: prove for Example 1 that each element in S is divisible by 4

Proof by induction

e S1 ... ne S e S

e S

P( )ii e


Function (& relation) definition by induction

• For each axiom , define f(e)

• For each rule

define f(e) in terms of

Example:

fact(0) = 1

fact(n+1) = (n+1)*fact(n)

Why is it called definition by induction?Hint: What is the domain, and the rules that define it?

and what is the form of the function definition

e S

1 ... ne S e S

e S

1f ( ), ..., f ( )ne e


Let rfact be the graph of fact , a binary relation.

Compare:

To fact(0) = 1

fact(n+1) = (n+1)*fact(n)

( 1) (0,1)

( , )( 2)

( 1,( 1)* )

f rfact

rfact n kf

rfact n n k

( 1) 0

( 2)1

N

nN

n

“Inductive” has a double meaning– the domain is inductively defined –The function is inductively defined (in a special syntax) and the two rules sets correspond


Function def. by induction:

function defined by clauses, one for each rule defining the domain

Significance :

One-to-one correspondence fact is total

Sometimes, we need a partial function;

we provide clauses for only some of the rules

that define the domain

Example: head and tail on lists


Consider:

Define

We have:

What went wrong? (think of proof trees)

“Function” is total, but ill-defined!

1 N

x N y N

x y N

(1) 0

( ) 1 max( ( ), ( ))

d

d x y d x d y

(4) 1 max( (3), (1)) 3

(4) 1 max( (2), (2)) 2

d d d

d d d


Function & relation definitions, summary

Over an inductively defined domain• Inductive definition of relation is always ok.

• Inductive function definition – Is always total

– may fail to be well-defined

a sufficient condition for well-defined-ness:

each value of domain has a unique proof tree


Generalizing rule form

It is allowed to combine membership formulas by• Conjunction in head• Disjunction in body

But,

Disjunction in head, negation in body/head are

forbidden

Why?

As for quantifiers --- see reading material

Documents

Winter 2004/5Pls – inductive – Catriel Beeri1 Inductive Definitions (our meta-language for specifications) Examples Syntax Semantics Proof Trees