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Coinductive semantics
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Date: -- (:)
From: Francisco J. Valverde Albacete <fva@t...>
Subject: Re: [Caml-list] Coinductive semantics

although I may be out on a limb here, I recall reading *somewhere* :( 
that while initial algebras where good models for stateless abstract 
data types (and (structural) induction the way to work over terms in the 
free algebra defined by constructors modulo the laws of the ADT), final 
algebras where good models for *stateful* datatypes (and coinduction the 
way to work over the finer "state descriptors" modulo the laws of state 
equivalence), hence they *might* be better models for *objects* (as 
stateful datatypes) than initial algebras.

Has anybody read anything proving or disproving this?


    Francisco Valverde - DTSC
    Univ. Carlos III de Madrid

Andrej Bauer wrote:

>Alessandro Baretta wrote:
>>Is there anyone around who  can help?
>Induction is about initial algebras. It says that an initial algebra
>does not have any non-trivial SUB-algebras. General induction principle
>for the initial algebra A with operations f_1, ..., f_n (some of which
>may be 0-ary operations, i.e., constants) goes as follows:
>Suppose S is a subset of A which is closed under operations f_i, meaning
>that if x_1, ..., x_{k_i} are in S then also f_i(x_1, ..., x_{k_i}) is
>in S (here k is the arity of operation f_i). Then S = A.
>If we apply this to the case when we have one constant f_0=Empty and one
>binary operation f_1=Tree, we get as A all finite binary trees and the
>induction principle:
>Suppose S is a set of finite binary trees such that the empty tree is in
>S, and whenever x and y are in S then also tree Tree(x,y) is in S. Then
>S is the set of all finite binary trees.
>Now coinduction is about final coalgebras. It says that a final
>coalgebra does not have any non-trivial QUOTIENT coaglebras. This is
>usually expressed as follows: if F is the final coalgebra for operations
>f_1, ..., f_n, and E is an equivalence relation on F which respects the
>operations, then E is the identity relation (because if it were not, we
>could form a nontrivial quotient coaglebra F/E). The actual coinduction
>principle may be stated for an arbitrary relation R (think of E as the
>equivalence relation generated by R):
>Suppose R is a relation on the final coalgebra F which respects the
>operations, i.e., if x_1 R y_1, ...., x_{k_i} R y_{k_i} then
>f_i(x_1, ..., x_{k_i}) R f_i(y_1, ..., y_{k_i}) for all x's, y's and
>f_i's. Then R(x,y) implies x = y.
>An R satisfying the above condition generates the trivial equivalence
>relation (equality) and so the quotient F/R is just F.
>We take again as example the final coalgebra F with one constant
>f_0=Empty and one binary operation f_1=Tree. This is the set of finite
>and infinite binary trees.
>Suppose R is a relation on trees such that:
>(1) Empty R Empty
>(2) if x R y and x' R y' then Tree(x,y) R Tree(x',y').
>Then x R y implies x = y.
>It takes some practice to get used to coinduction and to figure out how
>to prove properties of final coalgebras with it. If this was too terse,
>let me know (and tell me which bits to expand upon).
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