[rules-users] How complex can be an expression in given rule...

[Edson Tirelli]

   Yes, Drools is First Order Logic complete, i.e., any expression that
involves predicates and quantifiers over predicates can be expressed.

   Edson

2009/10/15 Costigliola Joel (EXT) <joel.costigliola-ext at natixis.com>

> Yes.
> Just have a look at the doc (Rule Language Conditional Elements section) :
>
> http://downloads.jboss.com/drools/docs/5.0.1.26597.FINAL/drools-expert/html/ch04.html#RuleLanguage-ConditionalElements
>
> -----Message d'origine-----
> De : rules-users-bounces at lists.jboss.org [mailto:
> rules-users-bounces at lists.jboss.org] De la part de Madhav Bhamidipati
> Envoyé : jeudi 15 octobre 2009 08:39
> À : rules-users at lists.jboss.org
> Objet : [rules-users] How complex can be an expression in given rule...
>
> Hi,
>
>
> Are the expressions in a rule can be as complex as a regular language
> support.
> For example can the expression be as complex as (a == b || (a <= d && a > e
> )?
>
> Where can I find the info. about what sort of expressions supported?
>
>
> Madhav
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-- 
 Edson Tirelli
 JBoss Drools Core Development
 JBoss by Red Hat @ www.jboss.com
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