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(a)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/ht...
-----Message d'origine-----
De : rules-users-bounces(a)lists.jboss.org [mailto:
rules-users-bounces(a)lists.jboss.org] De la part de Madhav Bhamidipati
Envoyé : jeudi 15 octobre 2009 08:39
À : rules-users(a)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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