Hi Alexandre,
Meanwhile, I have implemented SVA's first_match and ##[i:j].
The latter is just syntactic sugar: a ##[i:j] b = a:([*i:j];b) ##[i:j] b = [*i:j];b ##i = ##[i:i]
While you are at this, maybe you could also implement SVA's ##[*] and ##[+] (as sugar for ##[0:$] and ##[1:$]) for completeness? Please also note that the equality a ##[i:j] b = a:([*i:j];b) is wrong (despite being recommended in some textbooks!), e.g. {empty_word} ##[0:1] {a} =[by definition] = ({empty_word} ##0 {a}) | ({empty_word} ##1 {a}) = {} | {a} = {a} but {empty_word} : ([*0:1];{a}) = {} The reason is that ##0 always loses the empty word. Hence, it's best to split off ##0 (fusion) from ##[i>0:j] (union of concatenations) and handle them separately. Regards, Victor.