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Re: DTD and Illegal Construct

Its the fact you have 2 different models (A, B, D?) and the other (A?, B, C? ...)

that says if i have an A which one of these two models am I using? It could be
either one - it doesn't have anything to do with the other elements in the models. B has the same problem and so does D.


> Honestly, I'm not playing stump the experts, but what about the instance?
> ((A, B, D?) | (A?, B, C?, D?, E))
> In this instance 'C' can only be present if 'E' is present.
> -Jim
> -----Original Message-----
> (Excerpt)
> <!ELEMENT Request ((A, B, C?) | (A?, B, C?, D))> becomes
> <!ELEMENT Request ((A, B, C?, D?) | (B, C?, D))>