Non existing constraints: closed or open interpretation?

David Moner damoca at
Wed Jul 4 02:32:25 EDT 2018

Templates further constrain archetypes, that constrain the reference model.
In both cases, if the ANY constraint is used, that means that anything
defined in the underlying archetype (or in the underlying reference model)
is acceptable. Thus, you have to do an open interpretation.

> On a normal case where the full OPT is defined, from COMPOSITION to
> ELEMENT.value, how that OPT is interpreted? open or closed? Are extra
> ENTRIES and other nodes allowed even no definition for them is on the OPT?
> (open interpretation), or only OPT defined nodes are allowed? (closed
> interpretation).
If you mean the case when we already have an structure defined in the
template (or in the archetype), then the interpretation is closed. If you
already constrained an OBSERVATION inside a COMPOSITION, you removed the
ANY constraint, and thus the OBSERVATION is your only option. This is not
completely true in specialized archetypes, where you can expand the
definition and create new constraints, but that is a special case that is
not possible (as far as I know) at the template level.


David Moner Cano

Twitter: @davidmoner
Skype: davidmoner
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