Otherwise, if
is-derived-from-specialization-of<Q, quantity_spec>()
is true, and
the specialization of Q::quantity_spec has a template argument equal to is_kind,
returns q.
Otherwise, if DerivedQuantitySpec<Q> is true,
returns
expr-map<to-kind, derived_quantity_spec, struct dimensionless>(q)
where to-kind is defined as follows:
template<QuantitySpec Q>usingto-kind=decltype(get-kind-tree-root(Q{})); // exposition only