(j3.2006) derived types with type parameters are different

[Aleksandar Donev]

On Wednesday 30 July 2008 02:45, Lawrie Schonfelder wrote:
> This too was rejected and the current
> expression equivalence rule invented.
What rule is this?
> It was claimed that this form of 
> equivalence was already a feature of determining the characteristics of a
> procedure interface, so it was no big deal.
Can someone please explain this---I must have missed this in the standard.
Thanks,
Aleks