Why do "const inout" and "const inout shared" exist?

[Walter Bright via Digitalmars-d]

On 7/1/2017 3:12 PM, Timon Gehr wrote:
> const(const(T))     = const(T)
> const(immutable(T)) = immutable(T)
> const(inout(T))     = ?
> 
> It used to be the case that const(inout(T)) = const(T), but this is wrong, 
> because if we replace 'inout' by 'immutable', the result should be immutable(T), 
> not const(T). Hence const(inout(T)) cannot be reduced further.

If const(inout(T)) is reduced to inout(T), it works.