the :1 is short for :0001 basically and then just put that bit of the address at the very end and put the first bit of the address at the front, and then just fill each missing group inbetween with 0000

"just"

Yes, in fact "just". This isn't remotely hard.

These types of complaints are how I know the objection to v6 is not serious.

Well, okay, show us how to follow those instructions then.

"the :1 is short for :0001 basically" is easy enough: you get 2001::0001::0001.

Then "just put that bit at the very end" -- but which bit? If it means the ":0001", then there's two of them and they can't both go at the very end. If not, then it fails to specify which bit. Either way I don't see how these instructions are followable at all, let alone easily.

It’s a really complicated rule called “subtraction”. Addresses are always 128 bits long, or 8 groups of four hex digits. 2000::1 is two groups, so you need six groups in between to make 2000:0000:0000:0000:0000:0000:0000:1. But I don’t know why people always ask this, because it’s always the computer you are typing addresses in to that does the subtraction. You never ever have to type out the whole address. Just type the shortened version, because 2000::1 _is_ the whole address.

They were answering the question of why "2000::1::1" would be ambiguous if it was allowed.

My answer was too terse. IF there was two :: in the address, then the length of EACH :: denoted section is not known. It can be either longest left :: or longest right :: and that wasn't defined, because the rule is THERE IS ONLY ONE :: section.

Posed as a question, disingenuously.