In both instances, you still need to bruteforce 12 words to get the actual seed phrase... So, you're working with a 2028^12 search space... which is 5444517870735015415413993718908291383296.
I don't think we can call it the same. If the total words are 12 and you are missing all 12 then the search space is 2
128. But if the total words are 24 and you have 12 then you still have half of the entropy and depending on which half it could be a lot simpler. For example if you have the second half of the words then the
bulk of the search space is suddenly reduced by roughly 94% because of the checksum.
I think this is incorrect.
When someone knows the last 12 words of the 24th sentence, they know 132 bits. But since 8 bits are for the checksum, they gain only 124 bits of information. So, there are still 256 - 124 = 132 bits left to attack. Now, the brute attacker knows the checksum, so we need to understand how many combinations among the 2^132 generate the same checksum: those that don't can be immediately discarded. The combinations that yield the same CS are about 2^124 (2^132 / 2^8).
For an unknown 12-word phrase, the search space is 2^128. So, the former is slightly less secure, but negligibly so.