#theoretical-cs
1 messages · Page 1 of 1 (latest)
Assume SUBEXP = PSPACE and EXP = PSPACE
Then SUBEXP = EXP, contra time hierarchy, no? So you could prove anything in that case 😅
I think that's the main sticking point, SUBEXP and EXP are already separated
Exactly, that means that SUBEXP = PSPACE -> EXP != PSPACE
my idea was proof by contradiction showing that SUBEXP = PSPACE -> EXP != PSPACE and SUBEXP != PSPACE -> EXP != PSPACE, therefore EXP != PSPACE
the SUBEXP != PSPACE -> EXP != PSPACE almost, gets tantalizingly close to working
but that splits into 2 possibilities, either PSPACE contains SUBEXP or it doesn't contain SUBEXP
if PSPACE doesnt contain SUBEXP you get a trivial contradiction, because SUBEXP is contained strictly within EXP, that means that there is a problem in EXP not in PSPACE, so PSPACE != EXP
The thing is, we also need to acount for PSPACE containing SUBEXP which seems unlikely but a P vs NP type seperation
but like there should intuitionally be some problem in SUBEXPSPACE that needs at least SUBEXPSPACE that can also be solved in SUBEXPTIME
so this reduces to SUBEXP vs SUBEXPSPACE
if SUBEXP = SUBEXPSPACE then this is trivial
if SUBEXP != SUBEXPSPACE then uhh
fun