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Jul 30, 2021 at 15:48 vote accept Alexey Kholodkov
Jul 27, 2021 at 13:24 comment added idmean @AlexeyKholodkov Yes, I think so, but it's not really interesting without time bounds. A NTM $M$ can compute the answer during the reduction and output a tautology if $\forall (y_1...y_j) \exists (z_1...z_k) f(x_1...x_i, y_1...y_j, z_1...z_k)$ and a falsum otherwise. All possible assignments for $y_1 ... y_j$ are countable. There are $2^j$ such assignments, which means we need $O(j)$ bits to store our 'iteration pointer'. For each assignment the NTM $M$ simulates another NTM deciding SAT that runs in polynomial space (which exists because $SAT \in NP \subseteq PSPACE$) on the formula.
Jul 27, 2021 at 12:46 comment added Alexey Kholodkov That answers practical part of problem. But I'm still interested to know - is there polinomial SPACE reduction?
Jul 27, 2021 at 6:26 history answered idmean CC BY-SA 4.0