Suppose I have a language $L$ (over alphabet $\Sigma$), such that $$ w \in L \iff (\forall x \in \Sigma^*) (\exists y \in \Sigma^*) P(x,y,w). $$ and I can give a turing machine that decides $P(x,y,w)$ in polynomial space in $w$, where $x$ and $y$ are given as separate read-only, forward-only tapes. That is, on input $x$ in tape 1, $y$ in tape 2, and $w$ in tape 3, I use only a polynomial-in-$w$ amount of tape 3's space, and I only move forwards in tape 1 and tape 2.
Does this imply $L \in \text{PSPACE}$?
I feel it should, because if there were only one quantifier, say ($\exists y \in \Sigma^*) P(y,w)$, then $\text{NPSPACE} = \text{coNPSPACE} = \text{PSPACE}$ implies the result. But somehow for two quantifiers I can't figure out how to prove this. I should be able to eliminate the tape $y$ using $\text{NPSPACE} = \text{PSPACE}$, but I don't know if this result still holds in the presence of the additional tape $x$.