Quantified CNF: is not this a proof that NP = PSPACE?

Question

If boolean formula $f$ is in CNF it is known that:$$\forall x \exists y_1 \exists y_2 .. \exists y_n : f(x, y_1, y_2..y_n) :\iff \exists y_1 \exists y_2 .. \exists y_n : f(0, y_1, y_2..y_n) \land f(1, y_1, y_2..y_n) :\iff \exists y_1 \exists y_2 .. \exists y_n : f(y_1, y_2..y_n) :\iff f(y_1, y_2..y_n)$$ Last formula is non-quantified CNF. For multiple $\forall$-quantified variables same reduction is also possible. SAT is NP problem, QSAT is PSPACE-complete (even in CNF). Does this involve that NP = PSPACE?

Answer 1

Each of your purported equivalences is problematic:

In the first one, you seem to be moving $\forall x$ willy-nilly inside the $\exists$s. This does not preserve meaning: $$ \forall x\exists y (x=y) \qquad\text{and}\qquad \exists y(0=y\land 1=y) $$ are not equivalent -- the first one is obviously true and the second one just as obviously false.

In the second one, $f$ inexplicably changes from having $n+1$ parameters to having only $n$. What is going on there? Is it even the same $f$?

In the third one, where do the quantifiers suddenly go? You seem to be asserting that the formula $\exists y_1\cdots \exists y_n(f(y_1,\ldots,y_n))$ which has no free variables is equivalent to $f(y_1,\ldots,y_n)$ which has $n$ of them. Unless you imagine that the truth value of $f$ does not depend on its variables at all (which you don't seem to have any reason to), this can't be true.

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Finally, your plan seems to be to prove that PSPACE${}\subseteq{}$NP by reducing QSAT to SAT. However, even if your equivalences worked, if you have $n$ universal quantifiers repeating your first rewrite for each of them would blow up the formula by a factor of $2^n$, so it is not a polynomial-time reduction and would not say anything about QSAT being in NP.