Infinite-valued TQBF The TQBF problem is defined for two-valued logic. Consider a similar problem in a setting of infinitely many truth values.
This is to prevent propositional quantification from being trivially translated away. 
While it is well known that multi-valued Boolean algebra without quantification collapses to two-valued,
is this also the case for infinite-valued quantified boolean logics? Does the infinite-valued TQBF problem collapse into two-valued TQBF?
 A: The reduction of QBF to propositional logic only works if you interpret the quantifiers as quantifying over the two-element boolean algebra, $\Bbb{B} = \{\top, \bot\}$. Consider the following QBF that says any $x$ is either at the top or the bottom of the lattice:
$$\forall x [(\forall y (y \Rightarrow x)) \lor (\forall y (x \Rightarrow y))]$$
This holds in $\Bbb{B}$, but not in any Boolean algebra with more than two elements. E.g., in the four-element boolean algebra $\Bbb{B} \times \Bbb{B}$, with $x = (\top, \bot)$ and $y = (\bot, \top)$, neither  $x \Rightarrow y$ nor $y \Rightarrow x$ holds.
Without quantifiers, the language of propositional logic is too weak to distinguish between $\Bbb{B}$ and any other Boolean algebra, e.g., $(x \Rightarrow y) \lor (y \Rightarrow x)$ holds in $\Bbb{B} \times \Bbb{B}$ just as it does in $\Bbb{B}$. (To prove the general claim you use Stone's representation theorem for Boolean algebras to embed an arbitrary algebra in $\Bbb{B}^I$ for some index set $I$. A quantifier-free formula that is true in each factor of a product algebra is true in the product algebra.)
