AP equals PSPACE

To prove $\mathrm{P}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}\subseteq \mathrm{A}\mathrm{P}$, it is enough by [L2] to show that TQBF lies in AP. On input a quantified Boolean formula, an alternating machine reads the quantifier prefix from left to right: at an existential quantifier it enters an existential state and chooses the bit, and at a universal quantifier it enters a universal state and branches over both bits. After the prefix is exhausted, it deterministically evaluates the matrix. By [L3], this machine accepts exactly the true formulas, so TQBF is in AP.

For the reverse inclusion, let $L\in \mathrm{A}\mathrm{P}$ and choose an alternating machine $M$ for $L$ running in time $p(n)$. On input $x$, a deterministic procedure evaluates the acceptance of the computation tree depth-first: at an accepting or rejecting leaf it returns the corresponding truth value; at an existential configuration it ORs the recursive values of the children; at a universal configuration it ANDs them. This exactly matches the acceptance rule in [L1].

Now let $L\in \mathrm{P}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}$. By [L2], there is a polynomial-time reduction $f$ from $L$ to TQBF. An alternating machine for $L$ first computes $f(x)$ deterministically and then runs the TQBF machine from step 1.1 on $f(x)$. This still uses only polynomial time, so [L1] gives $L\in \mathrm{A}\mathrm{P}$. Thus $\mathrm{P}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}\subseteq \mathrm{A}\mathrm{P}$.

The recursion depth is at most $p(\mathrm{\mid }x\mathrm{\mid })$ because every branch halts within that many steps. One stack frame stores the current configuration and the remaining depth counter, both of polynomial size, so the whole depth-first evaluation uses polynomial space. Therefore $L\in \mathrm{P}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}$.

Steps 2.1 and 2.2 prove both containments, so $\mathrm{A}\mathrm{P}=\mathrm{P}\mathrm{S}\mathrm{P}\mathrm{A}\mathrm{C}\mathrm{E}$.