What's wrong with this proof that NP=Co-NP implies NP=PSPACE Below is a short informal proof that NP=co-NP implies NP=PSPACE. What's wrong with the proof? 

Assuming NP=co-NP, an instance F of TQBF can be solved by a polynomial NDTM this way:


  Non-deterministically guess the values for all the variables corresponding to existential quantifiers in F.
  The resulting formula F1 (with the guessed values substituted in the formula) will only have universal quantifiers, which makes F1 a Co-NP problem instance.
  Since NP=co-NP, there's a short certificate to verify F1's truth.
  Guess this short certificate too.


That is, the NDTM non-deterministically guesses two things: The values for all the variables corresponding to existential quantifiers in F, and the short certifcate for the intermediate (co-NP) instance F1.
A totally quantified boolean formula consists of an alternating sequence of quantifiers. This sequence is not of length 2, but arbitrarily long. That is the mistake you're making: an instance can be represented as a pair (P, phi) where phi is a propositional formula and L = (X_1, ..., X_k) is a partition of the set of variables occuring in phi, such that X_i are the variables in the i'th quantifier (which is an existential one if i is odd, and universal otherwise). You are assuming that this instance is equivalent to (L', phi), L' = (X_1', X_2') where X_1' is the union of the odd-indexed X_i, and X_2' is the union of the even-indexed X_i. Please verify for yourself that this is not necessarily true.