Now, suppose a language L is decided by alternating Turing machine M. Then that Turing machine performs some number of existential branching, followed by some number of universal branching.
Consider the sub-trees of the computation path whose roots are the first universal step along the path. For each such sub-tree, M is (by definition) performing a computation. Thus, it is deciding a language in (Note: it may possibly be a different language for each sub-tree, but it does not matter).
Now consider a nondeterministic machine S that behaves just as M does, but replaces each of the computation sub-trees discussed above with a deterministic computation by a machine in (this is feasible because ). Note that this machine only contains one run of existential branches, each extended with a deterministic computation that uses an SAT oracle and runs in polynomial time.
If assume be the maximum number of steps taken by the alternating machine before the universal branches start, and be the maximum number of steps taken by any of the machines which have substituted for the universal branching, then the running time of S is bounded by . Note that the term is a composition of functions, because the sub procedures are computing with inputs that may be longer than n (but must be smaller than or equal to, since only steps have been executed at the time the sub-procedures are used).
Therefore, both a and p polynomials, so is their composition. Hence, S is in