On Closure Properties of Bounded 2-sided Error Com- the Equivalence Problem for Regular Expres- Sions with Squaring Requires Exponential Space. In

17 Let L p d A via k-tt-condition f(x) = h; y 1 ;. . .; y k i, that is, for each x 2 6 3 , x 2 L if and only if Ass(f(x)) \ A 6 = ;. Without loss of generality, assume that the length of each y 2 Ass(f(x)) is exactly q(jxj), q 2 I Pol. Let A 2 9 1 SP 1 K. Then, there exist a set B 2 SP 1 K and a polynomial p such that for every y 2 6 3 , y 2 A if and only if there exists a string z, jzj = p(jyj), such that hy; zi 2 B. Let r(n) := p(q(n)) for n 0. 1 if hy; zi 2 B 0 otherwise. For each hx; zi, jzj = r(jxj), let m(hx; zi) := P y2Ass(f (x)) g(hy; zi). Then, by Proposition 2.2 and the above observation about the functions, h dened below is a GAP 1 K function: h(hx; zi) = 8 > < > : m(hx; zi) 0 P k l=1 (l 0 1) 1 k l (m(hx; zi)) if jzj = r(jxj) 0 otherwise. Furthermore, since h(hx; zi) = 1 if m(hx; zi) > 0 and h(hx; zi) = 0 otherwise, h witnesses that D := fhx; zi : jzj = r(jxj) ^ h(hx; zi) = 1g is in SP 1 K. Assume x 2 L. Then, there is a string y 2 Ass(f(x)) \ A. Thus, m(hx; zi) > 0 for some z, jzj = r(jxj), and hence, hx; zi 2 D. For x 6 2 L, Ass(f(x)) \ A = ;. That is, for every y 2 Ass(f(x)) and every z 2 6 r(jxj) , we have g(hy; zi) = 0 and thus m(hx; zi) = 0. It follows that hx; zi 6 2 D for every z, jzj = r(jxj). Thus, D witnesses that L 2 9 1 SP 1 K. 2 Acknowledgments The author is very grateful to J org Vogel and Arfst Nickelson for helpful discussions, and to J org Vogel for carefully proofreading the technical reports some results of which are presented in this paper. He thanks Lane Hemaspaandra for improving the English style of the paper and for his kind guidance, useful advice, and wonderful collaboration during the author's stay in Rochester. 16 loss of generality, suppose k = kAss(f(x))k to be odd for each …