Talk:Sipser–Lautemann theorem

lemma 1 looks strange
In Lemma 1 you need
 * $$\sum_r \frac{1}{2^{|x| \cdot |r|}} = \frac{1}{2^{|x|}}$$

for which there are $$2^r$$ summands. Though, this equality is just plain wrong, as it is equivalent to
 * $$2^{|r|} \frac{1}{2^{|x| \cdot |r|}} = \frac{1}{2^{|x|}}$$
 * $$2^{|r|} \cdot 2^{|x|} = 2^{|x| \cdot |r|}$$
 * $$2^{|r| + |x|} = 2^{|x| \cdot |r|}$$
 * $$|r| + |x| = |x| \cdot |r|$$ —Preceding unsigned comment added by 141.24.51.251 (talk) 13:44, 23 March 2011 (UTC)

I think both proofs are the same
The page gives two proofs for the theorem and claims the first is based on Sipser(-Gacs) and the second on Lautemann but (imo) both proofs are the same (based on Lautemann), Sipser used hash functions.

Ps. I really don't think it's correct to call it Sipser-Lautemann theorem as the first person to prove the result was Gacs, see e.g. Trevision calls it Gács-Sipser-Lautemann: http://lucatrevisan.wordpress.com/2010/04/27/cs254-lecture-5-the-polynomial-hierarchy/ — Preceding unsigned comment added by 80.98.160.141 (talk) 05:23, 2 February 2014 (UTC)