The name of the following function is motivated by the (again unproven) observation that for ,
PROOF: If , then and the claim is trivial. Let and . By Lemma 7.1.1, , which implies
PROOF: Let and . By Corollaries 7.1.2 and 7.2.2, . By Lemma 7.1.3, . Consequently, since is either or , , i.e., is -exact.
PROOF: We first consider the case . Let , , , and . We shall show that . Note that by Lemmas 7.1.2, 7.1.6, and 7.1.4, .
Case 1: and .Case 1: and .
. Since is -exact, Lemma 6.4.6 implies .
Case 2: and .
Since is -exact, Lemma 4.2.16 implies that is not -exact; similarly, since is -exact, is not -exact. Therefore,
Case 3: and .
By Lemma 7.1.3, is -exact and is -exact. By Lemma 7.1.6, , and it follows from Lemma 6.1.11 that . Thus, and by Lemma 6.4.6, .
Case 4: and .
In this case, and , which is not -exact. Thus,
The proof is completed by induction on . If , then by Lemma 6.4.11,
PROOF: Let and . Suppose . Then , , and hence . By Lemmas 6.4.7, 6.4.6, and 7.1.3,
To prove the second inequality, we note that if , then by Lemmas 6.4.7, 6.4.6, and 7.1.3,
PROOF: Let and . If , then by Lemma 7.1.4, , so that and by Lemma 4.2.16,
David Russinoff 2017-08-01