Algebra (Translation of Cours d'algèbre) - download pdf or read online

By Roger Godement

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For a a maximal 1-1 assignment let Ca be the first clause in 7r satisfying 1-3. For ,B a 1-1 assignment, not necessarily maximal, let C'5 be the first clause Ca for some maximal a extending P. Claim 2. Assume that fi is a 1-1 assignment which assigns (n /4) ones. Then there are at least (n /4) + 1 i 's for which either there is j such that pij E C'5 or there are at least (n/2) j s such that pij E Co. To prove the claim let C' Ca for a maximal 1-1 assignment a. Define: COL{i 12j, -'pij E CO} {i i E dom(a) A (Vj, -pi j 0 CO) A 3'-(n/2) j, Pij E CO} COL+ io := the i 0 dom(a) such that for exactly (n/2) j's, pij E CO I A:={pij I a(pij) = 1A,8(pij)=0} The sets COL-,COL+and(i0)are mutually disjoint (as COL-f1(n\dom(a)) _ 0 holds because CO is not satisfied by a).

Clearly the number of resolution inferences is bounded by the number inferences in ir; we use the assumption that it is treelike to maintain this bound during the simulation of a V : left inference. 4 every resolution refutation of -, PHP'+' must contain at least 22(n) clauses. Hence n must contain 2n0) inferences, which is 2""' for m=ISO=0(n3). 15 the sequent S has a size m 0(1) LK-proof. D. 3 has a much stronger lower bound for the sequent S. The next proposition will show that every treelike proof can be balanced with only a polynomial increase in size.

There is a sequent S of size I S1 = m such that every cut free, treelike LK proof of S has at least 2"(" sequents. Moreover, the sequent S has an L K proof of size m 00). 2. Let 7r be a cut-free, treelike LK-proof of S. ,q, Basic propositional logic 38 where Dr are disjunctions of literals p;j or -p;j and qs are atoms among pig. , D (we identify a disjunction of literals with the clause consisting of those literals) with at most k resolution inferences. If Z is an initial sequent in r or derived by a structural rule, then there is nothing to prove.

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Algebra (Translation of Cours d'algèbre) by Roger Godement

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