3.6AABA030109

3.6AABA030109 - 1 3.6. AABA. Recall the reduced AA table...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 3.6. AABA. Recall the reduced AA table from section 3.4. REDUCED AA 1. B . fA A . gA. INF. AL. ALF. FIN. NON. 2. B . fA A . gB. INF. AL. ALF. FIN. NON. 3. B . fA A . gC. INF. AL. ALF. FIN. NON. 4. C . fA A . gA. INF. AL. ALF. FIN. NON. 5. C . fA A . gB. INF. AL. ALF. FIN. NON. 6. C . fA A . gC. INF. AL. ALF. FIN. NON. Recall the reduced AB table from section 3.5. REDUCED AB 1. A . fA B . gA. INF. AL. ALF. FIN. NON. 2. A . fA B . gB. INF. AL. ALF. FIN. NON. 3. A . fA B . gC. INF. AL. ALF. FIN. NON. 4. C . fA B . gA. INF. AL. ALF. FIN. NON. 5. C . fA B . gB. INF. AL. ALF. FIN. NON. 6. C . fA B . gC. INF. AL. ALF. FIN. NON. The reduced BA table is obtained from the reduced AB table by switching A,B. We use 1-6 to avoid any confusion. REDUCED BA 1. B . fB A . gB. INF. AL. ALF. FIN. NON. 2. B . fB A . gA. INF. AL. ALF. FIN. NON. 3. B . fB A . gC. INF. AL. ALF. FIN. NON. 4. C . fB A . gB. INF. AL. ALF. FIN. NON. 5. C . fB A . gA. INF. AL. ALF. FIN. NON. 6. C . fB A . gC. INF. AL. ALF. FIN. NON. We consider all 36 pairs, arranged in cases according to the first clause of the ordered pair. The status of all of our proposition attributes are determined by the reduced AA table except AL and NON. Thus, we need only obtain the status of AL and NON. part 1. B . fA A . gA. 1,1. B . fA A . gA, B . fB A . gB. INF. AL. ALF. FIN. NON. 2 1,2. B . fA A . gA, B . fB A . gA. INF. AL. ALF. FIN. NON. 1,3. B . fA A . gA, B . fB A . gC. INF. AL. ALF. FIN. NON. 1,4. B . fA A . gA, C . fB A . gB. INF. AL. ALF. FIN. NON. 1,5. B . fA A . gA, C . fB A . gA. INF. AL. ALF. FIN. NON. 1,6. B . fA A . gA, C . fB A . gC. INF. AL. ALF. FIN. NON. LEMMA 3.6.1. There exists g ELG SD such that the following holds. Suppose A gB is cofinite and A gA = . Then A B. We can require that rng(g) 2N+1. Furthermore, we can require that for all X and n, 4n+3 gX n X. Proof: Define g ELG SD as follows. For all m > n, define g(n,4m 2 +4n+1) = 16m 2 +4n+1. For all other pairs p,q, define g(p,q) = 4|p,q|+3. Let A gB be cofinite and A gB = . Let n A\B. We derive a contradiction. Note that the last two requirements on g hold....
View Full Document

This note was uploaded on 08/05/2011 for the course MATH 366 taught by Professor Joshua during the Fall '08 term at Ohio State.

Page1 / 10

3.6AABA030109 - 1 3.6. AABA. Recall the reduced AA table...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online