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

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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.
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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. We first claim that m > n 4m 2 +4n+1 gB. To see this, let m > n, 4m 2 +4n+1 gB. Note that 4m 2 +4n+1 1 mod 4. Hence for some n’,m’ B, m’ > n’, we have 4m 2 +4n+1 = g(n’,m’) = 16m’ 2 +4n’+1. Since n B and n’ B, we have n n’. Also 16m’ 2 - 4m 2 = 4n - 4n’. 4m’ 2 - m 2 = n - n’. (2m’ - m)(2m’ + m) = n - n’. 2m’ - m 0. 2m’ + m > 2n’ + n. 2n’ + n < |(2m’ - m)(2m’ + m)| = |n - n’| n + n’.
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3 n’ < 0.
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