3.12ABBC030109

3.12ABBC030109 - 1 3.12. ABBC. Recall the following reduced...

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Unformatted text preview: 1 3.12. ABBC. Recall the following reduced table for AB 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 table for BC is obtained from the reduced table for AB via the permutation that sends A to B, B to C, and C to A. We use 1'-6' to avoid confusion. REDUCED BC 1. B . fB C . gB. INF. AL. ALF. FIN. NON. 2. B . fB C . gC. INF. AL. ALF. FIN. NON. 3. B . fB C . gA. INF. AL. ALF. FIN. NON. 4. A . fB C . gB. INF. AL. ALF. FIN. NON. 5. A . fB C . gC. INF. AL. ALF. FIN. NON. 6. A . fB C . gA. INF. AL. ALF. FIN. NON. This results in 36 ordered pairs, which we divide into six cases. We begin with two Lemmas. We will determine the status of all attributes INF, AL, ALF, FIN, NON, for all ordered pairs. LEMMA 3.12.1. C . fX B . gY, Z . fB C . gW has INF, FIN. Proof: Let f be as given by Lemma 3.2.1. Let g ELG be given by g(n) = 2n+1. Let C . fX B . gY, Z . fB C . gW, where A,B,C are nonempty. Clearly fB 2N C. By C B gY, we have fB 2N B. Hence by Lemma 3.2.1, fB is cofinite. Hence B is infinite. This establishes that FIN. Also Z is finite. This establishes that INF. QED LEMMA 3.12.2. C . fX B . gY, Z . fB C . gW, B fB = has NON. 2 Proof: We can continue the proof of Lemma 3.12.1. Using fB is cofinite and B is finite, we obtain an immediate contradiction from B fB = . QED We use Lemmas 3.12.1 and 3.12.2 in cases 5,6 below. part 1. A . fA B . gA. 1,1. A . fA B . gA, B . fB C . gB. INF. AL. ALF. FIN. NON. 1,2. A . fA B . gA, B . fB C . gC. INF. AL. ALF. FIN. NON. 1,3. A . fA B . gA, B . fB C . gA. INF. AL. ALF. FIN. NON. 1,4. A . fA B . gA, A . fB C . gB. INF. AL. ALF. FIN. NON. 1,5. A . fA B . gA, A . fB C . gC. INF. AL. ALF. FIN. NON. 1,6. A . fA B . gA, A . fB C . gA. INF. AL. ALF. FIN. NON. The following pertains to 1,4, 1,6. LEMMA 3.12.3. A . fA B . gA, A . fB C . gX has INF, ALF provided X {A,B}, even for EVSD. Proof: Let f,g EVSD. Let n be sufficiently large. By Lemma 3.2.5, let A [n, ) be infinite, where A is disjoint from f(A fA) g(A fA). Let B = (A fA)\gA, and C = (A fB)\gX....
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3.12ABBC030109 - 1 3.12. ABBC. Recall the following reduced...

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