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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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