2.2.A,fA071110

2.2.A,fA071110 - 1 2.2. EBRT, IBRT in A,fA. This section is...

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1 2.2. EBRT, IBRT in A,fA. This section is intended to be a particularly gentle introduction to BRT classification theory. It is wholly subsumed by section 2.3. Recall the five main BRT settings introduced at the beginning of this Chapter: (SD,INF), (ELG,INF), (MF,INF), (ELG SD,INF), (EVSD,INF). We begin with the BRT fragments α = EBRT in A,fA on these five BRT settings. As discussed in sections 1.1 and 2.1, classification of these BRT fragments amounts to a determination of the true α assertions, which take the form 1) ( f V)( A K)( ϕ ) where ϕ is an α equation (since we are in the environment EBRT). As discussed in sections 1.1 and 2.1, we work, equivalently, with the α statements, which take the form 1’) ( f V)( A K)(S) where S is an α format, interpreted conjunctively. Recall that in EBRT, S is correct if and only if 1') holds. S is incorrect if and only if 1') fails. In this case of EBRT in A,fA, the number of elementary inclusions is 4, and the number of formats is 16. Since 16 is so small, we might as well list all of the formats S. It is most convenient to list the formats S in increasing order of their cardinality – which is 0–4. The four A,fA elementary inclusions are as follows. See Definition 1.1.36. (These do not depend on the BRT environment or BRT setting). A fA = . A fA = U. A fA.
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2 fA A. According to Definition 1.1.13 of the universal set U in BRT settings, we see that on our five BRT settings, U is N. Before beginning this tabular EBRT classification, we organize the nontrivial mathematical facts that we will use. THEOREM 2.2.1. Let f EVSD and E A N, where E is finite, A is infinite, and E fE = . Also let D N be infinite. There exists infinite B such that E B A, B fB = , and neither A nor D are subsets of B fB. Moreover, this is provable in RCA 0 . Proof: Let f,E,A,D be as given. Let n D be such that n > max(E fE), and |x| n f(x) > |x|. Let t > n, t A. We define an infinite strictly increasing sequence n 1 < n 2 ... by induction as follows. Define n 1 = min{m A: m > t}. Suppose n 1 < . .. < n k have been defined, k 1. Define n k+1 to be the least element of A that is greater than n k and all elements of f(E {n 1 ,...,n k }). Let B = E {n 1 ,n 2 ,...} A. Clearly B fB = . Also n,t B, and so A,D are not subsets of B fB. QED In the applications of Theorem 2.2.1 to the tabular EBRT classification below, we can ignore E,A,D. We just use that for all f EVSD, there exists infinite B N such that B fB = . Here is the other fact that we need. COMPLEMENTATION THEOREM. For all f SD there exists A INF such that fA = N\A. We proved the Complementation Theorem in section 1.3 within
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2.2.A,fA071110 - 1 2.2. EBRT, IBRT in A,fA. This section is...

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