2.5.A,B,fA,fBELG062310

2.5.A,B,fA,fBELG062310 - 1 2.5. EBRT in A,B,fA,fB, on...

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1 2.5. EBRT in A,B,fA,fB, on (ELG,INF). In this section, we use the tree methodology described in section 2.1 to analyze EBRT in A,B,fA,fB, on (ELG,INF) and (EVSD,INF). We handle both BRT settings at once, as they behave the same way for EBRT in A,B,fA,fB, . In particular, we show that they are RCA 0 secure (see Definition 1.1.43). Some of this treatment is the same as for EBRT in A,B,fA,fB, on (SD,INF) given in section 2.4. However, many new features appear that makes this section considerably more involved than section 2.4. A key difference between EBRT in A,B,fA,fB, on (SD,INF) and on (ELG,INF) is that the Compelmentation Theorem holds on (SD,INF), yet fails on (ELG,INF). E.g., it fails for f(x) = 2x. Let f:N k N be partial. Define the following series of sets by induction i 1. S 1 = N. S i+1 = N\fS i . LEMMA 2.5.1. S 2 S 4 S 6 ... ... S 5 S 3 S 1 . I.e., for all i 1, S 2i S 2i+2 S 2i+1 S 2i-1 . Proof: We argue by induction on i 1. The basis case is S 2 S 4 S 3 S 1 . To see this, clearly S 3 S 1 . N\S 1 N\S 3 . S 2 S 4 . S 2 S 1 . N\S 1 N\S 2 . S 2 S 3 . fS 2 fS 3 . N\fS 3 N\fS 2 . S 4 S 3 . Now assume the induction hypothesis S 2i S 2i+2 S 2i+1 S 2i-1 .
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2 Then fS 2i fS 2i+2 fS 2i+1 fS 2i-1 . N\fS 2i-1 N\fS 2i+1 N\fS 2i+2 N\fS 2i . S 2i S 2i+2 S 2i+3 S 2i+1 . fS 2i fS 2i+2 fS 2i+3 fS 2i+1 . N\fS 2i+1 N\fS 2i+3 N\fS 2i+2 N\fS 2i . S 2i+2 S 2i+4 S 2i+3 S 2i+1 . QED LEMMA 2.5.2. Let f:N k N be partial, where each f -1 (n) is finite. Let A = S 2 S 4 ..., and B = S 1 S 3 ... . Then A B, A = N\fB, B = N\fA. Proof: Let A,B be as given. By Lemma 2.5.1, A B. Fix i 1. S2i = N\fS2i-1, S2i fS2i-1 = , S2i fB = . Since i 1 is arbitrary, A fB = . I.e., A N\fB. Since S 2i+1 = N\fS 2i , we see that for all j i, S 2i+1 fS 2j = . Hence S 2i+1 fA = . Since i 1 is arbitrary, B fA = . I.e., B N\fA. Now let n N\fB. We claim that for some j 0, n fS 2j+1 . Suppose that for all j 0, n fS 2j+1 . Since f -1 (n) is finite, there exists x f -1 (n) which lies in infinitely many S 2j+1 . Hence there exists x f -1 (n) such that x B. Therefore n fB. This establishes the claim. Fix j 0 such that n fS 2j+1 . Then n S 2j+2 , and so n A. This establishes that A = N\fB. Finally, let n N\fA. Then for all i, n fS 2i . Hence for all j, n S 2j+1 . Therefore n B. This estabslihes that B = N\fA. QED LEMMA 2.5.3. Let f:[0,n] k [0,n] be partial, n 0. There exist A B [0,n] such that A = [0,n]\fB and B = [0,n]\fA. Proof: Let n,f be as given. Obviously f:Nk
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2.5.A,B,fA,fBELG062310 - 1 2.5. EBRT in A,B,fA,fB, on...

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