my forty years on his shoulders - Department of Mathematics

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46These two theorems are official statements in BRT. In thecomplementation theorem, A is unique.We now write them in BRT form.THIN SET THEOREM. For all f ∈ MF there exists A ∈ INF such that fA ≠ N.COMPLEMENTATION THEOREM. For all f ∈ SD there exists A ∈ INF such thatfA = N\A.The thin set theorem lives in IBRT in A,fA. There are only 2 2^2 = 16statements in IBRT in A,fA. These are easily handled.The complementation theorem lives in EBRT in A,fA. There are only 2 2^2 =16 statements in IBRT in A,fA. These are easily handled.For EBRT/IBRT in A,B,C,fA,fB, fC,gA,gB,gC, we have 2 2^9 = 2 512statements. This is entirely unmanageable. It would take several major new ideasto make this manageable.DISCOVERY. There is a statement in EBRT in A,B,C,fA,fB, fC,gA,gB,gC that isindependent of ZFC. It can be proved in SMAH+ but not in SMAH, even withthe axiom of constructibility.Here SMAH+ = ZFC + (∀n)(∃κ)(κ is a strongly k-Mahlo cardinal). SMAH= ZFC + {(∃κ)(κ is a strongly k-Mahlo cardinal} k .The particular example is far nicer than any “typical” statement in EBRTin A,B,C,fA,fB,fC,gA,gB,gC. However, it is not nice enough to be regarded assuitably natural.
47Showing that all such statements can be decided in SMAH+ seems to betoo hard.What to do? Look for a natural fragment of full EBRT inA,B,C,fA,fB,fC,gA,gB,gC that includes the example, where we can decide allstatements in the fragment within SMAH+.We also look for a bonus: a striking feature of the classification that is itselfindependent of ZFC. Then we have a single natural statement independent ofZFC.In order to carry this off, we need to use the function class ELG offunctions ofs expansive linear growth.These are functions f:N k → N such that there exist constants c,d > 1 suchthatc|x| ≤ f(x) ≤ d|x|holds for all but finitely many x ∈ N k .TEMPLATE. For all f,g ∈ ELG there exist A,B,C ∈ INF such thatX ∪. fY ⊆ V ∪. gWP ∪. fR ⊆ S ∪. gT.Here X,Y,V,W,P,R,S,T are among the three letters A,B,C.Note that there are 6561 such statements. We have shown that all of thesestatements are provable or refutable in RCA 0 , with exactly 12 exceptions.These 12 exceptions are really exactly one exception up to the obvioussymmetry: permuting A,B,C, and switching the two clauses.The single exception is the exotic case:
Page 1: 1MY FORTY YEARS ON HIS SHOULDERSbyH
Page 4 and 5: 4theorem is demonstrably implied by
Page 6 and 7: 6At the outer limits, normal mathem
Page 8 and 9: 8“Thus, according to Gödel, the
Page 11 and 12: 11(Rosser 1936) is credited for sig
Page 13 and 14: 13(Davis 1973), (Matiyasevich 1993)
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Page 19 and 20: 19PROV[x 1 /#(A)] → PROV[x 1 /PR(
Page 21 and 22: 21FORMAL SECOND INCOMPLETENESS (PA(
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Page 37 and 38: 37P.J. Cohen proved that if ZF is c
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Page 57 and 58: 57Monographs, 24, Oxford: Clarendon
Page 59 and 60: 59Vol. 41, No. 3, September pp. 209
Page 61 and 62: 61Friedman, H., Robertson, N., and
Page 63 and 64: 63Society, 8:437-479, 1902.Hilbert,
Page 65 and 66: 65Mostowski, A. 1952. Sentences und
Page 67 and 68: 67Scott, D.S. 1961. “Measurable c
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my forty years on his shoulders - Department of Mathematics

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