Heller M, Woodin W.H. (eds.) Infinity. New research frontiers (CUP, 2011)(ISBN 1107003873)(O)(327s)_MAml_

interpretation of zf in mbt 161Lemma 8.35The interpretation of Infinity holds.proof By Lemma 8.17, let R be a prewellordering with no greatest element,where every nonempty “subset” of any {y: R(y,x)} has an R greatest element.Because of Lemmas 8.6 and 8.9, we are in the standard situation with the secondform of the Peano axioms for the natural numbers. Therefore, we can buildarithmetic and definitions by induction. It is then straightforward to build a copy(fld(R), S, = R ) of the set theorist’s (V (ω), ∈, =), where S is a binary “relation”on fld(R) that respects = R . We can put this in the form of a system (fld(R), = R ,S, x), which will witness Infinity under our interpretation.Lemma 8.36 The interpretation of all axioms of Z holds. In addition, theinterpretations of Foundation (scheme) and Transitive hold.proof By Lemmas 8.26, 8.28, 8.29, 8.30, 8.32, 8.33, 8.34, and 8.35.Theorem 8.37 Z is interpretable in B + SDE. In fact, Z + Foundation(scheme) + Transitive is interpretable in B + SDE.proof By Lemma 8.36 and 8.34.6.9 Interpretation of ZF in MBTWe now show that ZF is interpretable in B + SDE + UI.By Lemma 8.36, we have only to prove the interpretation of Collection in B + SDE +UI, using the same interpretation of L(>, ≫, =)inL(∈, =) as we used in Section 6.8.By Theorem 5.2, MBT proves B + SDE + UI. Hence, in this section we will haveestablished that the interpretation of Section 6.8 is also an interpretation of ZF in MBT.Before we prove the interpretation of Collection from ZF, we first show that B +SDE + UI proves a form of Collection in L(>, =).Lemma 9.1 Let n ≥ 2. Suppose x 1 ,...,x n is coded by c using y,a 1 ,...,a n .Then (∀u)((∃v)(u < ∗ v< ∗ c) ↔ u = x 1 ∨ ...∨ u = x n ∨ u = a 1 ∨ ...∨ u =a n ).proof Let n, x 1 ,..., x n , c, y, a 1 ,..., a n be as given. Then x 1 ,..., x n ≤y,A(y,a 1 ,..., a n ). Also, let u 1 > ex {x 1 ,a 1 }∧...∧ u n > ex {x n , a n } ∧ y> ex{u 1 ,...,u n }, where A(y,a 1 ,...,a n ).We claim that the u’s are incomparable under ≤. To see this, suppose u i ≤ u j .Then a i ex {x i , a i }and A(y,a 1 ,...,a n ).Collection (>, =). (∀y 1 ,..., y n