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A premouse inheriting strong cardinals from V

Farmer Schlutzenberg1

Abstract

We identify a premouse inner model L[E], such that for any coarsely iterablebackground universe R modelling ZFC, L[E]R is a proper class premouse of Rinheriting all strong and Woodin cardinals from R, and iteration trees on L[E]R

lift to coarse iteration trees on R.We also prove that a slight weakening of (k + 1)-condensation follows from

(k, 1 + 1)-iterability in place of (k, 1, 1 + 1)-iterability. We also prove thatfull (k+1)-condensation follows from (k, 1+1)-iterability and (k+1)-solidity.

In order to prove these results we introduce structures generalizing bicephali,and prove some general theorems regarding these structures.

Keywords: bicephalus, condensation, normal iterability, inner model, strongcardinal2010 MSC: 03E45, 03E55

1. Introduction

Consider fully iterable, sound premice M,N , such that M = = N .

Under what circumstances can we deduce that either M E N or N EM? Thisconclusion follows if is a cutpoint of both models. By [2, Lemma 3.1],1 theconclusion follows if is a regular uncountable cardinal and there is no premousewith a superstrong extender. We will show that if

M ||(+)M = N ||(+)N

and M,N have a certain joint iterability property, then M = N .The joint iterability required, and the proof that M = N , is motivated by

the bicephalus argument of [3, 9]. The bicephali of [3] are structures of theform B = (P,E, F ), where both (P,E) and (P, F ) are active premice. If B is aniterable bicephalus and there is no iterable superstrong premouse then E = F(see [3, 9] and [5]); the proof is by comparison of B with itself. In 3, we willconsider a more general form of bicephali, including, for example, the structureC = (,M,N), where ,M,N are as in the previous paragraph. Given that

Email address: farmer.schlutzenberg@gmail.com (Farmer Schlutzenberg)1The paper [2] literally deals with premice with Jensen indexing, whereas we deal with

Mitchell-Steel indexing. However, the same result still holds.

Preprint submitted to Elsevier June 15, 2015

C is iterable, a comparison of C with itself will be used to show that, in theabove case, M = N . We will also deal with bicephali (,M , N ) in which M

or N might fail to be fully sound. However, we will assume that both M , N

project to , are -sound, and M , N agree below their common value for()+. If such a bicehpalus is iterable, it might be that M 6= N , but we will seethat in this situation, M is an ultrapower of some premouse by an extender inE+(N ), or vice versa. We will also prove similar results regarding cephalanxes,a blend of bicephali and phalanxes. The presence of superstrong premice makescephalanxes somewhat more subtle than bicephali.

We will give two applications of these results. First, in 4, we considerproving condensation under a normal iterability hypothesis. Let k < , letH,M be k-sound premice, let : H M be a near k-embedding2, let Hk+1

< Hk , and suppose that H is -sound and cr(). We wish to prove(k+1)-condensation holds.3 Recall that the standard (phalanx-based) proof ofcondensation relies on the (k, 1, 1+1)-iterability of M , through its appeal tothe weak Dodd-Jensen property. We wish to reduce this assumption to (k, 1+1)-iterability. Given the latter, and the (k + 1)-solidity of M , we will deducethe usual conclusion of condensation. Also, without assuming any solidity ofM , we will show that a slight weakening of (k + 1)-condensation still followsfrom (k, 1 + 1)-iterability. (But note that the assumption that H is -soundentails that pHk+1\ is (k+1)-solid for H .) Since we do not have (k, 1, 1+1)-iterability, it is natural to consider the circumstance that M fail to be (k + 1)-solid. (Though on the other hand, the author believes that, at least if M hasno superstrong initial segments, then it is likely that the (k + 1)-solidity of Mfollows from (k, 1 + 1)-iterability; see 6.) Our proof makes use of bicephaliand cephalanxes in place of phalanxes, and avoids using (weak) Dodd-Jensen.4

Next, let N be the output of a typical fully backgrounded L[E]-construction.Assuming that various structures associated to the construction are sufficientlyiterable, every Woodin cardinal is Woodin in N . However, it seems that might be strong, but not strong in N . In [10], Steel defined the local Kc-construction, whose output M inherits both Woodin and strong cardinals. Butthis construction requires V to be a premouse, and (an important feature whichhelps ensure that strong cardinals are inherited is that) the background exten-ders used can be partial. As a consequence, when one lifts iteration trees onM to iteration trees U on V , the tree U might have drops. In 5, we identifya new form of L[E]-construction C, uniquely definable in any ZFC universe Vwhich is coarsely iterable in some larger universe W . Letting L[E] be the final

2Actually we will work with the more general class of k-lifting embeddings; see 2.1.3Approximately, that is, the version . . . with H

k+1replacing H in [3, pp. 8788], or [2,

Lemma 1.3], though this uses Jensen indexing, or [12, Theorem 9.3.2], though this uses Jensenindexing and -fine structure.

4The way we have presented our proof, we do make use of the standard proof of condensa-tion, in proving 2.13, but in circumstances in which Dodd-Jensen is not required. This appealto the standard proof can, however, be removed, by arranging things more inductively andusing the main structure of the proof of 4.2 to prove 2.13.

2

model of C, (a) L[E] is a proper class premouse, (b) if is strong (Woodin),then is strong (Woodin) in L[E], and (c) working in W , L[E] is iterable, withiteration trees on L[E] lifting to (coarse) trees on V . Thus, we achieve many ofthe properties of the the local Kc-construction, but with the advantages thatV need not be a premouse, and that trees U on V resulting from lifting treeson L[E] are such that for all + 1 < lh(U), EU is a total extender in M

U . (In

the case of the local Kc-construction, locally strong cardinals are also inherited,but this does not seem to hold for C.)

Conventions & Notation.General : The reverse of a finite sequence = (x0, . . . , xn1) is

=(xn1, . . . , x0).

The universe N of a first-order structure M = (N, . . .) is denoted M.Regarding premice and fine structure, we mostly follow [3] and [11], with

some modifications as described below. We also make use of generalized soliditywitnesses; see [12, 1.12].

Premice: We deal with premice and related structures with Mitchell-Steelindexing, but with extenders of superstrong type permitted on their extendersequence. That is, a super-fine extender sequence ~E is a sequence such that foreach dom( ~E), ~E is acceptable at , and if E 6= then either:

E is a (, ) pre-extender over J~E and E is the trivial completion of

E (E) and E is not type Z, or

J~E has largest cardinal and E is a (, ) pre-extender over J

~E and

iE() = = (E),

and further, properties 2 and 3 of [11, Definition 2.4] hold. We then define pre-mouse in terms of super-fine extender sequences, in the usual manner. Likewisefor related terms, such as segmented-premouse (see [5, 5]). See [9, 2.12.6,2.14] for discussion of the modifications of the general theory needed to dealwith these changes.

Let P be a segmented-premouse with active extender F 6= . We say that F ,or P , has superstrong type iff iF (cr(F )) < lh(F ). (So if F has superstrong typethen iF (cr(F )) is the largest cardinal of P , and then P is a premouse iff the initialsegment condition holds for P .) In [5], all premice are assumed to be belowsuperstrong type, but certain results there (in particular, [5, 2.17, 2.20]) hold inour context (allowing superstrong type), by the same proofs, and when we citethese results, we literally refer to these generalizations. (However, [5, Theorem5.3] does not go through as stated at the superstrong level; Theorem 3.32 of thepresent paper generalizes that result at the superstrong level.) At certain pointswe will explicitly restrict our attention to premice below superstrong type.

Let P be a segmented-premouse. We write FP = F (P ) for the active ex-tender of P (possibly FP = ), EP = E(P ) for the extender sequence of P , ex-cluding FP , and EP+ = E+(P ) = E

P FP . If FP 6= we write lh(FP ) = ORP .(So lh(FP ) is the length of FP when FP is not of superstrong type.) Given ORP , we write P | for the Q E P such that ORQ = , and write

3

P || = (Q ,EQ, ). (We use the same notation for cephals P , given that P ; see 3.5.) If P has a largest cardinal , lgcd(P ) denotes . If P isactive, (P ) and (FP ) both denote max(lgcd(P ), (FP )). So if P is an activepremouse then (P ) = (FP ). A premouse extender is an extender FP for someactive premouse P .

Given two segmented-premice P,R and an ordinal min(ORP ,ORR),define

(P R)| P | = R|.

We also use the same notation with more than two structures, and also with|| replacing |. (We use the same notation for cephals; see 3.5.)

Fine structure: We officially use Mitchell-Steel fine structure, (a) as modifiedin [8], and (b) as further modified by using k-lifting embeddings in place of weakk-embeddings. (Modification (a) involves dropping the objects un, and definingstandard parameters without regard to these objects. The reader who prefersthe original Mitchell-Steel fine structure simply need put the relevant uns intovarious hulls and theories. Modification (b) is described in 2.) Let m < and

let Q be an m-sound premouse. For i m+1 we write ~pQi = (pQ1 , . . . , p

Qi ). Let

q OR(Q) .See [5, 2.1] for the definition of semi-close (extender).Ultrapowers : Let E be a (possibly long) extender over a segmented-premouse

M . We write Ult(M,E) for the ultrapower formed by using functions in M ,without squashing (so FUlt(M,E) is defined as when M is a type 2 premouse).A ultrapower of M formed in this way is simple.

For M an n-sound premouse, we write Ultn(M,E) for the degree n ultra-power, with Ultn(M,E) = Ultn(M

sq, E)unsq if M is type 3.ForM an active segmented-premouse, UltM and Ult(M) both denote Ult(M,FM ),

and UltMk = Ultk(M) denotes Ultk(M,FM ).

For a type 3 premouseM , let C1(M) = C0(M), and for an extender E overC0(M), let and Ult1(M,E) = Ult0(M,E).

Embeddings : Given structuresX,Y , if context determines an obvious naturalembedding i : X Y we write iX,Y for i.

Let M,N be segmented-premice. A simple embedding : M N is a func-tion with dom() = M and cod() = N, such that is r0-elementary.(Note that if M is active then (lgcd(M)) = lgcd(N), because the amenablepredicates for FM and FN specify the largest cardinal.) If M,N are type3 premice, a squashed embedding : M N is, literally, a function with

4

dom() = C0(M) and cod() = C0(N), such that is r0-elementary (morecorrectly, q0-elementary, but we just use the notation r for both cases).

Let : M N be simple. If M is passive then then denotes . If M isactive then

: Ult(M,FM ) Ult(N,FN )

denotes the simple embedding induced by (using the Shift Lemma).Let : M N be squashed. Then

: Ult0(M,FM ) Ult0(N,F

N )

denotes the squashed embedding induced by .So in all cases.We say :M N is -preserving iff eitherM,N are passive or ((FM )) =

(FN ). We say is -preserving iff either M,N are passive or ((FM )) =

(FN ). We say is c-preserving iff for all , if is a cardinal of M then ()is a cardinal of N . We say is pj-preserving iff (p

Mj ) = p

Nj . We say is

~pj-preserving iff (~pMj ) = ~p

Nj .

Iteration trees : Let T be an iteration tree. Then T is maximal iff it isk-maximal for some k .

2. Fine structural preliminaries

2.1 Definition. Let H,M be k-sound premice with Hk , Mk > . Let L be the

language of H (here if H is passive, we take L = {,E}). We say an embedding : H M is k-lifting iff is r0-elementary with respect to L, and if k > 0then THk T

Mk .

A k-lifting embedding is similar to a (k)0 -preserving embedding of [12].

2.2 Lemma. Let H,M, k,L be as in 2.1 and let : H M . Then:

1. is k-lifting iff for every rk+1 formula and x H, if either L ork > 0 then

H |= (x) = M |= ((x)).

2. If is k-lifting and H,M have different types then k = 0, H is passiveand M is active.

3. If k > 0 and is k-lifting then is rk elementary, (k 1)-lifting andc-preserving.

4. If k > 1 and is rk elementary then is pk2-preserving and k2-preserving, and if Hk1 <

H0 then

(pHk1) = pMk1\(

Hk1) and

supHk1 Mk1 (

Hk1).

5

5. If k > 0 and is rk elementary and pk1-preserving, (pHk ) p

Mk .

6. The Shift Lemma holds with weak k- replaced by k-lifting, or by k-liftingc-preserving.

Proof. Parts 13 are straightforward. For part 4, use (k 1)-solidity witnessesfor pk1. For part 5 use the fact that if t is a k-solidity witness for (H, p

Hk ),

then (t) is a generalized k-solidity witness for (M,(pHk )).Part 6: We adopt the notation of [3, Lemma 5.2], but with n replaced by

k. Let F = F N and U = Ultk(M, F ) and U = Ultk(M,FN ). Define

: C0(U) C0(U)

as there. It is straightforward to see that is rk-elementary. Suppose k > 0.Let us observe that T Uk T

Uk . Let t T

Uk . Let x U and <

Uk be such

thatt = ThUrk( {x}).

Let y M and a (F )

in place of 0, when T does not drop in model along [0, + 1]T . In all otherrespects, the details are as usual. Moreover, if (i) [0, ]T drops in model or (ii)degT () k 2 or (iii) degT () = k 1 and is pk1-preserving, then isa near degT ()-embedding; this uses the argument in [4].

2.4 Lemma. Let k 0, let : H M be k-lifting, let Hk+1 Hk . Then:

1. If pMk1, pMk rg() and

Mk = sup

Hk then is a k-embedding.

2. If H is -sound and ( ) M and is not a k-embedding, then H, ( Hk ) M .

Proof. Part 1: This is fairly routine. By 2.2, we have (pHk1) = pMk1. The

rk+1 elementarity of follows from this, together with the facts that is k-lifting, pMk rg() and

Hk is unbounded in

Mk . Now let (q) = p

Mk . Then

pHk q by 2.2, and q pHk by rk elementarity.

Part 2: If Hk is bounded in Mk , use a stratification of rk+1 truth like

that described in [3, 2]. Given the reader is familiar with this, here is a sketch.Let = supHk . Then the theory

t = ThMrk( (~pHk ))

is in M . Moreover, for any rk+1 formula and ~ 1 and (pHk1) 6= pMk1. By 2.2, we therefore have

(pHk1) < pMk1. Let = sup

Hk .

Claim. Let be an rk formula, let x H and ~ max(~). Let

v = ThHrk( {x} ~pHk1).

7

Note then that for all ~ 0, let

Ck(Q) = (Q||k(Q), T),

where T = ThQrk(k ~pQk ), and T

is given from T by substituting ~pQk for aconstant symbol c.

2.6 Definition. Let k 0. Let Q be a k-sound premouse with Qk > . Wesay that (U, ) is k-suitable for Q iff:

U, Q||Qk ,

U is a k-sound premouse with Uk > , and

: Ck(U) Ck(Q) is 0-elementary.

2.7 Remark. Clearly, if (U, ) is k-suitable for Q then extends uniquelyto a ~pk-preserving k-lifting : U Q, and moreover,

supUk < Qk .

Conversely, if : U Q is ~pk-preserving k-lifting and supUk < Qk and

= (U ||Uk ) is in Q, then (U, ) is k-suitable for Q.

2.8 Lemma. Let k 0. Then there is an rk+1 formula k such that for allk-sound premice Q with < Qk , and all U,

Q,

Q |= k(U, , ~pQk )

iff (U, ) is k-suitable for Q.

8

Proof. We assume k > 0 and leave the other case to the reader.The most complex clause of k says There is <

Qk such that letting

t = ThQrk( ~pQk ),

then for each < Uk , letting

u = ThUrk( ~pUk ),

and letting t, u be given from t, u by substituting ~pQk , ~pUk for the constant c,

we have (u) t, and this is rk+1. The rest is clear.

2.9 Definition. Let m 0 and let M be a segmented-premouse. Then M ism-sound iff either m = 0 or M is an m-sound premouse.

2.10 Definition. Let r 0 and let R be an r-sound premouse. Then we saythat suitable condensation holds at (R, r) iff for every (H, ), if (H, ) isr-suitable for R, H is (r + 1)-sound and

cr() = Hr+1,

then either H R, or R| is active with extender F and H Ult(R|, F ).Let m 0 and let M be an m-sound segmented-premouse. We say that

suitable condensation holds below (M,m) iff for every R E M and r < such that either R M or r < m, suitable condensation holds at (R, r). We saythat suitable condensation holds through (M,m) iffM is a premouse5 andsuitable condensation holds below and at (M,m).

2.11 Lemma. Let m 0. Then there is an rmax(m,1) formula m suchthat for all m-sound segmented-premice M , suitable condensation holds below(M,m) iff M |= m(~pMm1), where p

m1 = . Moreover, if M is a premouse,

then suitable condensation holds through (M,m) iff M |= m+1(~pMm ).6

Proof. This follows easily from 2.8.

2.12 Remark. Our proof of condensation from normal iterability (see 4.2) willuse our analysis of bicephali and cephalanxes (see 3). This analysis will, inturn, depend on the premice involved satisfying enough condensation, at lowerlevels (that is, lower in model or degree). We will only have normal iterabilityfor those premice, so we cant appeal to the standard condensation theoremfor this. One could get arrange everything inductively, proving condensationand analysing bicephali and cephalanxes simultaneously. However, it is simplerto avoid this by making use of the following lemmas, which are easy to provedirectly. We will end up generalizing them in 4.2.

5We could have formulated this more generally for segmented-premice, but doing so wouldhave increased notational load, and we do not need such a generalization.

6This clause only adds something because we do not assume that M is (m + 1)-sound.

9

2.13 Lemma (Condensation for -sound mice). Let h m < and let H,Mbe premice. Suppose that:

H is (h+ 1)-sound.

M is (m+ 1)-sound and (m,1 + 1)-iterable.

Either Mm+1 = or m h+ 5.

There is an h-lifting ~ph-preserving embedding : H M with cr() = Hh+1.

Then either

H = M , or

H M , or

M | is active and H Ult(M |,G).

Proof. Let , etc, be a counterexample. Let = (H ||Hh ).

Claim. H M .

Proof. Suppose not. By 2.4, is an h-embedding, and Mh+1. Note that

(pHh+1) pMh+1\ (using generalized solidity witnesses). If (p

Hh+1) < p

Mh+1\

then we are done, so suppose otherwise. Then Mh+1 pMh+1 rg(), so H =M ,

contradiction.

Now we may assume that Mm+1 = , by replacing M with cHullMm+1(~pm)

if necessary: all relevant facts pass to this hull because cr() and by theclaim and by 2.2(1). We can now run almost the usual proof of condensation.However, in the comparison (T ,U) of the phalanx (M,H, ) with M , we forman (m,h)-maximal tree on (M,H, ), and an m-maximal tree on M . BecauseH M , and using the fine-structural circumstances in place of the weak Dodd-Jensen property, this leads to contradiction.

2.14 Lemma (Suitable condensation). Let M be an m-sound, (m,1 + 1)-iterable segmented-premouse. Then suitable condensation holds below (M,m),and if M is a premouse, through (M,m).

Proof. If M is not a premouse this follows from 2.13. So suppose M is apremouse. By 2.11, we may assume that Mm+1 = , by replacing M with

cHullMm+1(~pMm ) if necessary. So we can argue as at the end of the proof of

2.13.

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3. The bicephalus & the cephalanx

3.1 Definition. An exact bicephalus is a tuple B = (,M,N) such that:

1. M and N are premice.

2. < min(ORM ,ORN ) and is a cardinal of both M and N .

3. M ||(+)M = N ||(+)N .

4. M is -sound and for some m {1} , we have Mm+1 . Likewisefor N and n {1} .

We say B is non-trivial iffM 6= N . Write B = andMB =M and NB = N ,and mB, nB for the least m,n as above. Let (+)B be (+)M = (+)N . We sayB has degree (mB , nB). We say that B is sound iff M is mB + 1-sound andN is nB + 1-sound.

From now on we will just say bicephalus instead of exact bicephalus. Inconnection with bicephali of degree (m,n) with min(m,n) = 1, we need thefollowing:

3.2 Definition. The terminology/notation (near) (1)-embedding, (1)-lifting embedding, Ult1, C1, and degree (1) iterability are defined byreplacing 1 with 0. For n > 1 and appropriate premice M , the coreembedding Cn(M) C1(M) is just the core embedding Cn(M) C0(M).

3.3 Definition. Let q < . A passive right half-cephalanx of degree q isa tuple B = (, ,Q) such that:

1. Q is a premouse,

2. is a cardinal of Q and (+)Q = < ORQ,

3. Q is -sound,

4. Qq+1 < Qq .

An active right half-cephalanx (of degree q = 0) is a tuple B = (, ,Q)such that:

1. Q is an active segmented-premouse,

2. is the largest cardinal of Q and < = ORQ.

A right half-cephalanx B is either a passive, or active, right half-cephalanx.We write B, B, QB, qB for , ,Q, q as above. If B is active, we write SB =RB = Ult(Q,FQ). If B is passive, we write SB = Q.

Note that if B = (, ,Q) is a right-half cephalanx, then B is active iff Q|is active. So it might be that B is passive but Q is active.

11

3.4 Definition. Let m {1} and q < . A cephalanx of degree (m, q)is a tuple B = (, ,M,Q) such that, letting B = (, ,Q), we have:

1. (, ,Q) is a right-half cephalanx of degree q,

2. M is a premouse,

3. = (+)M < ORM ,

4. M ||(+)M = SB

||(+)M ,

5. M -sound,

6. Mm+1 < Mm .

We say that B is active (passive) iff B is active (passive). 7 We writeB, B, etc, for , , etc. We write RB for RB

, if it is defined, and SB for SB

.

We say B is exact iff (+)SB

= (+)M .Suppose B is active. Let R = RB. We say B is non-trivial iff M R. If B

is non-exact, let NB denote the N R such that (+)N = (+)M and N = ,and let nB denote the n {1} such that Nn+1 = <

Nn .

Now suppose B is passive. We say B is non-trivial iff M 5 Q. Let NB

denote the N E Q such that (+)N = (+)M and N . Let nB be the

n {1} such that NB

n+1 < NB

n .A pm-cephalanx is a cephalanx (, ,M,Q) such that Q is a premouse.

3.5 Definition. A cephal is either a bicephalus or a cephalanx. Let B be acephal, and let M = MB.

A short extender E is semi-close to B iff cr(E) < B and E is semi-close8

to M .For B, let B|| = M ||, and for < B, let B| =M | and (+)B =

(+)M . We write P B iff P B||B. Let C, be such that B, and eitherC is a segmented-premouse and ORC , or C is a cephal and C . Thenwe define

(B C)|| B|| = C||.

If also < B and either C is a segmented-premouse or < C , we use thesame notation with | replacing ||. We also use the same notation with morethan two structures.

3.6 Remark. Because of the symmetry of bicephali and the partial symmetriesof cephalanxes, we often state facts for just one side of this symmetry, eventhough they hold for both.

The proofs of the next two lemmas are routine and are omitted. In 3.73.13below, the extender E might be long.

7Note that a passive cephalanx (, ,M,Q) might be such that M and/or Q is/are active.8See [5].

12

3.7 Lemma. Let Q be an active segmented-premouse. Let E be an extenderover Q with ms(E) cr(FQ) + 1. Let R = UltQ and Q = Ult(Q,E) and

R = UltQ

. Then R = Ult(R,E) and the ultrapower embeddings commute.Moreover, iRE = iQ

E.

3.8 Lemma. Let Q be an active segmented-premouse. Let E be an extenderover Q with (cr(FQ)+)Q < cr(E). Let R = UltQ and R = Ult(R,E) and Q =Ult(Q,E). Then Ult(Q,FQ

) = R and the ultrapower embeddings commute.9

Let : R R be given by the Shift Lemma (applied to id : Q Q and iQE).Then iRE = .

3.9 Definition. Let E be a (possibly long) extender over a segmented-premouseM . We say that E is reasonable (for M) iff either M is passive, or letting = cr(FM ), iME is continuous at (

+)M , and if M |=++ exists then iME iscontinuous at (++)M .

Given a bicephalus B = (,M,N), an extender E is reasonable for B iffE is over B||, if mB 0 then E is reasonable for M , and if nB 0 then E isreasonable for N .

Given a cephalanx B = (, ,M,Q), an extender E is reasonable for Biff E is over B||, if qB 0 then E is reasonable for Q, if mB 0 then E isreasonable for M , and if NB is defined and nB 0 then E is reasonable forNB.

3.10 Lemma. Let Q be an active segmented-premouse and let E be an extender

reasonable for Q. Let Q = Ult(Q,E) and R = UltQ and R = UltQ

andR = Ult(R,E). Let = cr(FQ) and = (++)Q. If < ORQ then let

= iQ,R(), = iRE(),

= iQE();

then = iQ,R(). If = ORQ then let = ORR, = ORR

and =

ORQ

. Then in either case, (R R)| and

iRE iQ,R (Q|) = iQ,R iQE (Q|).

Moreover, let : R| R| be induced by the shift lemma applied to iQ,Q (Q|) and iQ,Q . Then = iRE (R|).

Proof. LetG be the extender derived fromE, of length iE(). Let j : Ult(Q,G)

Ult(Q,E) be the factor embedding. Then cr(j) > (iQG()++)UG since E is rea-

sonable. Apply 3.7 to G, and then 3.8 to the extender derived from j.

3.11 Definition. Let M be a type 3 premouse. The expansion of M is theactive segmented-premouse M such that M|cr(FM) = M |cr(FM ), and FM

is the Jensen-indexed version of FM . That is, let F = FM , let = cr(F ), let = (+)M , let = iF (), let R = Ult

M ; then M||OR(M) = R|, and FM

is the length iF () extender derived from iF .

9Note that in the conclusion, it is Ult(Q,FQ

), not UltQ

.

13

The calculations in [3, 9] combined with a simple variant of 3.10 give thefollowing:

3.12 Fact. Let Q be a type 3 premouse. Let E be an extender over Qsq,reasonable for Q. Let Q be the expansion of Q, let U = Ult(Q, E) andU = Ult0(Q,E). Suppose U is wellfounded. Then U is wellfounded and U isits expansion. Moreover, let i : Q U and i0 : Q U be the ultrapowerembeddings (so literally, dom(i) = Q and dom(i0) = Q

sq). Then i0 = i Qsq,

and i = i0 Q.

3.13 Remark. We will apply 3.10 and 3.12 when E is the extender of aniteration map iT,, and if is a successor, the map i

T, i

T , where (, ]T

does not drop and degT () = 0.

3.14 Definition (Ultrapowers of bicephali). Let B = (,M,N) be a bicephalusof degree (m,n) and let E be an extender reasonable for B. Let

iME :M Ultm(M,E)

be the usual ultrapower map, and likewise iNE and n. Let = sup iME and

defineUlt(B,E) = (,Ultm(M,E),Ultn(N,E)).

We say that Ult(B,E) is wellfounded iff both Ultm(N,E) and Ultn(N,E) arewellfounded.

3.15 Definition. Let B be a bicephalus. The associated augmented bi-cephalus is the tuple

B = (,M,N,M, N)

where ifm 0 thenM =M , and otherwiseM is the expansion ofM ; likewisefor N. (Note that if m = 1 then M is type 3 and = (FM ).)

Let E reasonable for B. If m 0 let M = Ultm(M,E); otherwise letM = Ult(M, E). Likewise for N . Then we define

Ult(B, E) = Ult(B,E) M, N

.

We say that Ult(B, E) is wellfounded iff Ult(B,E), M , N are all wellfounded.

3.16 Lemma. Let B = (,M,N) be a bicephalus. Let E be reasonable for B.Let U = Ult(B,E) and U = Ult(B, E) = (

U ,MU , NU , M , N). Suppose thatU is wellfounded. Then:

(1 ) U is a bicephalus of degree (m,n) and U = U.

(2 ) U is trivial iff B is trivial.

(3 ) iME (pMm+1\) = p

MU

m+1\U .

14

(4 ) iME (+)B = iNE (

+)B and iME and iNE are continuous/cofinal at

(+)B.10

(5 ) iME = iME M.

(6 ) Suppose E is short and semi-close to B. Then MU is m+ 1-sound iff M

is m + 1-sound and cr(E) < Mm+1. If MU is m + 1-sound then M

U

m+1 =

sup iME Mm+1 and p

MU

m+1 = iME (p

Mm+1).

Likewise regarding N,n,E.

Proof. Part (6) is by [5, 2.20], (3) is a standard calculation using generalizedsolidity witnesses (see [12]), and (5) is by 3.12 ((5) is trivial when m 0).

Consider (4). Let W = Ult(B||(+)B , E) and j : B||(+)B W be theultrapower map. We claim that (): j = iME (

+)B, and letting = j(),

M ||(+)M = W.

If m 0 this is immediate. If m > 0, then because (+)B Mm , by [3, 6],all functions forming the ultrapower MU with codomain (+)B are in fact inB||(+)B , which gives ().

Now (4) follows from (). Consider (1). By 3.12, M is the expansion of MU .Now we have U and by (),

M ||(+)M = N ||(+)N .

Also if m 0 then < m(MU ). The rest of the proof of (1) is routine.Now let us prove (2). Assume M 6= N . We may assume m = n. Because

M 6= N and by -soundness, there is some rm+1 formula and < suchthat

M |= (pMm+1\, ) N |= (pNm+1\, ).

Now iME and iNE are rm+1-elementary, and by (), i

ME () = i

NE (); let

=iME (). So by (3),

MU |= (pMU

m+1\U , ) NU |= (pN

U

m+1\U , ),

and therefore MU 6= NU .

3.17 Definition (Ultrapowers of cephalanxes). Let B = (, ,M,Q) be acephalanx of degree (m, q) and let E be reasonable for B. Let iME be the degreem ultrapower map and let = iME () and

= sup iME . If B is active thenwe define

Ult(B,E) = (, ,Ultm(M,E),Ult(Q,E)).

10That is, if (+)B dom(iME

) then iME

is continuous there; if m 0 and (+)B =

M0

then MU

0= sup iM

E(+)B ; if m = 1 and (+)B = OR(M) then OR((MU )) =

sup iME

(+)B .

15

(Recall that the ultrapower Ult(Q,E) is simple; it might be that Q is type 3.)If B is passive then we define

Ult(B,E) = (, ,Ultm(M,E),Ultq(Q,E)).

3.18 Lemma. In the context of 3.17, suppose that B is passive, and that U =Ult(B,E) is wellfounded. Let = (+)M . Then:

(1 ) U is a passive cephalanx of degree (m, q).

(2 ) iME = iQE .

(3 ) If C0(M) then = iME (); otherwise = 0(M

U ). Likewise for

Q, iQE , QU .

(4 ) iME() = iQ

E() = .

(5 ) If (+)M dom(iME) then iM

Eis continuous at (+)M ; otherwise M is

passive, ORM = (+)M and OR(MU ) = sup iME ORM .

(6 ) iME(+)M =

iQE(+)M .

(7 ) iME (pMm+1\) = p

MU

m+1\.

(8 ) Suppose E is short and semi-close to B. Then MU is (m + 1)-sound iffM is (m + 1)-sound and cr(E) < Mm+1. If M

U is (m + 1)-sound then

MU

m+1 = sup iME

Mm+1 and p

MU

m+1 = iME (p

Mm+1).

(9 ) If B is non-exact then U is non-exact.

(10 ) If B is exact (so NB = Q) but U is not, then 0 nB < q.

(11 ) Suppose that B is non-trivial and that suitable condensation holds below(Q, q). Let N = NB and n = nB. Then:

(i) U is non-trivial,

(ii) NU = UltnB (N,E) and nU = n,

(iii) Parts (2 )(8 ) hold with M replaced by N and m by n.

We also have iQE(pQq+1\) = p

QU

q+1\U , but we wont need this.

Proof. Parts (2)(8) are much as in the proof of 3.16. (For (6), note that givenA P() M , the value of iM

E(A) is determined by the values of iM

E(A )

for < ; likewise for iQE(A).) So (1) follows. Part (9) follows from (5) and

(6); part (10) is easy.Now consider (11). We first deal with the case that B is exact, so assume

this. Part (iii) is just as forM , so consider (i) and (ii). Since B is exact, N = Q.

16

By the proof of 3.16, we have Ultn(Q,E) 6= Ultm(M,E), so it suffices to seethat

Un = Ultn(Q,E) E QU = Ultq(Q,E) = Uq.

We may assume that n < q. If n = 1 then we easily have Un = Uq, so alsoassume n 0, and so C0(Q). We have

Qq+1 < = Qq =

Qn+1 <

Qn .

Let : Un Uq be the natural factor map. Let in : Q Un and iq : Q Uqbe the ultrapower maps. Then in = iq, is ~pn+1-preserving n-lifting and

cr() > . Also, Un, Uq are (n+ 1)-sound and Unn+1 =

= Uqn+1.

Suppose (()+)Un = (()+)Uq < cr(). Then

Uqn = supUnn ,

since otherwise, using the previous paragraph and as in the proof of 2.4, other-wise Un Uq, collapsing (()+)Uq in Uq. So by 2.4, is an n-embedding, andin particular, is rn+1-elementary. Since

Uqn+1 p

Uqn+1 rg(),

therefore Un = Uq, which suffices.Now suppose that (()+)Un < (()+)Uq . Then much as in the previous

case,Uqn > sup

Unn .

Let = (Un||Unn ). By 2.4 we get Un, Uq and (Un, ) is n-suitable

for Uq. Since suitable condensation holds below (Q, q), and by 2.11, and sinceUq| is passive, it follows that Un Uq, which suffices.

Now consider the case that B is non-exact. So N Q. Let Un = Ultn(N,E),consider the factor embedding

: Un iQE(N)

and argue that Un E iQE(N), like before. This completes the proof.

3.19 Lemma. In the context of 3.17, suppose that B is active, and that U =Ult(B,E) and RU are wellfounded. Let = (+)M . Then:

(1 ) U is an active cephalanx of degree (m, 0).

(2 ) If C0(M) then = iME (); otherwise = 0(M

U ).

(3 ) 3.18(2 ), (4 )(8 ) hold.

(4 ) U is exact iff B is exact.

(5 ) Suppose that B is non-exact and non-trivial and that suitable condensationholds below (Q, 0). Let N = NB and n = nB. Then:

17

(i) U is non-trivial,

(ii) NU = UltnB (N,E) and nU = n,

(iii) Parts (2 )(3 ) hold with M replaced by N and m by n.

Proof. This follows from 3.10, 3.12 and the proof of 3.18.11

3.20 Lemma. Let C be a cephal of degree (m, k). If C is a bicephalus letB = C, and otherwise let B = C. Let E

3. B,M ,N are disjoint and for each < , either

(a) B and B = (,M, N) is a bicephalus of degree (m,n) =deg(), or

(b) M and B = M is a segmented-premouse and N = , or

(c) N and B = N is a segmented-premouse and M = .

4. For each + 1 < :

(i) Either E E+(M) or E E+(N).

(ii) cr = cr(E) and = (E) and lh = lh(E).

(iii) For all < we have lh lh.

(iv) predT (+ 1) is the least such that cr < .

Fix + 1 < and = predT (+ 1) and = cr.

5. Suppose B and < and E is total overB|| . Then deg(+1) =(m,n) and

(+1,M+1, N

+1) = B

+1 = B

andB+1 = Ult(B

+1, E)

andi+1 :M

+1 M+1

is the ultrapower map, and likewise j+1, and i,+1 and j,+1 are definedfor T + 1 in the obvious manner.

6. Suppose that E E+(M). Suppose that either / B, or < andE is not total over B|| . Then we set N+1 = N+1 = , and j

+1,

etc, are undefined. We setM+1 EM and deg(+1), etc, in the mannerfor maximal trees. Let k = deg(+ 1). Then

M+1 = Ultk(M+1, E)

and i+1, etc, are defined in the usual manner. We set B+1 = M

+1 and

B+1 =M+1.

7. Suppose that E / E+(M) (so E E+(N)) and B+1 is not definedthrough clause 5. Then we proceed symmetrically to clause 6 (interchang-ing M with N).

8. + 1 D iff either 6= M+1 M or 6= N+1 N .

9. For every limit < , D [0, )T is bounded in , and B iff [0, )T B; the models M, etc, and embeddings i,, etc, are defined via directlimits.

For < lh(T ), B() denotes max(B [0, ]T ).

19

3.22 Lemma. Let T be an iteration tree on a bicephalus of degree (m,n) andlet < lh(T ). We write B = BT , etc. Then:

1. If + 1 < lh(T ) then E is semi-close to B+1.

2. If + 1 < lh(T ) and + 1 / B then E is close to B+1.

3. B is closed downward under

Suppose B. Then:

If E E+(M) and either < or E is not total over M thenM+1 EM and Q+1 = .

If E / E+(M) and either < or E is not total over Q thenQ+1 E Q and M+1 = .

Now suppose that < and E is total over B || (so ). Then:

Suppose either < or E E+(M). Then B+1 = B.13

If = and E / E+(M)14 then M+1 = and Q+1 = Q.15

The remaining details are like in 3.21.

3.24 Lemma. Let T be an iteration tree on a cephalanx B = (, ,M,Q) ofdegree (m, q) and let +1 < lh(T ). Then parts 16 of 3.22, replacing N withQ, hold. Parts 5 and 6, replacing M with Q, M with Q, m with q,and with , also hold.

Proof. This is proved like 3.22.

3.25 Definition. Let T be an iteration tree on a cephal B and + 1 < lh(T ).We write P T for the active segmented-premouse P such that E

T = F

P andeither

B is a bicephalus and P EMT or P E NT , or

B is a cephalanx and P EMT or P E QT .

3.26 Definition. Let B be a cephal. A potential tree on B is a tuple

T =(

The next lemma is easy:

3.27 Lemma. Let T be a potential tree on a cephal B. Then T is an iterationtree. Moreover, if < < lh(T ) and BT then we can apply 3.20 to B, Band the sequence of extenders used along (, ]T . Further, assume that if B is

an active cephalanx and lgcd(QB) < (FQB

) then QB is a premouse. Thenevery model of T is either a cephal or a premouse.

3.28 Definition (Iterability for cephals). Let B be a bicephalus and OR.The length iteration game for B is defined in the obvious way: givenT + 1 with + 1 < , player I must choose an extender E, and given T for a limit < , player II must choose [0, ]T . The first player to break one ofthese rules or one of the conditions of 3.21 loses, and otherwise player II wins.

The iteration game for cephalanxes is defined similarly.We say that a cephal B is -iterable if there is a winning strategy for player

II in the length iteration game for B.

3.29 Lemma. Let B be an (1 + 1)-iterable cephal of degree (m, k). Let T bean iteration tree on B and < lh(T ). Then:

Suppose MT 6= . If BT let d = m; otherwise let d = degT (). Then

suitable condensation holds through (MT ,max(d, 0)).

Suppose B is a cephalanx and QT 6= . If BT let d = k; otherwise

let d = degT (). Then suitable condensation holds below (QT , k), andif either [0, ]T drops or Q,Q

T are premice, then suitable condensation

holds through (QT , k).

Proof. If T is trivial, use 2.14 (for example, MB is (m,1 + 1)-iterable). Thisextends to longer trees T by 2.11 and the elementarity of the iteration maps.

3.30 Definition. Let m < and let M be a -sound premouse, where Mm+1

Mm , and let < ORM . We say that M has an (m, )-good core at iff

< and and lettingH = cHullMm+1( ~p

Mm+1),

H is -sound andH ||(+)H =M ||(+)M ,

and letting : H M be the uncollapse map, cr() = and () and

(pHm+1\) = pMm+1\.

In this context, let HMm, = H and let GMm,, be the length extender derived

from .

3.31 Remark. Note that if M has an (m, )-good core at then, with ,H asabove, we have Mm+1 , M is not (m+1)-sound, G = G

Mm,, is semi-close to

H , M = Ultm(H,G) and iHG = .

22

We can now state and prove a restriction on iterable bicephali.

3.32 Theorem. Let B = (,M,N) be an (1 + 1)-iterable non-trivial bi-cephalus. Then B is not sound. Let m = mB and n = nB. Then exactlyone of the following holds:

(a) N is active type 1 or type 3 with largest cardinal , and letting = cr(FN ),then m 0 and M has an (m, )-good core at , and GMm,, = F

N .

(b) Vice versa.

Proof. Let B be a counterexample. We may assume that B is countable. Wemimic the self-comparison argument used in [3, 9]. That is, fix an (1 + 1)-iteration strategy for B. We form a pair of padded iteration trees (T ,U) onB, each via , by comparison. We will ensure that we never use compatibleextenders in the process, and use this to show that the comparison terminates,using the ISC and an extra argument. Assuming that B is sound, we willreach a contradiction by showing that the comparison cannot terminate. If B isunsound, we will reach the desired conclusion by examining the circumstancesunder which the comparison must terminate.

Regarding padding, for each we will have either ET 6= or EU 6= . If

= predT ( + 1) is such that ET = , then = . Likewise for U .At stage of the comparison, given BT , we may set ET = , and

simultaneously declare that, if T is to later use a non-empty extender, thenletting > be least such that ET 6= , we will have E

T E+(M

T ) =

E+(MT ). Or instead, we may declare that ET E+(N

T ). Toward this, we

define non-empty setsM

T {M

T , N

T }\{}.

We will require that if ET 6= , then ET E+(P ) for some P M

T . All

models in MT will be non-empty.

We also define sets ST tT {0, 1} for convenience. Let 0 t

T iffM

T 6= ,

and 1 tT iff NT 6= . Let 0 S

T iff M

T M

T , and 1 S

T iff N

T M

T .

(We will explicitly define either MT or ST , implicitly defining the other.)

The preceding definitions also extend to U .We now begin the comparison. We start with BT0 = B = B

U0 and S

T0 =

{0, 1} = SU0 .Suppose we have defined (T ,U) for some limit . Then (T ,U) + 1 is

determined by , and ST = lim

Subcase 1.1. For some choice of Y, Z witnessing the choice of , Y | and Z|are both active and (FY |) = (FZ|) = .

Fix such Y, Z. We set ET = FY | and EU = F

Z|. This determines (T ,U)+ 2. Also set ST+1 = t

T+1 and S

U+1 = t

U+1.

Subcase 1.2. Otherwise.Then take Y, Z witnessing the choice of and such that either:

Y | is active, (FY |) = , and if Z| is active then (FZ|) > ; or

vice versa.

Say Y | is active with (FY |) = . Then we set ET = FY | and EU = . This

determines (T ,U) + 2. Set ST+1 = tT+1. Now suppose there is X M

U

with X | active and (FX|) = . Then X | = Y |, so we must be careful toavoid setting EU = F

X| at some > . So we set MU+1 = {Z}, and set

SU+1 accordingly. If there is no such X then set SU+1 = S

U . (In any case, later

extenders used in U will be incompatible with ET .) The remaining cases arecovered by symmetry.

Case 2. Otherwise.Then we stop the comparison at stage .

This completes the definition of (T ,U). For < lh(T ,U), let ST () be thelargest T such that S

T = {0, 1}; here if B

T then BT = BT . Let

SU() be likewise.

Claim 1. The comparison terminates at some stage.

Proof. This follows from the ISC essentially as in the proof that standard com-parison terminates (using the fact that we observe the restricting setsMT+1,M

U+1

as described above).

So let be such that the comparison stops at stage .

Claim 2. card(ST ) = card(SU ) = 1 and M

T = M

U .

Proof. If BT then BT is non-trivial, by 3.27; likewise for U . So becauseCase 2 attains at stage , we do not have ST = S

U = {0, 1}.

It is not true that () Q P or P Q for some Q MU and P MT . For

suppose () holds; we may assume QP . Then Q is sound, so by 3.22, BU ,so by () and Case 2 hypothesis, card(SU ) = 1. Say S

U = {0}. Let = S

U().Then BU = B

U and for all [, ), E

U = , and E

T E+(N

U ). Let =

U .

Then lhT (+)B

U

. So P() P = P() BU , contradicting the fact that

MU = Q P .

Now suppose that ST = {0, 1} but card(SU ) = 1. Let be least such

that MT | 6= NT |. Let Q M

U . Then Q M

T || = N

T ||, so () holds,

contradiction. So card(ST ) = card(SU ) = 1, and because () fails, M

T =

MU .

Claim 3. BTBU .

24

Proof. By Claim 2, / BT BU , so assume that / BT BU . Thenstandard calculations using 3.22 give that T ,U use compatible extenders, acontradiction.

Using the previous claims, let us assume that BT \BU , ST = {0} andSU = {1}, so B = B

T is a bicephalus, N

U , and MT = NU ; the other

cases are almost symmetric. We will deduce that conclusion (a) of the theoremholds; under symmetric assumptions (b) can hold instead. Let = ST (). Let = (B). Then B = BT and for all [, ), we have E

T = 6= E

U and

(+)B lhU .

Claim 4. = + 1 and lhU = (+)B and EU is type 1 or type 3.

Proof. Suppose the claim fails. Then by 3.22, NU is not -sound (recall that if > + 1 and lhU+1 = lh

U then E

U+1 is type 2). But by 3.22, M

T is -sound.

So MT 6= NU , contradiction.

Let B = (, M , N) = BT = BT . Since E

U E+(N), and lh

U = (

+)B,

N |(+)B projects to , so ORN = (+)B and F N = EU . Let F = FN and

= cr(F ). It follows that (a) of the theorem holds regarding B; using theiteration embeddings we will deduce that B is not sound, and (a) holds regardingB. Note that either OR(M) > OR(N), or OR(M) = OR(N), N has superstrongtype and M is type 2; in either case m 0. Also ORN = (+)B and N is activewith F = FN , a preimage of F . Let = cr(F ); so < .

Claim 5. M is not m+ 1-sound, so B is not sound.

Proof. Suppose M is m + 1-sound. Let z = zMm+1 and = Mm+1. By [5, 2.17],

z = pMm+1 and = Mm+1 . So

HullMm+1( z ~pMm ).

Let z = zMm+1 and = Mm+1. By [5, 2.20], z = i

T0,(z) and = sup i

T0,, so

and

iT0,() HullMm+1( z ~p

Mm ). (2)

Let H = NU . Then M = NU = Ultm(H, F ) and = sup i

HFH , and since

, therefore . Also, z = iHF(zHm+1). But / rg(i

HF), so

/ HullMm+1( z ~pMm ). (3)

But iT0, = jT0, , so i

T0,() = , contradicting lines (2) and (3).

We can now complete the proof:

Claim 6. Conclusion (a) of the theorem holds.

25

Proof. Suppose N is type 1. Let p = pMm+1\ and

H = cHullMm+1( p ~pMm )

and let : H M be the uncollapse. Then H = NU , = jU , H is -sound

and letting q = pHm+1\, we have (q) = p,

H ||(+)H = M ||(+)M = N ||(+)N

and Mm = sup Hm.

We have ,H, defined as in (a); let p = pMm+1\. Let M 0 use the stratification of rm+1 truth described in [3, 2].) Let = ()and let E be the (short) extender E of length

derived from . Then

H , E M . Let : H M and and E be defined likewise over M .

We have (H M)||(+)H and E F N for each ; the former is because

by 2.13, N |=Lemma 2.13 holds for my proper segments.Let i = iT0,. Since i is an m-embedding and i(, p) = (, p), for each

< Mm , i(H) = Hi(), and i() = and i(E) = Ei(). Also

Mm = sup i

Mm

and ORN = sup jORN and i, j are continuous at (+)N and jFN F N .It follows easily that (H M)||(

+)H and E FN for each < Mm .

Therefore H ||(+)H =M ||(+)M and FN is derived from .It follows that FN is semi-close to H , M = Ultm(H,F

N ), and = iMFN

(because we can factor the embedding : H M through Ultm(H,FN ), and(FN ) = ). So by [5], (zHm+1) = z

Mm+1, but z

Mm+1\ = p

Mm+1\, and therefore

zHm+1\ = pHm+1\, so H is -sound. This completes the proof assuming that

N is type 1.If instead, N is type 3, then almost the same argument works.

This completes the proof of the theorem.

We now move on to analogues of 3.32 for cephalanxes.

3.33 Definition. Let B be a passive cephalanx of degree (m, q) and let N =NB. We say that B has a good core iff m 0 and N is active and lettingF = FN , = cr(F ) and = (F ), we have:

ORN = (+)M and N is type 1 or 3,

M has an (m, )-good core at ,

GMm,, = F , and

if N is type 1 then HMm, = Q and m = q.

26

3.34 Theorem. Let B = (, ,M,Q) be an (1+1)-iterable, non-trivial, passivecephalanx. Then B is not sound, and B has a good core.

Proof. The proof is based on that of 3.32. The main difference occurs in therules guiding the comparison, so we focus on these.

We may assume that B is countable. We define padded iteration trees T ,Uon B, and sets ST , S

U ,M

T ,M

U , much as before. We start with S

T0 = S

U0 =

{0, 1}. At limit stages, proceed as in 3.32. Suppose we have defined (T ,U)+1,ST and S

U and if card(S

T ) = card(S

U ) = 1 then B

T 5 B

U 5 B

T (otherwise

the comparison has already terminated). We just consider enough cases thatthe rest are covered by symmetry.

Case 1. card(ST ) = card(SU ) = 1.

Choose extenders as usual (as in 3.32).

Case 2. ST = {0, 1} and if SU = {0, 1} then

T

U .

So BT is a cephalanx; let BT = (T , T ,MT , QT ) = BT . Let B

U = BU .We will have by induction that for every < , lhT

T and lhU T . Since

B is passive, BT |T and BU |T are well-defined premice.

Subcase 2.1. BT |T 6= BU |T .Choose extenders as usual.

Suppose BT |T = BU |T . We say that in T we move into MT to meanthat we either set ET 6= and E

T E+(M

T ), or set ET = and ST+1 = {0}.

Likewise for move into QT , and likewise with regard to U if SU = {0, 1}. In eachcase below we will move into some model in T . In certain cases we do likewise forU . These choices will produce two premice R,S from which to choose ET , E

U ,

in the usual manner, given that R 5 S 5 R (for example, if SU = {1} and inT we move into MT , then R = MT and S = QU). If R E S or S E R, thenwe terminate the comparison, and say that the comparison terminates early. IfBU is a cephalanx and we do not move into any model in U and EU = thenwe set SU+1 = {0, 1}.

Subcase 2.2. card(SU ) = 1 and BT |T = BU |T .

Let P MU .If QT E P then in T we move into MT .If QT 6E P then in T we move into QT .

Subcase 2.3. ST = SU = {0, 1} and B

T |T = BU |T .Let (U , U ,MU , QU ) = BU . So T U .Suppose QT = QU . Let X {0, 1} be random. Then:16

If T < U or X = 0 then in T we move into MT , and if also MT |U =BU |U then in U we move into QU .

If T = U and X = 1 then in U we move into MU and in T we move intoQT .

16We use the random variable X just for symmetry.

27

If QT QU , then in T we move into MT . (Note that T < U and QT BU ||U , so we do not need to move into any model in U .)

If QU QT then in T we move into QT and in U we move into MU . (Notethat T < U .)

Suppose QT 5 QU 5 QT . Then in T we move into QT . If also QT |U =BU |U , then in U we move into QU .

The remaining cases are determined by symmetry.The comparison terminates as usual. We now analyse the manner in which

it terminates.

Claim 1. Let < lh(T ,U). Then (i) the comparison does not terminate early atstage ; (ii) if at stage , in T we move into R, then for every (, lh(T ,U)),R S for any S MU .

Proof. By induction on . Suppose for example that Subcase 2.2 attains atstage . We have P MU .

Suppose QT E P , so in T we move into MT . We have MT |T = P |T andNT E QT E P and MT 6= NT and

MT ||((T )+)MT

= NT ||((T )+)NT

and both MT , NT project T . So MT 5 P and letting be least such thatMT | 6= NT |, we have

T < min(OR(MT ),OR(NT )).

So the comparison does not terminate early at stage , and since MT projects T , for no > can we have MT S MU .

Now suppose QT 5 P , so in T we move into QT . If / BU then P = BU isunsound. Otherwise there is < such that at stage , in U we moved into P .In either case (by induction in the latter), P QT . So again, the comparisondoes not terminate early at stage . Let be least such that QT | 6= P |. ThenT < and since QT projects T , there is no > such that QT S MU .

The proof is similar in the remaining subcases.

So let + 1 = lh(T ,U). As in the proof of 3.32, we have card(ST ) =card(SU ) = 1 and B

TBU . We may assume that BT , so BT =(, ,M , Q) is a cephalanx and BU is not. Then B

U is not sound, so letting

P MT , we have P E BU . But by Claim 1, P B

U , so P = B

U . Let

= ST ().

Claim 2. ST = {0}.

Proof. Suppose ST = {1}, so Q = P = BU is

-sound. At stage , in T wemove into Q. For all [, ), ET = , so E

U 6= , and

< lhU , because

B| = BU |, and therefore U , because

is a cardinal of Q. But then

BU is not -sound, contradicting the fact that Q = BU .

So M = P = BU . Let N = NT .

28

Claim 3. We have:

OR(N ) = (()+)M

,

N is active type 1 or type 3,

= + 1,

EU = FN ,

if N is type 1 then BU = Q.

Proof. Assume, for example, that Subcase 2.2 attains at stage . So N E Q EBU . We have M

6= N , both M , N project to , and

M ||(()+)M

= N ||(()+)N

.

We have ET = , so EU 6= and note that E

U E+(N

) and lhU > . Since

M = BU is -sound it follows that = + 1 and U =

, so EU is type 1 or

type 3. Therefore N |lhU projects to , so OR(N ) = lhU .

Now suppose further that N is type 1; we want to see that BU = Q. We

have Q E BU and cr(FN ) = and (Q

) and

P() Q = P() N .

So it suffices to see that predU() = . We may assume that lhU = for

some < . Then is a cardinal of BU , so Q BU , so Q

= BU . So BU is

-sound, so there is a unique such that lhU = , and moreover, EU is type 3

and = + 1. Therefore predU () = , as required.

To complete the proof, one can now argue like in Claim 6 of 3.32.

3.35 Remark. We next proceed to the version of 3.34 for active cephalanxesB = (, ,M,Q). Here things are more subtle for two reasons. First, if Qis type 3 then we can have such that QT or Q

U is not a premouse, and in

particular, its active extender can fail the ISC; this complicates the proof thatthe comparison terminates. Second, if Q is superstrong then the comparisontermination proof is complicated further, and more importantly, it seems thatwe need not get the analogue of a good core (see 3.44), and moreover, in this casewe do not see how to rule out the possibility that B is exact andM is sound withMm+1 = . In fact, it is easy enough to illustrate how the latter might occur.Let Q be a sound superstrong premouse and = cr(FQ) and let J be a soundpremouse such that J ||(++)J = Q|(++)Q and Jm+1 = (

+)Q = (+)J < Jm.

Let M = Ultm(J, FQ) and B = (, ,M,Q), where = ORQ and is the

largest cardinal of Q. Suppose that M is wellfounded. Then B is an exact,sound bicephalanx. (We have Mm+1 = <

Mm and M is m + 1-sound, and B

is exact because iJFQ

and iQFQ

are both continuous at (++)J .) It seems thatreasonable that such a pair (J,Q) might arise from as iterates of a single model,

29

and so it seems that B might also be iterable. Conversely, we will show that thekind of example illustrated here is the only possibility (other than that givenby good cores).

3.36 Definition. Let T be an iteration tree on an active cephalanx B and+ 1 < lh(T ). We say is T -special iff BT and ET = F (Q

T ).

3.37 Lemma. Let T be an iteration tree on an active cephalanx B and +1. If < lhT+1 it is straightforward, so suppose = lh

T+1. Then since

the lemma held for = + 1, either ET+1 is type 2, in which case things are

straightforward, or +1 is T -special, so letting = crT+1 and = predT (+2),

we have that BT+2 = BT is a cephalanx and < (B

T ), which implies that

< (BT+2) and is a cardinal of BT+2. The rest is clear.

3.39 Definition. Let B = (, ,M,Q) be an active cephalanx of degree (m, 0).We say that B is exceptional iff

B is active and exact,

Q has superstrong type, and

30

either Mm+1 = or M is not -sound.

3.40 Lemma. Let M be an m-sound premouse and let Mm+1 < Mm . Then

M is -sound iff M = HullMm+1( zMm+1 ~p

Mm ).

Proof. This follows from [5, 2.17].

3.41 Lemma. Let B,B be active cephalanxes such that B is an iterate of B.Then B is exceptional iff B is exceptional.

Proof. By 3.27, 3.37(a) and 3.40 and [5, 2.20].

3.42 Definition. Let B = (, ,M,Q) be an active cephalanx of degree (m, 0).We say that B has a exceptional core iff Q has superstrong type and lettingF = FQ, = cr(F ), X = iQF (

+)M , m = max(m, 0),

H = cHullMm+1(X zMm+1 ~p

Mm ),

and : H M be the uncollapse, then (+)H = X and

H ||(++)H = M |(++)M .

3.43 Lemma. Let B = (, ,M,Q) be an active pm-cephalanx of degree (m, 0).Suppose B has an exceptional core. Let F, ,m, H, be as in 3.42. Then:

1. M = Ultm(H,F ) and = iHF is an m-embedding.

2. (zHm+1) = zMm+1 and (p

Hm+1\(

+)H) = pMm+1\.

3. Hm+1 (+)H < Hm and H is (

+)H-sound.

4. If Hm+1 = (+)H then Mm+1 = and H,M are (m+ 1)-sound.

5. If Hm+1 then Hm+1 =

Mm+1 and M is not (m+ 1)-sound.

6. If M = HullMm+1( zMm+1 ~p

Mm ) where < and is least such, then

rg().

Proof. Parts 14 are standard. Part 5: Because Hm+1 , we have m 0.Since Q is a type 3 premouse and M ||(++)M = H ||(++)H , F is close to H ,so Mm+1 =

Hm+1 . Suppose M is (m+ 1)-sound, so

M = HullMm+1( ~pMm+1).

It follows thatM = HullMm+1(rg() q)

for some q

Since is continuous at (+)H and is rm+1-elementary and (zHm+1) =

zMm+1, this fact easily reflects to H ; i.e. there is < (+)H such that

H = HullHm+1( zHm+1 ~p

Hm).

Let be least such. Then = () is least such that line (4) holds.

3.44 Definition. Let B = (, ,M,Q) be an active cephalanx of degree (m, 0),with m 0. We say that B has a good core iff the following holds. Either:

(i) B is exact; let F = FQ; or

(ii) B is not exact and letting N = NB, we have ORN = (+)M and N isactive type 1 or 3; let F = FN .

Let = cr(F ) and = (F ). Then:

1. M has an (m, )-good core at , and GMm,, = F .

2. Suppose case (ii) holds and N is type 1; so = . Then:

If Q is type 2 then HMm, = Q.

Suppose Q is not type 2, nor superstrong. Let = cr(FQ). Then Mhas an (m, )-good core at , and GMm,, = F

Q.

3.45 Remark. It seems that B might have an exceptional core but not a goodcore.

3.46 Theorem. Let B = (, ,M,Q) be an (1+1)-iterable, non-trivial, activepm-cephalanx, of degree (m, q). Then B is not sound. If B is non-exceptionalthen m 0 and B has a good core. If B is exceptional then B has an exceptionalcore.

Proof. Suppose first that B is exact and Q is superstrong, but B is not excep-tional. Then Mm+1 and M is -sound, as is Q. So C = (,M,Q) is anon-trivial bicephalus, and note that C is (1 + 1)-iterable. So by 3.32, B hasa good core. So we may now assume that:

If B is exact and Q is superstrong, then B is exceptional. (5)

Under this assumption, the proof is based on that of 3.34. The main differ-ences occur in the rules guiding the comparison, the proof that the comparisonterminates, and when B is exceptional.

Assume B is countable. We define T ,U on B and sets ST , SU ,M

T ,M

U ,

much as before. Suppose we have defined (T ,U) + 1, ST and SU , but if

card(ST ) = card(SU ) = 1 and R M

T and S M

U then R 6E S 6E R. We will

implicitly specify two segmented-premice from which to select ET , EU ; however,

we minimize on (E), rather than (E), when selecting these extenders. (Forexample, if ET 6= 6= E

U then

T =

U .) We use the terminology terminates

early as before. Let BT = BT , MT =MT , etc.

32

Case 1. card(ST ) = card(SU ) = 1.

Do the obvious thing.

Case 2. ST = {0, 1} and if SU = {0, 1} then

T U .We will have by induction that () for every < , if ET 6= then lh

T

T

and T T , and if EU 6= then lh

U

T and U T . We leave the

maintenance of () to the reader. We may assume that BT ||T = BU ||T .We use the terminology in T we move into MT as before. In T , we will not

move into QT (but we might set ET = F (QT )). Likewise with regard to U if

SU = {0, 1}.We say that is (T ,U)-unusual iff either (i) there is < such that

F (QT )(F (QT )) = EU U ;

or (ii) there are < < such that

= + 1,

ST = {0, 1} and ET = and E

U = F (Q

T ) and S

T+1 = {0},

SU = {0, 1} and EU = and E

T = F (Q

U ) and S

U+1 = {0},

crT = (BT ).

In case (i) (respectively, (ii)) we say that is type (i) (respectively, (ii)). Wedefine (U , T )-unusual symmetrically.

Subcase 2.1. is not (T ,U)-unusual and card(SU ) = 1.Let P MU . We have B

T ||

T = P ||T .If QT E P then in T we move into MT (so ET = and E

U = F (Q

T )).If QT 6E P then we select extenders from QT and P (but do not move into

QT in T ).17

Subcase 2.2. is not (T ,U)- or (U , T )-unusual and ST = SU = {0, 1}.

If QT E QU then in T we move into MT , and set EU = F (QT ).

If QT 5 QU then we select extenders from QT and QU .

Subcase 2.3. is (T ,U)-unusual.In T we move into MT .Suppose also that SU = {0, 1} (otherwise we are done); so is type (i). If

QU E MT then in U we move into MU ; otherwise select extenders from MT

and QU .

The remaining rules for the comparison are determined by symmetry. Wewill observe in Claim 2 below that no ordinal is both (T ,U)-unusual and (U , T )-unusual, so the definition of (T ,U) is reasonably symmetric (although not com-pletely). Now if B is active and Q is type 3, for some , QT might fail the ISC.So the next claim needs some argument.

17It might be that P |OR(QT ) is active with extender E and (F (QT )) > (E), in whichcase ET = and E

U = E. In this case we keep S

T+1 = {0, 1}. This is because if E is

superstrong, we could end up with F (QT ) active on some S MU+1.

33

Claim 1. For all + 1, + 1 < lh(T ,U), if ET 6= 6= EU then

ET T 6= E

U

U .

Proof. Suppose otherwise and let (, ) be the lexicographically least counterex-ample. Let = lhT .

Suppose that lhU = . So ET = E

U , so 6= ; so suppose < . Note that

there is [, ) such that EU 6= ; let be least such and let G = EU . Then

lh(G) = , and since lhU = , therefore G has superstrong type. So T = (G)

and = and ET 6= G. Let = predU ( + 1). By 3.38, + 1 BU and

cr(G) = (BU ). So is not U-special, so G is a premouse extender. Standardarguments (for example, see [5, 5]) now show that there is < such thatET = G. But (

, )

QT = PU and E

U 6= F , contradiction. So let be least such and let G = E

T .

Then as in the proof of Claim 1, G is a superstrong premouse extender also usedin U , contradicting Claim 1.

Since lhU < T , QT is not a premouse, so Q is type 3. It easily follows that

lhU < T , since if is a successor cardinal in Q then QT is a premouse.

Now suppose Q is superstrong. Then because QT is not a premouse, thereis T , and so

BU+1|T = QT ||T . Now suppose there is [, ) such that ET 6= . Fix such

a with + 1 [0, ]T . Let = predT ( + 1). So BT and = crT

T .

If < (F (QT )) then easily (F ) > U , contradiction. So (F (Q

T )). But

then standard arguments show that ET is a premouse extender used in bothT ,U , a contradiction.

It follows that + 1 = and ET = , so we are done.

Now suppose that is type (ii). Let F = F (QT ) = EU , = cr(F ) and

= crT = (BT ).

Suppose that Q is not superstrong. Let [, ) be such that +1 [0, ]U .Let G = EU and = cr(G). Then < < iG(), so / rg(j

U0,), but

= cr(F (QU )), so rg(jU0,), contradiction.

We now observe that = predT (). Let < be such that ET 6= . Then 6= . If < then

T (F (QT )) =

(using (); see the begining of Case 2). Since ST+1 = {0}, if > then T

T .

This suffices.We now prove that B is exact. Let = predU(+ 1). Then we claim that

+ 1 U & cr(F (QU )) = . (6)

For BU and cr(F (QU )) = = U . So suppose line (6) fails; then one can

show that + 1 BU and U = and EU+1 = F (Q

U+1); here

U+1 = .

Then is not (U , T )-unusual, by induction and since U = = cr(F (QT )).

So ET = F (QT ). But then + 1 is (U , T )-unusual, so E

U+1 6= F (Q

U+1),

contradiction.Using line (6) and since EU is total over B

U , (

++)QT

(++)BU

and

(QT BU )||(

++)QT

.

Since k = iQTF is continuous at (

++)QT

, therefore

(++)Ult(QT

,F ) (++)BT

+1

35

and(Ult(QT , F ) B

U+1)||(

++)Ult(QT

,F ).

Now suppose that B is not exact. By 3.27, neither is BT . So

(++)MT

< (++)Ult(QT

,F )

and(MT Ult(Q

T , F ) B

U+1)||(

++)MT

,

but by non-triviality, MT Ult(QT , F ). So M

T B

U+1. We have E

T =

and ST+1 = {0} and MT+1 = M

T . So

(++)MT

= (++)PT

= (++)BU

= (++)BU

+1 > (++)MT

,

contradiction. (For similar reasons, (++)BU

= (++)QT

.)We leave to the reader the proof that for any > , if BT then

cr(F (QT )) 6= T , and if B

U then cr(F (QU )) 6= T .

Finally suppose that is both (T ,U)-unusual and (U , T )-unusual. Theneither is type (ii) with respect to both, or type (i) with respect to both, sincethis depends on whether or not Q is superstrong. But then = + 1 andET = = E

U , contradiction.

Claim 3. Let < be such that SU = {0, 1} and EU = and E

T = F (Q

U )

and + 1 BT and letting = predT ( + 1), then crT = (BT ). Then + 1

is (T ,U)-unusual of type (ii).

Proof. Suppose not. Then note that is (T ,U)-unusual. But cr(F (QU )) = T ,

contradicting Claim 2.

Claim 4. The comparison terminates at some countable stage.

Proof. We may assume that B is active and Q is type 3, since otherwise everyextender used in (T ,U) is a premouse extender, and so the usual argumentworks.

Suppose that (T ,U) reaches length 1 + 1. Let OR be large and let : X V be elementary with X countable and transitive, and everythingrelevant in rg(). Let = cr(). Let W = BT1 ||1 = B

U1||1. Standard

arguments show that either iT,1 or jT,1

is defined, and if iT,1 is defined then

MT ||(+)M

T

= W ||(+)W

andiT,1 (W ||(

+)W ) = (W ||(+)W ),

and likewise if jT,1 is defined. Likewise for U .

36

Let +1 = min((, 1]T ) and +1 = min((, 1]U ). Let us assume that ; in the contrary case the proof is essentially18 the same. Let = min(T ,

T ).

Then ET = EU .

Subclaim 4.1. We have:

(a) The trivial completion of ET T is a premouse extender.

(b) < and T < U and

T <

U and lh

T < lh

U .

(c) EU T / P

U , so P

U is not a premouse and is U-special.

Proof. Part (a): We have T U because

T

U and by compatibility. So

part (a) follows from a standard argument (i.e. otherwise we get some premouseextender, which factors into ET , used in both T ,U ; cf. [5, 5]).

Part (b): We have T U . If

T =

U then E

T = E

U and < ,

contradicting Claim 1. So T < U , and therefore < .

We have T U . But we cant have

T =

U , by Claim 1 and compatibility.

So T < U .

We have lhT lhU . Suppose lh

T = = lh

U . Let P = P

U and =

lgcd(P ) = lgcd(P T ). Then ET / P . Since

T <

U and by part (a), therefore

P is not a premouse. So is U-special, so T < = U . But

T as

= lgcd(P T ), a contradiction. So lhT < lh

U .

Part (c): lhT is a cardinal of P = PU and E

T / P , by 3.38 and agreement

between models of T and U . Since T < lhT < lh

U and by part (a), P fails the

ISC, and so is U-special.

Let and T > U and lh

T > lh

U and

T >

U .

(c) ET U / P

T , so P

T fails the ISC and is T -special.

18Only essentially because our definition of (T ,U) was not quite symmetric.

37

Proof. This follows from the proof of Subclaim 4.1 and the fact that is notU-special.

Since is T -special and

= crT > crT =

and + 1 = predT (+ 1), by 3.37(c), BT and cr(F (QT )) and

= jT,+1().

But then by line (7), iF () = U . So iF () = U , so Q

U and Q have

superstrong type.It is now easy to see that is (U , T )-unusual of type (ii), and therefore

EU 6= F (QU ), a contradiction. This completes the proof of the claim.

Now that we know the comparison terminates, we must analyse the mannerin which it does. Let + 1 = lh(T ,U). Let BT = BT , etc.

Claim 5. Suppose that B is non-exceptional and the comparison does not ter-minate early. Then:

BTBU and card(ST ) = card(SU ) = 1 and MT = MU .

m 0 and the cephalanx C {BT , BU} has a good core.

Proof. We have:

Either B is non-exact or Q is non-superstrong, (8)

because B is non-exceptional and by line (5).We will show later that card(ST ) = card(SU ) = 1; assume this for now. Let

Y MT and Z MU . So Y, Z are the final models of the comparison and eitherY E Z or Z E Y . As in the proof of 3.34, because card(ST ) = card(SU ) = 1and the comparison does not terminate early, we have BTBU ; we mayassume BT \BU . So Z is unsound and MT = Y E Z. We need to see thatBT has a good core. We have ST = {0}. Let = BT (). We will show laterthat MT = Z; assume this for now.

Case I. is (T ,U)-unsual.By line (8) and Claim 2, is type (i), = + 1, and EU is equivalent to

F (QT ), and Q is type 3 but not superstrong. At stage , in T we move intoMT = M

T . We have ET = 6= EU and (M

T BU )|T and T < lhU and

T is a successor cardinal of BU . So T U . Since M

T is T -sound, it followsthat there is exactly one ordinal such that and + 1 U , and in fact + 1 = . So PU is a premouse, as / B

U . Since MT is T -sound, therefore

= and EU is type 1 or type 3, with lhU = ((

T )+)MT

. It follows that m 0

and BT is non-exact, and letting F = F (NBT

) and = cr(F ), we have

EU = F, and MT has an (m, T )-good core at , and GM

T

m,,T= F T

38

and HMT

m, = BU . Also, if F

is type 1 then predU () = and / BU and

BU = BU , and M

T has an (m, T )-good core at cr(F (QT )), etc. So BT has agood core.

Case II. is not (T ,U)-unusual.So EU = F

where F = F (QT ).

Subcase II.1. Q is not superstrong.So F does not have superstrong type. Things work much as in the previous

case, but there are a couple more possibilities, which we just outline. If B isexact then = + 1, and F is the last extender used in U . If B is non-exact

then = +2 and like above, F = F (NBT

) is type 1 or type 3 and is the last

extender used in U . In the latter case, if NBT

is type 1 and QT is type 2 thenQT = BU .

Subcase II.2. Q is superstrong.So F has superstrong type, and by line (8), B is non-exact. Things work

much as before, but there are some extra details. We just give the details inone example case. Let = predU ( + 1). So = cr(F ) < U and

(()+)QT

lhU .

Suppose for example that (()+)QT

= lhU . Then EU is type 2 and B

U+1 = P

U ,

and OR(BU+1) = OR(QT ) and BU+1 is active type 2, so E

U+1 = F (B

U+1). Note

that(QT BU+1)||((

)++)QT

,

and so by 3.8,

(Ult(QT , F ) BU+2)||((T )++)Ult(Q

T ,F ),

and so EU+2 = F (NBT ), etc. We leave the remaining details to the reader.

This completes the proof under the assumptions made above. Now supposethe assumptions fail. Then we either have 6= MT Z for all Z MU , orthe reflection of this; suppose the former. Suppose B is non-exact. Arguing as

above, if is (T ,U)-unusual then NBT

BU ; otherwise EU = F (Q

T ) and either

NBT

BU+1 or NBT BU+2. But because M

T 6= NBT

this contradicts the fact

that MT Z. So B is exact, so Q is not superstrong. But then arguing again

as above, ((T )+)Z = ((T )+)MT

, again contradicting that MT Z.

The next claim follows immediately from the rules of comparison. It appliesin particular if any is (T ,U)- or (U , T )-unusual.

Claim 6. If Q is type 1 or type 3 then for all , ST 6= {1} 6= SU .

Claim 7. Suppose the comparison terminates early (so is either (T ,U)- or(U , T )-unusual). Then:

B is exact.

39

If is (T ,U)-unusual then SU = {0}, and vice versa.

The final models of the comparison are MT =MU .

Proof. We use Claim 2 in what follows.Suppose is (T ,U)-unusual of type (i). Since QT does not have superstrong

type and BT is non-trivial, it is easy to see thatMT BU , and BT , B are exact.But then if SU = {0, 1}, the comparison in fact does not terminate early at stage. So SU = {0}. Also = + 1 and EU 6= , so / B

U . So MU is unsound,

so MT = MU .Now suppose is (T ,U)-unusual of type (ii). We adopt the notation used

in the comparison rules and in the proof of Claim 2. Let = (BT ). We have

T = U = (+)M

T

= (+)MU

and MU = MU and either MT EMU or MU EMT and MT ,MU both project

T . So it suffices to see that

(++)MT

= (++)MU

.

Now Q is superstrong and B is exact. As mentioned in the proof of Claim 2,

(++)QT

= where = (++)BU

, and (QT BU )|. Let iF = i

QTF . So by

exactness, (letting be)

= (++)MT

= (++)BU

+1 = (++)BU

= (++)BT

= sup iF

(recall that F = F (QT )) and

(MT BU+1 B

U B

T )|

and letting G = F (QU ) and iG = iQUG ,

(++)MU

= sup iG

and letting H = F (QT ) and iH = iQT

H ,

(++)MT

= sup iH.

We claim thatiH = iG iF ,

which completes the proof. This claim holds because H = G F and (so)

iH (+)Q

T

= iG iF (+)Q

T

and iH and iG iF are both continuous at (+)QT

.

40

Claim 8. Suppose that the comparison terminates early andB is non-exceptional.Then BT has a good core.

Proof. Because B is non-exceptional, by Claim 7 and line (5), is either (T ,U)-or (U , T )-unusual of type (i). Now argue like in the proof of Claim 5, using Claim7.

We now move to the case that B is exceptional.

Claim 9. Suppose B is exceptional and the comparison does not terminateearly. Then one of BT , BU is a cephalanx with an exceptional core.

Proof. We consider a few cases:

Case I. Either (a) ST = {0} = SU and MU MT , or (b) ST = {0, 1}.This case is covered by the next case and symmetry.

Case II. Either (a) ST = {0} = SU and MT MU , or (b) SU = {0, 1}.If (b) holds, then because the comparison does not terminate early, ST = {0}

andMT BU . So given either (a) or (b), MT BU . SoMT is sound, so BT .Let = ST (). So EU 6= and lh

U

T . BecauseMT projects T , it follows

that lhU = T , so is not (T ,U)-unusual and EU = F

where F = F (QT ).Also, if > + 1 then + 1 is not (U , T )-unusual, by Claim 2 and sinceMT BU . Let = cr(F ) and = predU ( + 1). We split into two subcases:

Subcase II.1. BU+1|T is active.

By the discussion above:

BU+1 is type 2 and T = OR(BU+1),

(+)QT

= OR(BU+1),

EU = F (BU+1) and E

U+1 = F (B

U+1),

= + 2.

Let R = BU+1 and G = FR = EU . Then

(+)Ult0(R,G) = (+)QT

= lh(G)

and(Ult0(R,G) Q

T )||(++)QT

,

but because BT is exact and MT BU and by 3.10,

(++)Ult0(R,G) > (++)QT

.

Let H Ult0(R,G) and h {1} be such that (++)H = (++)QT

and

Hh+1 = lh(G) < Hh . Let H

= iUlt0(R,G)F (H); so

H Ult0(Ult0(R,G), F).

41

We claim thatUlth(H,F

) E H. (9)

If h = 1 this is by the ISC, so suppose h 0. Let

: Ulth(H,F) H

be the factor map. Arguing as in the proof of 2.13, we get thatH, Ult0(R,G)and the hypotheses of 2.13 hold (for H,, h,H). By 2.13, R |=Lemma 2.13holds for my proper segments. If 6= id then Hh+1 < cr(), and so line (9)holds.

So Ulth(H,F) BU+2. But

((T )+)Ulth(H,F) = ((T )+)M

T

,

so h = m and MT = Ulth(H,F). It easily follows that BT has an exceptional

core, and with X as in 3.42,

H = cHullMT

m+1(X zMT

m+1 ~pMT

m ).

Subcase II.2. BU+1|T is passive.

Then = + 1, so MT BU+1 = BU . Let R = BU . If B

U then

< (R), so (++)R is well-defined. In any case, is not the largest cardinal

of R. We have (+)R = (+)QT

and

(R Q)||(++)QT

.

If (++)R > (++)QT

then a simplification of the argument in the previous

subcase works. Suppose then that (++)R = (++)QT

. Because MT BU , it iseasy enough to see that / BU , so R is a premouse. If R is active type 3, then

(+)R < (FR), because if (+)R = (FR) then OR(BU+1) = ((T )+)M

T

, a

contradiction. Let d = degU(+1). Then iU+1 is discontinuous at (++)R, and

so (+)R = Rd , so d > 0. Let r < d be such that Rr+1 = (

+)R < Rr . Thenarguing like in the previous subcase, but using 3.29 instead of 2.13,

Ultr(R,F) BU+1,

so Ultr(R,F) =MT , and like before, BT has an exceptional core (and m = r).

Case III. ST = {0} = SU and MT = MU .Then BTBU ; assume BT \BU . Let = ST ().

Subclaim 9.1. is not (T ,U)-unusual.

Proof. Suppose otherwise. Let < < witness this. Since the comparisondoes not terminate early, and MT is T -sound, it follows that = + 1 and

T = U < lhU = ((

T )+)MT

.

42

So T is not the largest cardinal in MT , and so nor in MU . So PU M

U , so

((T )+)MT

< ((T )+)MU

.

This contradicts Claim 2.

So EU = F = F (QT ).

Subclaim 9.2. + 1 is not (U , T )-unusual.

Proof. This is like the proof of Subclaim 9.1.

By the subclaim and Claim 3, one of the following holds:

(a) = + 1.

(b) = + 2 and lhU+1 = T and EU+1 is type 2.

(c) = + 2 and lhU+1 = ((T )+)M

T

and EU+1 is either (i) type 1 or (ii)type 3.

(d) = + 3 and lhU+1 = T and EU+1 is type 2 and lh

U+2 = ((

T )+)MT

and EU+2 is either type (i) 1 or (ii) type 3.

The same general argument works in each case, but the details vary. We justdiscuss cases (a), (b), (c)(i) and sketch (d)(i). In each case let = predU (+1)and R = BU+1 and = cr(F

).Consider case (a). We first observe that

Rm+1 = (+)R < Rm.

For if Rm+1 > (+)R then

m+1(MU ) > T m+1(M

T );

if Rm+1 then m+1(MU ) < T and MU is T -sound, so BT is not excep-

tional, contradicting 3.41.

Let d = degU(). Note that (++)R = (++)BT

and

((T )+)Ultm(R,F) = ((T )+)M

T

,

so arguing like in the proof of 2.13, it follows that

Ultm(R,F) = Ultd(R,F

) = MT

and the factor map between the ultrapowers is the identity. (We dont need touse any condensation here.) Letting = iRF and H = R, then H, are as in3.42.

Now consider case (b). Note that R = PU , so R is active type 2, and

ORR = (+)BT

. Note that degU (+2) = m and cr(FR) = crU+1, so predU (+

43

2) = predU( + 1) and BU+1 = BU+2 and deg

U ( + 1) = m. Let H = BU+1.Then

Ultm(H,F) =MU

and letting = iHF , then H, are as in 3.42.Now consider case (c)(i).

Subclaim 9.3. In case (c)(i), EU is the preimage of EU+1 and

lhU = (++)B

T

.

Proof. We have PU E R and

(+)BT

= (+)PU

= (+)PU

= (+)R < lhU .

We also have (++)R (++)BT

and

if + 1 BU then (++)R > (++)BT

;

the latter is because PU+1 6EMT and PU+1 projects

T .

Let P E R and p {1} be such that (++)BT

= (++)P and

Pp+1 (+)B

T

< Pp .

Then like before, using condensation or the ISC, we have

Ultp(P, F) EMU+1.

But((T )+)Ultp(P,F ) = ((T )+)M

T

,

and since U+1 = T , therefore Ultp(P, F

) = PU+1. So P is type 1 and ORP =

(++)BT

. Therefore EU = FP . Now iU+1 is continuous at (

+)BT

. So if

P R then iU+1 is continuous at ORP , and so iU+1(P ) = P

U+1. If P = R then

Ultp(P, F) = MU+1 (even if p < deg

U ( + 1)).

Since EU = FP and cr(FP ) = cr(F ), predU( + 1) = and BU+1 = R and

degU (+1) = degU (+1). Also, predU (+2) = +1 and m = degU (+2) =degU (+ 1). Using this, and letting H = BU+1, we get

Ultm(H,F) = MT

and letting = iHF , then H, are as in 3.42.Finally consider case (d)(i). For illustration, assume that + 2 / BU . Let

= predU ( + 2) and S = BU+2 and j = degU( + 2). A combination of the

preceding arguments gives the following:

PU is the type 2 preimage of PU+1,

44

predU (+ 1) = and BU+1 = S and degU (+ 1) = j,

PU+1 is the type 1 preimage of PU+2, under the map defined below,

= predU (+ 2) and degU (+ 2) = 0,

+ 1 = predU ( + 3) and m = degU ( + 3) = 0.

Let J = BU+1 and H = BU+2. Those arguments also give that

Ultj(J, F) = BU+2

and letting = iJF , then (PU+1) = P

U+2 (as mentioned above), and

Ult0(H,F) =MT ,

and etc.Cases (b)(ii) and (c)(ii) are fairly similar to the preceding cases.

There is just one case left:

Claim 10. Suppose that B is exceptional and the comparison terminates early.Then BT BU and one of BT , BU has an exceptional core.

Proof. We may assume that is (U , T )-unusual. Let < < witness this.So EU = F

= F (QT ). We have either MT EMU or MU E MT . By Claim 2,

therefore MT = MU . Let H =MU . Then

Ultm(H,F) = MT ,

and etc.

If B is non-exceptional (exceptional) then by some of the preceding claims,we have an iterate B of B such that B has a good (exceptional) core. In thenon-exceptional case, the proof of Claim 6 of 3.32 then shows that B has a goodcore. So the following claim completes the proof of the theorem:

Claim 11. Suppose that B is exceptional and let B be a non-dropping iterateof B. Then B has an exceptional core iff B does.

Proof. The proof similar to 3.32, but with some extra argument. We assumethat m 0 and leave the other case to the reader (the main distinction inthat case is that even though m = 1, all ultrapower embeddings are at leastr1 elementary). Fix H,, F,X as in 3.42. Let B

= (, ,M , Q) and fixH , , F , X as in 3.42 with respect to B. Let i :M M and j : Q Q bethe iteration maps. So i(B||) = j (B||). Note that for each < , we haveX B|| and

i(X ) = X i(). (10)

Also, i(, zMm+1) = (, zM

m+1). Because B has an exceptional core, we have

HullM

m+1(X zM

m+1 ~pM

m ) = X .

45

From these facts, it follows easily that

HullMm+1(X zMm+1 ~p

Mm ) = X. (11)

It remains to see that

H ||(++)H = M |(++)M .

LetY = HullMm+1(X z

Mm+1 ~p

Mm ) (

+)M ,

let = iF (++)M , and Z = rg(). It suffices to see that Y = Z. Let Y , , Z

be defined analogously from B. Because B has an exceptional core, Y = Z .So the subclaims below immediately give that Y = Z, completing the proof.They are proven by breaking Y and Z into unions of small pieces.

Subclaim 11.1. For any < (+)M , Y iff i() Y .

Proof. If Y then i() Y because iX X and i(zMm+1) = zM

m+1.Suppose / Y . For < and < Mm let Y, be the set of all < (

+)M

such that HullMm+1((X ) z

Mm+1 ~p

Mm ),

as witnessed by some theory below

ThMrm( {~pMm }).

(See [3, 2], in particular, the stratification of rm+1 described there, for moredetails. If m = 0 or this should be modified appropriately; for example, ifm = 0 and M is passive and ORM is divisible by 2 then instead, the r1 factshould be true in M |.) Then Y, M . Define Y , analogously over M

. Let

I = Mm . Using line (10), we have

i(Y,) = Yi(),i(),

and an easy calculation gives

Y =

(,)I

i(Y,).

The fact that i() / Y follows easily.

Subclaim 11.2. For any < (+)M , Z iff i() Z .

Proof. Suppose Z. Let = 1(). We claim that (i()) = i(), whichsuffices. For let C (+)M be a wellorder of (+)M in ordertype , with C M .Then (C) M and is a wellorder of B in ordertype . Therefore i(C), i()and i((C)), i() are likewise. So it suffices to see that (i(C)) = i((C)). Butfor any D P() M , we have (i(D)) = i((D)), and so the continuity ofthe various maps at (+)M then easily gives what we want.

46

Now suppose / Z; we want to see that i() / Z . The proof is similar innature to the proof of Subclaim 11.1; we just describe the decomposition of Z.For < (++)M let f : (

+)M be a surjection in M . For < (++)M

and < letZ, = (f)(X ).

Then Z, M . Now is continuous at (+)M and note that

rg() =,

Z,.

Define f and Z, analogously from B

; we may assume that i(f) = fi().

Then we havei(Z,) = Z

i(),i(),

and it follows that i() / Z . This completes the proof of the theorem.

4. Condensation from normal iterability

Standard (k + 1)-condensation19 gives the following. Let H,M be (k + 1)-sound premice, where M is (k, 1, +1)-iterable, and let : H M be a neark-embedding, with cr() = Hk+1. Then either:

H EM , or M | is active with extender F and H Ult(M |, F ).

(12)

As discussed in the introduction, the standard proof uses the (k, 1, 1 + 1)-iterability of M . We now give a proof of the above statement, reducing theiterability hypothesis to just (k, 1+1)-iterability. In our proof, we will replacethe phalanx used in the standard proof with a cephal, and avoid any use ofDodd-Jensen. Much as in [12, 9.3.2], we will also weaken the fine structuralassumptions on ,H,M somewhat from those stated above. In particular, asdiscussed earlier, we will not assume that M is (k + 1)-solid. Because we dropthis assumption, it seems that we need to weaken a little the conclusion ofcondensation in the case that H / M (cf. 4.2(1)), compared to the versionstated in [12]. So in this sense we cannot quite prove full condensation.20 Inorder to state the weakened conclusion, we need the following definition.

4.1 Definition. Let M be a k-sound premouse and let [Mk+1, Mk ]. The

-solid-core of M is

H = cHullMk+1( zMk+1 ~p

Mk ),

and the -solid-core map is the uncollapse map : H M .

19Cf. [3, pp. 8788] or [12, Theorem 9.3.2].20But see 6.

47

Note that the -solid-core map is a k-embedding, since H /M and by 2.4.

4.2 Theorem (Condensation). Let M be a k-sound, (k, 1 + 1)-iterable pre-mouse. Let H be a -sound premouse where is a cardinal of H and Hk+1

< Hk . Suppose there is a k-lifting embedding : H M such that cr() .

Let = cardM (). Then either :

1. H /M and :

(a) Hk+1 = Mk+1 and (z

Hk+1) = z

Mk+1,

(b) H is the -solid-core of M and is the -solid-core map,

(c) Hk+1 / [, ),

(d) if Hk+1 = and (+)H < (+)M thenM | is active with a superstrong

extender with critical point and Mk+1 (+)M < ,

(e) Hk+1 Mk+1,

(f ) if M is (k + 1)-solid then Hk+1 = Mk+1,

(g) if Hk+1 = Mk+1 then H is the -core of M , is the -core map and

(pHk+1) = pMk+1;

or

2. H M and exactly one of the following holds :

(a) H M , or

(b) M | is active with extender F and H Ult(M |, F ), or

(c) M | is passive, M |(+)H is active with a type 1 extender G and H =

Ultk(Q,G), where QM is such that (+)Q = and Qk+1 = <

Qk ,

or

(d) k = 0 and H,M are active type 2 and M | is active with a type 2extender F and letti