Definition: p(x,b) divides l if there exists a c-indiscernible sequence (bi) in P(b) such that Λp(x,bi) is inconsistent. There exists a k-consistency witness y such that p(x,b) is k-inconsistent. If we have negations, then we can say: Λ¬p(x,bi) is k-consistent. And apply ext/ext to get (bi) indiscernible such that k is k-consistent. Definition: Let ψ(x,y) be a formula (x,y) tuples of variables. ψ(x,y0) is another formula such that each yi has the same length as y and each yi is in the sort of y. Then y is a k-inconsistency witness for ψ if: T ⊢ ψ(x,y0) T ⊢ ¬ψ(x,y) Definition: A formula ψ(x,b) divides l with respect to a k-inconsistency witness y if there exists a sequence (bi) in P(b) satisfying: Λi≤k ψ(x,bi) = ψ(x,y0) Λi≤k ψ(x,yi) = Λi≤k ψ(x,bi) Proposition 1: If p(x,b) divides l, then there exists a k-inconsistency witness y such that there exists a c-indiscernible sequence (bi) in tp(b|c) satisfying Λ(bi,bk-i). Proof: Assume p(x,b) divides l. There exists an indiscernible sequence (bi) in tp(b|c) such that Λp(x,bi) is inconsistent. By compactness, there exists a k < ω such that Λi≤k p(x,bi) is inconsistent. Let g(y0,...,yk) = tp(bk|c). g(y) Λi≤k p(x,yi) is inconsistent. There exists a y(g) ∈ tp(bk|c) such that p(x,y) Λi≤k p(x,yi) is inconsistent. Proposition 2: If we have a sequence (bi) in tp(b|c) satisfying y, then there exists a sequence (bi') in tp(b|c) satisfying y and being indiscernible. Proof: Since compactness applies to y but not to tp(b|c), we can apply extension/extraction to get a sequence (bi') indiscernible and having the same properties. Proposition 3: Λ(bi,bk-i) is inconsistent because ψ(b,b) is true. We prove by induction that if g ∈ D(P,=) then g ∈ D(Q,=) (converse is clear since p ≤ q). To come later... Extension is true. Proof: We are given a, b, c. Let g ∈ D(tp(a|c),=) be maximal. Since p is over b as a partial type, p(x) ∧ div_b,c(x) is consistent. Let a' ⊢ p(x) ∧ div_b,c(x). Then a' ∈ c and a' ∪ b since (tp(a|bc),=). D(tp(a|c),=) contains a maximal element of D(tp(a|c),=). Now since we have extension, we have symmetry, transitivity, etc. Still have independence theorem. Lemma: Assume (ai,bw) is c-indiscernible. Then a ∈ w. Proof: Let (cj: j < ω) be an ω-indiscernible sequence in tp(c|aω). Let ψ(x) ∈ acnc ∈ tp(aω/acw). Then there exists a ψ(an,cn,cj) (since ψ(an,cn) ∈ c). ⇒ ψ(x,acn,c) ∈ dnd/acw. Lemma: Let (ai: i < 2ω) be a c-indiscernible sequence over c,aω. Then (aωti: i < ω) is a Morley sequence over c,aω. Proof: (ai: i < ω) is an indiscernible over c ∪ {aj: j < ω}. Notice: tp(aωi,aωj/c) = tp(aωi,aωj+1/c). ⇒ aωti ≡ aωt(j+1) By induction, want to prove: aωti ≡ aωt(j+1) for n ≥ 0.

18.996A Simplicity Theory: Lecture 5 Notes

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