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5 Nov 2009: Then there is a subset B A of size β such that all theunordered γ-tuples from it are in the same piece of the partition. ... b) the property of relative largeness which plays such a central rôle inParis -Harrington is ill-typed in at least two senses.

21 Jan 2009: p] B1, and( iv ) there exist homogeneous elements y1,. , yt B such that. ... Lemma. Let (A, A1) be a Frobenius pair such that B and B1 are Gorenstein, andlet M be a finitely generated A-module.

31 Jul 2009: Lemma 44. There exist strictly positive constants A, A′ and B such that. ... Show that there is a θ R such that. |a b(δ exp iθ)N | = |a| |b|δN.

8 Oct 2009: Theorem B. Let F be a Finsler metric on S2 such that K δ > 0. ... Since v nevervanishes we may choose a smooth angle ϕ(t) such that.

4 Aug 2009: Theorem 5.12. If f : [a,b] R is continuous then we can find x1,x2 [a,b]such that. ... If f : [a,b] R is a continuousfunction with f differentiable on (a,b), then we can find a c (a,b) such that.

13 Aug 2009: Givenxn1 X and δn > 0, we can find xn X such that xn B(xn1,δn/4), butxn / En. ... Since g issmooth, there exists a constant B such that. |ĝ(r)| B|r|2.

16 Oct 2009: g0,W) such that the (r 1)-jet ofPg(Σ,γ) belongs to Q. ... Then there exists aresidual subset O Gr such that for all g O:. •

9 Oct 2009: One way to think of such a relation is. that if xrb then the tokens in the multiset x trigger the token b. ... maximal but many minimal relations r : M A B such that r = f.

6 Nov 2009: We say such a formula is maximal.). That feature invitesthe simplification that we showed there. ... What’s the problem? The problem isthat there might be no such set of rules.