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5.5comp010911

# 5.5comp010911 - 1 5.5 Comprehension indiscernibles We fix M...

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1 5.5. Comprehension, indiscernibles. We fix M = (A,<,0,1,+,-,•, ,log,E,c 1 ,c 2 ,...) and terms t 1 ,t 2 ,... of L(E) be given as in Lemma 5.4.17. We now consider unbounded quantifiers. Below, Q indicates either or . All formulas of L(E) are interpreted in M. LEMMA 5.5.1. Let n,m 0, r 1, and ϕ (v 1 ,...,v n+m ) be a quantifier free formula of L(E). Let x n+1 ,...,x n+m E [0,c r ]. Then (Q n x n E)...(Q 1 x 1 E)( ϕ (x 1 ,...,x n+m )) (Q n x n E [0,c r+1 ])...(Q 1 x 1 E [0,c r+n ])( ϕ (x 1 ,...,x n+m )). Proof: We prove the following statement by induction on n 0. Let m 0, r 1, ϕ (x 1 ,...,x n+m ) be a quantifier free formula in L(E), Q 1 ,...,Q n be quantifiers, and x n+1 ,...,x n+m E [0,c r ]. Then (Q n x n E)...(Q 1 x 1 E)( ϕ (x 1 ,...,x n+m )) (Q n x n E [0,c r+1 ])...(Q 1 x 1 E [0,c r+n ])( ϕ (x 1 ,...,x n+m )). The basis case n = 0 is trivial. Assume this is true for a given n 0. Let m 0, r 1, and ϕ (x 1 ,...,x n+1+m ) be a quantifier free formula in L(E). Let x n+2 ,...,x n+1+m E [0,c r ]. We wish to verify that (Q n+1 x n+1 E)...(Q 1 x 1 E)( ϕ (x 1 ,...,x n+1+m )) (Q n+1 x n+1 E [0,c r+1 ])...(Q 1 x 1 E [0,c r+n+1 ]) ( ϕ (x 1 ,...,x n+1+m )). By duality, we may assume that Q n+1 is . Thus we wish to verify that 1) ( x n+1 E)(Q n x n E)...(Q 1 x 1 E)( ϕ (x 1 ,...,x n+1+m )) ( x n+1 E [0,c r+1 ])(Q n x n E [0,c r+2 ])... (Q 1 x 1 E [0,c r+n+1 ])( ϕ (x 1 ,...,x n+1+m )). Let x n+1 E [0,c r+1 ] witness the right side of 1). I.e., 2) (Q n x n E [0,c r+2 ])...(Q 1 x 1 E [0,c r+n+1 ])( ϕ (x 1 ,...,x n+1+m )).

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2 According to the induction hypothesis applied to m+1,r+1, ϕ (x 1 ,...,x n+1+m ),Q 1 ,...,Q n , and x n+1 ,...,x m+1+m , E [0,c r+1 ], we have 3) (Q n x n E)...(Q 1 x 1 E)( ϕ (x 1 ,...,x n+1+m )) (Q n x n E [0,c r+2 ])...(Q 1 x 1 E [0,c r+n+1 ])( ϕ (x 1 ,...,x n+1+m )). By 2),3), (Q n x n E)...(Q 1 x 1 E)( ϕ (x 1 ,...,x n+1+m )), which is the left side of 1) instantiated with x n+1 . Finally, let x n+1 E witness the left side of 1). I.e., 4) (Q n x n E)...(Q 1 x 1 E)( ϕ (x 1 ,...,x n+1+m )). Let x n+1 c s , s r+1. According to the induction hypothesis applied to m+1,s, ϕ (x 1 ,...,x n+1+m ),Q 1 ,...,Q n , and x n+2 ,...,x n+1+m , E [0,c s ], we have 5) (Q n x n E)...(Q 1 x 1 E)( ϕ (x 1 ,...,x n+1+m )) (Q n x n E [0,c s+1 ])...(Q 1 x 1 E [0,c s+n ])( ϕ (x 1 ,...,x n+1+m )). By 4),5), ( x n+1 E [0,c s ])(Q n x n E [0,c s+1 ])... (Q 1 x 1 E [0,c s+n ])( ϕ (x 1 ,...,x n+1+m )). By Lemma 5.4.17 vii), since x n+2 ,...,x n+1+m E [0,c r ], we have ( x n+1 E [0,c r+1 ])(Q n x n E [0,c r+2 ])... (Q 1 x 1 E [0,c r+n+1 ])( ϕ (x 1 ,...,x n+1+m )) which is the right side of 1). QED Note that Lemmas 5.4.17 and 5.5.1 concern only the E formulas of L(E). I.e., all of the quantifiers are relativized to E. This is clear for Lemma 5.4.17 by Definitions 5.4.4, 5.4.5. Lemma 5.5.1 only involves quantifier free formulas which are inside quantifiers relativized to E. DEFINITION 5.5.1. Let k 1 and R A k . We say that R is M,E definable if and only if R E k , and R is definable by an E
3 formula of L(E) with parameters from E. I.e., there exists m 1, an E formula ϕ (x 1 ,...,x k+m ), and x k+1 ,...,x k+m E, such that R = {(x 1 ,...,x k ) E k : ϕ (x 1 ,...,x k+m )}.

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