5.5comp071010

# 5.5comp071010 - 1 5.5 Comprehension indiscernibles We fix M...

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Unformatted text preview: 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 by 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 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, it suffices to 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 )). 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 ])......
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5.5comp071010 - 1 5.5 Comprehension indiscernibles We fix M...

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