Logic1111 - More difficult results T 1 2 Knigs lemma...

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T° °°° °° More difficult results 1. König’s lemma 2. Compactness
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HW #8 will be assigned on 11/16. Read Ch.17 for the next week. In HW #6, Q2, T∀x(Kx ° x(Sx v~Kx)) T(Ka ° ° (Sx v~Kx)), replace all free occurrences of x with any parameter a. Then apply the tableau rule for 6 .
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Tree Level 0: origin a0 Level 1: successors of a0 – finite or denumerably many Level 2: successors of successors … ( descendents of the above) Path : finite or denumberable sequence (a0, a1, a2, …), where each an+1 is a successor of an at level n+1. A length of a path (a0, a1, a2, …, an) is
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(1) F∀x(Kx ° (∀xKx v ∀x~Kx)) => ∀xKx Level 0 (origin) (2) T∀x(Kx Y (∀xKx v ∀x~Kx)) (1) Level 1 (3) F∀xKx (1) (4) FKa (3) Level 2 (5) TKa Æ (∀xKx v ∀x~Kx) (2) (6) Tka (7) FKa (5) Level 3 (8) F∀xKx v ∀x~Kx (5)
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16.9 For each n, at least one path of length n. Is there necessarily an infinite path? Cf. the tree on p.185. The origin has de- numerably many successors. However, for any path, its length is finite. The set N of all natural numbers is infinite, but any of its members is a finite number.
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Finitely generated tree: each ele- ment (also called point) has only fi- nitely many successors. 16.10: In a finitely generated tree, must each level contain finitely many points?
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