Lecture_21_09 - Lecture 21 Lecture 21 Definable Sets...

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Unformatted text preview: Lecture 21 Lecture 21 Definable Sets Definable Sets XM = {0,1,2,3,4} HM = {3,4} RM = {(0,1),(0,2),(1,3),(3,3),(2,4)} cM = 4 Find the set of pairs defined by R(x2,x1) ∧¬(x1 ≈ c) ∧ H(x1) Definable Sets Definable Sets XM = {0,1,2,3,4} HM = {3,4} RM = {(0,1),(0,2),(1,3),(3,3),(2,4)} cM = 4 Define the set {0,2,4}. Distinguishing structures Distinguishing structures XM = N RM = {(x,y) : x < y} XM’ = N × N RM’ = {((q,r),(s,t)) : q < s or (q = s and r < t)} Find a sentence that is true in M and false in M’. Logical Implication Logical Implication Γ╞ ϕ iff there is no pair (M,g), where M is an interpretation and g is a variable assignment, such that (M,g) satisfies Γ but fails to satisfy ϕ. Note: If we restrict our attention to sentences, Note If then of course we can avoid the references to variable assignments. variable Counterexamples Counterexamples Show that the following do not hold: ∀x∃ y(y ≤ x)╞ ∀w∀x∃ y(y ≤ w ∧y ≤ x ) y(y y(y ∃ xH(x) ∧∃ yG(y) ╞ ∃ x ∃ y¬(x ≈ y) xH(x) yG(y) (x ∀x(H(x) → G(x)) ╞ ¬∃ x(¬H(x) ∧G(x)) x(H(x) H(x) ¬∃ Implication Laws for FOL Implication Laws for FOL Substitution of Free Variables by New Constants Γ╞⊥ iff Sv,c Γ╞⊥ iff v,c where c is any individual constant new for Γ. where Implication Laws for FOL Implication Laws for FOL Universal and Existential Quantification (∀,╞) Γ,∀xϕ ╞ ⊥ iff Γ,∀xϕ , Sx,cϕ ╞ ⊥ where c is any individual constant. (∃ ,╞) Γ,∃ xϕ ╞ ⊥ iff Γ,Sx,cϕ ╞ ⊥ where c is any individual constant new for Γ and ∃ xϕ . Implication Laws for FOL Implication Laws for FOL Negated Quantification (¬∀,╞) Γ,¬∀xϕ ╞ ⊥ iff Γ, ¬Sx,cϕ ╞ ⊥ where c is any individual constant new for Γ and ∀xϕ . (¬∃ ,╞) Γ,¬∃ xϕ ╞ ⊥ iff Γ,¬∃ xϕ ,¬Sx,cϕ ╞ ⊥ where c is any individual constant. ...
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