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Unformatted text preview: PRACTICE FINAL 1. True or False? (2 pts each) L is a firstorder language (with equality). M and M are interpreta tions of L . X M is the domain of M . X M is the domain of M . ψ ( x ) is a formula in L that has x as its only free variable. θ [ x : t ] is obtained from θ by replacing every free occurrence of x by t . • If X M has more elements than X M and X M is finite, then there is a φ ∈ L such that M  = φ and M  = ¬ φ . True : As we discussed, for each natural number k there is a sentence λ k such that M  = λ k iff X M has at least k elements. For example, we can take λ 2 to be ∃ x ∃ y ¬ ( x ≈ y ). • If A and B are Ldefinable subsets of X M , then A × B is an Ldefinable subset of X M × X M . True : If φ ( x ) defines A and ψ ( y ) defines B , then φ ( x ) ∧ ψ ( y ) defines the relation A × B . • If M  = ∀ xψ ( x ), then X M is the subset of X M that is defined by ψ ( x ). True : This follows directly from the definitions. • If M  = ∃ xα ∧ ∃ xβ , then M  = ∃ x ( α ∧ β ). False : Suppose L is built from a unary relation symbol H . Let X M = { , 1 } . Let H M = { } . M  = ∃ xH ( x ) ∧ ∃ x ¬ H ( x ) but it is not the case that M  = ∃ x ( H ( x ) ∧ ¬ H ( x )). • If a and b are constant symbols in L and M  = θ [ x : a ] ↔ θ [ x : b ] for all θ , where θ has x as its only free variable, then M  = a ≈ b . True : Suppose that M  = θ [ x : a ] ↔ θ [ x : b ] for all θ , where θ has x as its only free variable. It follows that M  = ( x ≈ a )[ x : a ] ↔ ( x ≈ a )[ x : b ] but since M  = ( x ≈ a )[ x : a ] it follows that M  = ( x ≈ a )[ x : b ]. Thus, M  = a ≈ b . 2. Explain (5 pts each) Explain two of your answers from Section 1. 1 2 PRACTICE FINAL 3. Inductive Definition (5 pts each) Using the successor function, give an inductive definition of the set { ( q,r,s )  q,r,s ∈ N and q + r ≤ s } ....
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This note was uploaded on 02/27/2011 for the course PHILOSOPHY 101 taught by Professor H during the Spring '11 term at Columbia College.
 Spring '11
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