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practice_final_2010_solutions

# practice_final_2010_solutions - PRACTICE FINAL 1 True or...

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PRACTICE FINAL 1. True or False? (2 pts each) L is a first-order language (with equality). M and M 0 are interpreta- tions of L . X M is the domain of M . X M 0 is the domain of M 0 . 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 0 and X M is finite, then there is a φ L such that M | = φ and M 0 | = ¬ 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 L -definable subsets of X M , then A × B is an L -definable 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 | = ( 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 | = ∧ ∃ , then M | = x ( α β ). False : Suppose L is built from a unary relation symbol H . Let X M = { 0 , 1 } . Let H M = { 0 } . 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

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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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