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e.g.: Q(x, y): x + y = 0, then (  x)(  y)Q(x, y):

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e.g.: Q(x, y): x + y = 0, then (  x)(  y)Q(x, y):

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  1. Section 1.4 Predicate LogicInference rules for predicate logic:1-7. All inference rules in propositional logic8. (x)P(x)  P(t) where t is a variable or a constant (universal instantiation ui)9. (x)P(x)  P(t) where t is a variable or a constant not previously used in proof sequence (existential instantiation ei)10. P(x)  (x)P(x) if (a) P(x) has not been deduced from any hypotheses in which x is a free variable, and (b) P(x) has not been deduced by ei from any wff in which x is a free variable (universal generalization ug)11. P(x) or P(a)  (x)P(x) (existential generalization eg)

  2. e.g.: Q(x, y): x + y = 0, then (x)( y)Q(x, y): for every integer x there is a negative integer y s.t. x + y = 0 Note that it is not true that (x)Q(x, t) where t is a particular element in domain, by adding that same integer t to every x will not always produce 0. In general, (x)( y)Q(x, y) (x)Q(x, t)

  3. ·   e.g. Prove (x)[P(x)Q(x)]  (x)P(x) (x)Q(x) • 1.      (x)[P(x)Q(x)] (hypothesis) • 2.      P(x)Q(x) (ui) • 3.      P(x) (2, simplification) • 4.      Q(x) (2, simplification) • 5.      (x) P(x) (3, ug) • 6.      (x) Q(x) (4, ug) • 7.      (x) P(x)  (x) Q(x) (5,6, conjunction)

  4. e.g. (x)[ P(x)Q(x) ]  (x)P(x)  (x)Q(x) first change it to: ( by rewriting implication.) (x)[ P(x)Q(x) ]  ( [(x)P(x)] (x)Q(x) ) 1. (x)[ P(x)Q(x) ] (hypothesis) 2.      [ (x)P(x) ] (hypothesis) 3.      (x)P(x) (2, negation) 4.      P(x) (3, ui) 5.      P(x)Q(x) (1, ui) 6.      Q(x) (4, 5, disj. Syllo) 7.      (x)Q(x) (6, ug)

  5. (x)P(x)  (x)Q(x)  (x)[ P(x)Q(x) ] a. Find an interpretation to prove this wff is not valid. Ans: P(x): x is even, Q(x): x is odd b. What’s wrong in the following proof sequence? 1.      (x)P(x) (hypothesis) 2.      P(a) (1, ei) 3.      (x)Q(x) (hypothesis) 4.      Q(a) (3, ei) ( ** WRONG!! a is a constant previously used ** ) 5.      P(a)Q(a) (2, 4 conjunction) 6.      (x)[ P(x)Q(x) ] (5, eg)

  6. note: (x)(x)  (x)(x)  (x)( (x)(x) ) YES BUT  won’t work (x)(x)  (x)(x)  (x)( (x)  (x) ) NO e.g.(x)(x): every ball in the bag is red (x)(x): every ball in the bag is blue Note that (x) excludes (x) (x)( (x)(x) ) (x)(x)  (x)(x) which is wrong !

  7. (x) (x)  (x)(x)  (x) ( (x) (x) )YES BUT  won’t work (x) (x)  (x)(x)  (x) ( (x)  (x) )NO e.g. (x) (x): some ball in the bag is red (x) (x): some ball in the bag is blue (x) (x) (x)(x):  a red ball and  a blue ball but  a ball that is red and blue is not true (x) (x)  (x)(x)  (x) ( (x)  (x) ) which is wrong !

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