Solution 1 let pw f be w has taken f and qf a be

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Unformatted text preview: the statement “x + y = 0.” Assume that domain is the real numbers. Then ∀x ∃y P(x, y) is true, but ∃y ∀x P(x, y) is false. Ques$ons on Order of Quan$fiers Example 1: Let domain be the real numbers, Define P(x, y) : x ·ȡ y = 0 What is the truth value of the following: 1.  ∀x ∀y P(x, y) Answer: False 2.  ∀x ∃y P(x, y) Answer: True 3.  ∃x ∀y P(x, y) Answer: True ∃x ∃y P(x, y) Answer: True 4.  Ques$ons on Order of Quan$fiers Example 2: Let domain be the real numbers, Define P(x, y) : x / y = 1 What is the truth value of the following: 1.  ∀x ∀y P(x, y) Answer: False 2.  ∀x ∃y P(x, y) Answer: True 3.  ∃x ∀y P(x, y) Answer: False 4.  ∃x ∃y P(x, y) Answer: True Quan$fica$ons of Two Variables Statement When True? When False P(x,y) is true for every pair x,y. There is a pair x, y for which P(x,y) is false. For every x there is a y for There is an x such that which P(x,y) is true. P(x,y) is false for every y. There is an x for which P(x,y) is true for every y. For every x there is a y for which P(x,y) is false. There is a pair x, y for which P(x,y) is true. P(x,y) is false for every pair x,y Transla$ng Quan$fiers into English Example 1: ∀x (C(x ) ∨ ∃y (C(y) ∧ F(x, y))) where C(x) is “x has a computer,” and F(x, y) is “x and y are f...
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