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Introduction to Predicates and Quantified Statements II

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1 Introduction to Predicates and Quantified Statements II
Lecture 10 Section 2.2 Fri, Feb 2, 2007

2 Negation of a Universal Statement
What would it take to make the statement “Everybody likes me” false?

3 Negation of a Universal Statement
What would it take to make the statement “Somebody likes me” false?

4 Negations of Universal Statements
The negation of the statement x  S, P(x) is the statement x  S, P(x). If “x  R, x2 > 10” is false, then “x  R, x2  10” is true.

5 Negations of Existential Statements
The negation of the statement x  S, P(x) is the statement x  S, P(x). If “x  R, x2 < 0” is false, then “x  R, x2  0” is true.

6 Example Are these statements equivalent?
“Any investment plan is not right for all investors.” “There is no investment plan that is right for all investors.”

7 The Word “Any” We should avoid using the word “any” when writing quantified statements. The meaning of “any” is ambiguous. “You can’t put any person in that position and expect him to perform well.”

8 Negation of a Universal Conditional Statement
How would you show that the statement “You can’t get a good job without a good edikashun” is false?

9 Negation of a Universal Conditional Statement
The negation of x  S, P(x)  Q(x) is the statement x  S, (P(x)  Q(x)) which is equivalent to the statement x  S, P(x)  Q(x).

10 Negations and DeMorgan’s Laws
Let the domain be D = {x1, x2, …, xn}. The statement x  D, P(x) is equivalent to P(x1)  P(x2)  …  P(xn). It’s negation is P(x1)  P(x2)  …  P(xn), which is equivalent to x  D, P(x).

11 Negations and DeMorgan’s Laws
The statement x  D, P(x) is equivalent to P(x1)  P(x2)  …  P(xn). It’s negation is P(x1)  P(x2)  …  P(xn), which is equivalent to x  D, P(x).

12 Evidence Supporting Universal Statements
Consider the statement “All crows are black.” Let C(x) be the predicate “x is a crow.” Let B(x) be the predicate “x is black.” The statement can be written formally as x, C(x)  B(x) or C(x)  B(x).

13 Supporting Universal Statements
Question: What would constitute statistical evidence in support of this statement?

14 Supporting Universal Statements
The statement is logically equivalent to x, ~B(x)  ~C(x) or ~B(x)  ~C(x). Question: What would constitute statistical evidence in support of this statement?

15 Algebra Puzzler Find the error(s) in the following “solution.”


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