Introduction to Predicates and Quantified Statements II

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Presentation transcript:

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

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

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

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.

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.

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.”

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.”

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?

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).

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).

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).

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).

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

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?

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