# Introduction to Predicates and Quantified Statements II - PowerPoint

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```					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.”
x                2     2
3              
x2                x     3
x2 1
             x     3
x    3             
2     2
2 1
1                x  3
x 3

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