Lecture Notes, March 30, 2011
Mathematical Logic
Logical Inference
Let A and B be two logical conditions, like A="it's sunny today" and B="the light
outside is very bright"
A
B
A implies B, if A then B
A
B
A if and only if B, A implies B and B implies A,
A and B are equivalent conditions
Proof
s
Just like in high school geometry.
Concept of Proof by contradiction
:
Suppose we want to show that
A
B.
Ordinarily, we'd
like to prove this directly.
But it may be easier to show that [not (
A
B)] is false.
How?
Show that [A & (not B)] leads to a contradiction.
A: x = 1,
B:x+3=4.
Then
[A & (not B)] leads to the conclusion that 1+3
4 or equivalently 1
1, a contradiction.
Hence [A & (not B)] must fail so A
B.
(Yes, it does feel backwards, like your pocket
is being picked, but it works).
Set Theory
Definition of a Set
{ }
{x | x has property P}
{1, 2, ..., 9, 10}
=
{ x | x is an integer,
1
x
10 }.
Elements of a set
x
A ;
y
A
x
{ x }
x
{ x }
the empty set (
null set), the set with no elements.
Subsets
if x
A
x
B
A
B
or
A
B
.
A
A
and
A
Set Equality
A = B if A and B have precisely the same elements
A = B if and only if
.
A
B
and
B
A
Economics 113
Prof. R. Starr
UCSD
Spring 2011
March 30, 2011
1

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