Transcript Final Review List of Logical Equivalences pT  p; pF  p Identity Laws pT  T; pF  F Domination Laws pp  p; pp  p Idempotent.

Final Review
List of Logical Equivalences
pT  p;
pF  p
Identity Laws
pT  T;
pF  F
Domination Laws
pp  p;
pp  p
Idempotent Laws
(p)  p
Double Negation Law
pq  qp; pq  qp
Commutative Laws
(pq) r  p (qr); (pq)  r  p  (qr)
Associative Laws
List of Equivalences
p(qr)  (pq)(pr)
p(qr)  (pq)(pr)
Distribution Laws
(pq)(p  q)
(pq)(p  q)
De Morgan’s Laws
p  p  T
p  p  F
(pq)  (p  q)
Miscellaneous
Or Tautology
And Contradiction
Implication Equivalence
pq(pq)  (qp)
Biconditional Equivalence
The Proof Process
Assumptions
Logical Steps
-Definitions
-Already-proved equivalences
-Statements (e.g., arithmetic
or algebraic)
Conclusion
(That which was to be proved)
Prove: (pq)  q  pq
(pq)  q
 q  (pq)
 (qp)  (q q)
 (qp)  T
 qp
 pq
Left-Hand Statement
Commutative
Distributive
Or Tautology
Identity
Commutative
Begin with exactly the left-hand side statement
End with exactly what is on the right
Justify EVERY step with a logical equivalence
Predicate Calculus: Quantifiers
Universe of Discourse, U: The domain of a variable in
a propositional function.
Universal Quantification of P(x) is the
proposition:“P(x) is true for all values of x in U.”
Existential Quantification of P(x) is the proposition:
“There exists an element, x, in U such that P(x) is
true.”
Universal Quantification of P(x)
xP(x)
“for all x P(x)”
“for every x P(x)”
Defined as:
P(x0)  P(x1)  P(x2)  P(x3)  . . . for all xi in U
Example:
Let P(x) denote x2  x
If U is x such that 0 < x < 1 then xP(x) is false.
If U is x such that 1 < x then xP(x) is true.
Existential Quantification of
P(x)
xP(x)
“there is an x such that P(x)”
“there is at least one x such that P(x)”
“there exists at least one x such that P(x)”
Defined as:
P(x0)  P(x1)  P(x2)  P(x3)  . . . for all xi in U
Example:
Let P(x) denote x2  x
If U is x such that 0 < x  1 then xP(x) is true.
If U is x such that x < 1 then xP(x) is true.
Quantifiers
xP(x)
•True when P(x) is true for every x.
•False if there is an x for which P(x) is false.
xP(x)
•True if there exists an x for which P(x) is true.
•False if P(x) is false for every x.
Negation (it is not the case)
xP(x) equivalent to xP(x)
•True when P(x) is false for every x
•False if there is an x for which P(x) is true.
 xP(x) is equivalent to xP(x)
•True if there exists an x for which P(x) is false.
•False if P(x) is true for every x.
Examples 2a
Let T(a,b) denote the propositional function “a
trusts b.” Let U be the set of all people in the
world.
Everybody trusts Bob.
xT(x,Bob)
Could also say: xU T(x,Bob)
 denotes membership
Bob trusts somebody.
xT(Bob,x)
Examples 2b
Alice trusts herself.
T(Alice, Alice)
Alice trusts nobody.
x T(Alice,x)
Carol trusts everyone trusted by David.
x(T(David,x)  T(Carol,x))
Everyone trusts somebody.
x y T(x,y)
Quantification of Two Variables
(read left to right)
xyP(x,y) or yxP(x,y)
•True when P(x,y) is true for every pair x,y.
•False if there is a pair x,y for which P(x,y) is false.
xyP(x,y) or yxP(x,y)
True if there is a pair x,y for which P(x,y) is true.
False if P(x,y) is false for every pair x,y.
Quantification of Two
Variables
xyP(x,y)
•True when for every x there is a y for which P(x,y) is true.
(in this case y can depend on x)
•False if there is an x such that P(x,y) is false for every y.
yxP(x,y)
•True if there is a y for which P(x,y) is true for every x.
(i.e., true for a particular y regardless (or independent) of x)
•False if for every y there is an x for which P(x,y) is false.
Note that order matters here
In particular, if yxP(x,y) is true, then xyP(x,y) is true.
However, if xyP(x,y) is true, it is not necessary that yxP(x,y)
is true.
Examples 3a
Let L(x,y) be the statement “x loves y” where U for both
x and y is the set of all people in the world.
Everybody loves Jerry.
xL(x,Jerry)
Everybody loves somebody.
x yL(x,y)
There is somebody whom everybody loves.
yxL(x,y)
Examples 3b1
There is somebody whom Lydia does not love.
xL(Lydia,x)
Nobody loves everybody. (For each person there is at
least one person they do not love.)
xyL(x,y)
There is somebody (one or more) whom nobody loves
y x L(x,y)
Basic Number Theory Definitions
•
•
•
•
•
from Chapter 2
Z = Set of all Integers
Z+ = Set of all Positive Integers
N = Set of Natural Numbers (Z+ and Zero)
R = Set of Real Numbers
Addition and multiplication on integers
produce integers. (a,b  Z)  [(a+b)  Z]
 [(ab)  Z]
Number Theory Defs (cont.)
•
•
•
•
•
 = “such that”
n is even is defined as k  Z  n = 2k
n is odd is defined as k  Z  n = 2k+1
x is rational is defined as a,b  Z  x =
a/b, b0
x is irrational is defined as a,b  Z 
x = a/b, b0 or a,b  Z, x  a/b, b0
p  Z+ is prime means that the only
positive factors of p are p and 1. If p is
not prime we say it is composite.
Methods of Proof
p q (Example: if n is even, then n2 is even)
• Direct proof: Assume p is true and use a
series of previously proven statements to
show that q is true.
• Indirect proof: Show q p is true
(contrapositive), using any proof technique
(usually direct proof).
• Proof by contradiction: Assume negation of
what you are trying to prove (pq). Show
that this leads to a contradiction.
Example of an Indirect Proof
Prove: If n3 is even, then n is even.
Proof: The contrapositive of “If n3 is even, then n is
even” is “If n is odd, then n3 is odd.” If the
contrapositive is true then the original statement
must be true.
Assume n is odd. Then kZ  n = 2k+1. It follows
that n3 = (2k+1)3 = 8k3+8k2+4k+1 =
2(4k3+4k2+2k)+1. (4k3+4k2+2k) is an integer.
Therefore n3 is 1 plus an even integer. Therefore
n3 is odd.
Assumption, Definition, Arithmetic, Conclusion
Example: Proof by Contradiction
Prove: The sum of an irrational number and
a rational number is irrational.
Proof: Let q be an irrational number and r be a
rational number. Assume that their sum is
rational, i.e., q+r=s where s is a rational
number. Then q = s-r. But by our previous
proof the sum of two rational numbers must be
rational, so we have an irrational number on
the left equal to a rational number on the right.
This is a contradiction. Therefore q+r can’t be
rational and must be irrational.
Structure of Proof by Contradiction
• Basic idea is to assume that the opposite of
what you are trying to prove is true and show
that it results in a violation of one of your initial
assumptions.
• In the previous proof we showed that assuming
that the sum of a rational number and an
irrational number is rational and showed that it
resulted in the impossible conclusion that a
number could be rational and irrational at the
same time. (It can be put in a form that implies n
 n is true, which is a contradiction.)
Approaches to Set Proofs
• Membership tables (similar to truth tables)
• Convert to a problem in propositional logic,
prove, then convert back
• Use set identities for a tabular proof
(similar to what we did for the propositional
logic examples but using set identities)
• Do a logical argument (similar to what we
did for the number theory examples)
Prove (AB)  (AB) = B
(AB)  (AB)
= {x | x(AB)(AB)}
Set builder notation
= {x | x(AB)  x(AB)}
Def of 
= {x | (xA  xB)  (xA  xB)} Def of  x2 and
Def of complement
= {x | (xB  xA )  (xB  xA )} Commutative x2
= {x | (xB  (xA  xA )}
Distributive
= {x | (xB  T }
Or tautology
= {x | (xB }
Identity
=B
Set Builder notation
Prove (AB)  (AB) = B
(Using Set Identities)
(AB)  (AB) =
(BA)  (BA)
=B  (A  A)
=B  U
=B
Commutative Law x2
Distributive Law
Definition of U
Identity Law
Prove (AB)  (AB) = B
Proof: We must show that (AB)  (AB)
 B and that B  (AB)  (AB) .
First we will show that (AB)  (AB)  B.
Let e be an arbitrary element of (AB) 
(AB). Then either e (AB) or e
(AB). If e (AB), then eB and eA. If
e (AB), then eB and eA. In either
case e B.
Prove (AB)  (AB) = B
Now we will show that B  (AB)  (AB).
Let e be an arbitrary element of B. Then
either e AB or e AB. Since e is in
one or the other, then e  (AB) 
(AB).
Functions: One-to-one function
A function f is said to be one-toone, or injective, if and only if
f(x) = f(y) implies that x=y for
all x and y in the domain of f.
f
a1
a2
f
a3
A
f
f
a1
a2
b1
f
a3
One-to-one?
b2
b3
A
f
b4
One-to-one?
b2
b1
b3
B
a0,a1  A
f(a0) = f(a1)  a0 = a1
OR
a0  a1  f(a0)  f(a1)
B
Onto Function
A function f from A to B is called
onto, or surjective, if and only
if for every element bB there
is an element aA with f(a) = b.
f
f
a1
a2
f
a3
A
f
a1
a2
b1
f
a3
b2
A
bB  aA such that f(a) =
b
f
b1
b2
b3
B
B
One-to-one Correspondence
The function f is a one-to-one
correspondence or a
bijection, if it is both one-toone and onto.
f
f
a1
a2
f
a3
A
f
a1
a2
b1
f
a3
Bijection?
A
Bijection?
b2
B
f
b1
b2
b3
B
Correspondence Diagrams: Oneto-One or Onto?
a
b
c
a
b
c
d
1
2
3
4
One-to-one, not
onto
a
b
c
d
1
2
3
Onto, not one-toone
1
2
3
4
Neither one-to-one
nor onto
a
b
c
Not a function!
a
b
c
d
One-to-one, and
onto
1
2
3
4
1
2
3
4
Inverse Function, f-1
Let f be a one-to-one correspondence from the set A to the set B. The
inverse function of f is the function that assigns to an element b belonging
to B the unique element a in A such that if f(a) = b, then f-1(b) = a.
Example:
f
b
a
f(a) = 3(a-1)
f-1(b) = (b/3)+1
f-1
Examples
Is each of the following (on the real numbers): a function? one-to-one? Onto?
Invertible?
f(x) = 1/x
not a function f(0) undefined
f(x) = x
not a function since not defined for x<0
f(x) = x2
is a function, not 1-to-1 (-2,2 both go to 4), not onto since no way to get to
the negative numbers, not invertible
Sequence
• A sequence is a discrete structure used to represent an
ordered list.
• A sequence is a function from a subset of the set of
integers (usually either the set {0,1,2,. . .} or {1,2, 3,. . .}to
a set S.
• We use the notation an to denote the image of the
integer n. We call an a term of the sequence.
• Notation to represent sequence is {an}
Examples
• {1, 1/2, 1/3, 1/4, . . .} or the sequence {an}
where an = 1/n, nZ+ .
• {1,2,4,8,16, . . .} = {an} where an = 2n, nN.
• {12,22,32,42,. . .} = {an} where an = n2, nZ+
Summations
• Notation for describing the sum of the terms
am+1, . . ., an from the sequence, {an}
a m,
n
am+am+1+ . . . + an =  aj
j=m
• j is the index of summation (dummy variable)
• The index of summation runs through all integers
from its lower limit, m, to its upper limit, n.
Summations follow all the rules
of multiplication and addition!
n
n
j 1
j 1
c  j   cj  c(1+2+…+n) = c + 2c +…+ nc
n
n
r  ar   ar
j
j 0
j 0
j 1
n1
  ar 
k
k 1
n
ar
n1
  ar  ar
k
k 1
n
n1
 a   ar
k0
k
Telescoping Sums
n
 (a
 a j 1 )  (a1  a0 )  (a2  a1 ) 
j
j 1
(a3  a2 )  ...  (an  an 1 )  an  a0
Example
4
2
2
[k

(k

1)
]

k 1
(1  0 )  (2  1 )  (3  2 )  (4  3 )
2
2
2
4 2  16  0  16
2
2
2
2
2
Closed Form Solutions
A simple formula that can be used to calculate a sum without
doing all the additions.
Example:
n(n  1)
k  2
k 1
n
Proof: First we note that k2 - (k-1)2 = k2 - (k2-2k+1) = 2k-1.
Since k2-(k-1)2 = 2k-1, then we can sum each side from k=1 to
k=n
n
n
 [k
k 1
2
 k 1 ]   2k 1
2
k 1
Proof (cont.)
n
 [k
2
k 1
n
 [k
k 1
2
n
 k 1 ]   2k 1
2
k 1
n
n
k 1
k 1
 k 1 ]   2k  1
2
n
n  0  2 (k )  n
2
2
k 1
n
n 2  n  2 (k)
k 1
n2  n
k  2
k 1
n
Big-O Notation
• Let f and g be functions from the set of integers or the
set of real numbers to the set of real numbers. We say
that f(x) is O(g(x)) if there are constants CN and kR
such that
|f(x)|  C|g(x)| whenever x > k.
• We say “f(x) is big-oh of g(x)”.
• The intuitive meaning is that as x gets large, the values
of f(x) are no larger than a constant time the values of
g(x), or f(x) is growing no faster than g(x).
• The supposition is that x gets large, it will approach a
simplified limit.
Show that
3
2
3x +2x +7x+9
is
O(x3)
Proof: We must show that  constants CN and kR such
that |3x3+2x2+7x+9|  C|x3| whenever x > k.
Choose k = 1 then
3x3+2x2+7x+9  3x3+2x3+7x3+9x3 = 21x3
So let C = 21.
Then 3x3+2x2+7x+9  21 x3 when x  1.
Show that n! is O(nn)
Proof: We must show that  constants CN and kR such
that |n!|  C|nn| whenever n > k.
n! = n(n-1)(n-2)(n-3)…(3)(2)(1)
 n(n)(n)(n)…(n)(n)(n)
n times
=nn
So choose k = 0 and C = 1
General Rules
• Multiplication by a constant does not change the rate
of growth. If f(n) = kg(n) where k is a constant, then f
is O(g) and g is O(f).
• The above means that there are an infinite number of
pairs C,k that satisfy the Big-O definition.
• Addition of smaller terms does not change the rate of
growth. If f(n) = g(n) + smaller order terms, then f is
O(g) and g is O(f).
Ex.: f(n) = 4n6 + 3n5 + 100n2 + 2 is O(n6).
General Rules (cont.)
• If f1(x) is O(g1(x)) and f2(x) is O(g2(x)), then f1(x)f2(x) is
O(g1(x)g2(x)).
• Examples:
10xlog2x is O(xlog2x)
n!6n3 is O(n!n3)
=O(nn+3)
Example: Big-Oh Not
Symmetric
• Order matters in big-oh. Sometimes f is O(g)
and g is O(f), but in general big-oh is not
symmetric.
Consider f(n) = 4n and g(n) = n2. f is O(g).
• Can we prove that g is O(f)? Formally, 
constants CN and kR such that |n2| 
C|4n| whenever n > k?
• No. To show this, we must prove that
negation is true for all C and k. CN,
kR, n>k such that n2 > C|4n|.
CN, kR, n>k such that n2 >
4nC.
• To prove that negation is true, start with
arbitrary C and k. Must show/construct an
n>k such that n2 > 4nC
• Easy to satisfy n > k, then
• To satisfy n2>4nC, divide both sides by n
to get n>4C. Pick n = max(4C+1,k+1),
which proves the negation.