Final Review
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Transcript Final Review
Final Review
List of Logical Equivalences
pT p;
pF p
Identity Laws
pT T;
pF F
Domination Laws
pp p;
pp p
Idempotent Laws
(p) p
Double Negation Law
pq qp; pq qp
Commutative Laws
(pq) r p (qr); (pq) r p (qr)
Associative Laws
List of Equivalences
p(qr) (pq)(pr)
p(qr) (pq)(pr)
Distribution Laws
(pq)(p q)
(pq)(p q)
De Morgan’s Laws
p p T
p p F
(pq) (p q)
Miscellaneous
Or Tautology
And Contradiction
Implication Equivalence
pq(pq) (qp)
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: (pq) q pq
(pq) q
q (pq)
(qp) (q q)
(qp) T
qp
pq
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 xP(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 xP(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: xU 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)
xyP(x,y) or yxP(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.
xyP(x,y) or yxP(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
xyP(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.
yxP(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 yxP(x,y) is true, then xyP(x,y) is true.
However, if xyP(x,y) is true, it is not necessary that yxP(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.
yxL(x,y)
Examples 3b1
There is somebody whom Lydia does not love.
xL(Lydia,x)
Nobody loves everybody. (For each person there is at
least one person they do not love.)
xyL(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, b0
x is irrational is defined as a,b Z
x = a/b, b0 or a,b Z, x a/b, b0
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 (pq). 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 kZ 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 (AB) (AB) = B
(AB) (AB)
= {x | x(AB)(AB)}
Set builder notation
= {x | x(AB) x(AB)}
Def of
= {x | (xA xB) (xA xB)} Def of x2 and
Def of complement
= {x | (xB xA ) (xB xA )} Commutative x2
= {x | (xB (xA xA )}
Distributive
= {x | (xB T }
Or tautology
= {x | (xB }
Identity
=B
Set Builder notation
Prove (AB) (AB) = B
(Using Set Identities)
(AB) (AB) =
(BA) (BA)
=B (A A)
=B U
=B
Commutative Law x2
Distributive Law
Definition of U
Identity Law
Prove (AB) (AB) = B
Proof: We must show that (AB) (AB)
B and that B (AB) (AB) .
First we will show that (AB) (AB) B.
Let e be an arbitrary element of (AB)
(AB). Then either e (AB) or e
(AB). If e (AB), then eB and eA. If
e (AB), then eB and eA. In either
case e B.
Prove (AB) (AB) = B
Now we will show that B (AB) (AB).
Let e be an arbitrary element of B. Then
either e AB or e AB. Since e is in
one or the other, then e (AB)
(AB).
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?
b1
b2
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 bB there
is an element aA with f(a) = b.
f
f
a1
a2
f
a3
A
f
a1
a2
b1
f
a3
b2
A
bB aA 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, nZ+ .
• {1,2,4,8,16, . . .} = {an} where an = 2n, nN.
• {12,22,32,42,. . .} = {an} where an = n2, nZ+
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
n1
ar
k
k 1
n
ar
n1
ar ar
k
k 1
n
n1
a ar
k0
k
Telescoping Sums
n
(a
j
a j 1 ) (a1 a0 ) (a2 a1 )
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 CN and kR
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 CN and kR 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 CN and kR 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 CN and kR such that |n2|
C|4n| whenever n > k?
• No. To show this, we must prove that
negation is true for all C and k. CN,
kR, n>k such that n2 > C|4n|.
CN, kR, 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.