Title: Chapter 2. Logic
1Chapter 2. Logic
- Weiqi Luo (???)
- School of Software
- Sun Yat-Sen University
- Emailweiqi.luo_at_yahoo.com OfficeA309
2Chapter two Logic
- 2.1. Propositions and Logical Operation
- 2.2. Conditional Statements
- 2.3. Methods of Proof
- 2.4. Mathematical Induction
- 2.5. Mathematical Statements
- 2.6. Logic and Problem Solving
32.1. Propositions and Logical Operation
- Statement (Proposition)
- A statement or proposition is a declarative
sentence that is either true or false but not
both - Example 1
- (a) The earth is round.
- (b) 235
- (c) Do you speak English?
- (d) 3-x5
- (e) Take two aspirins.
- (f) The temperature on the surface of the
planet Venus is 800 F -
- (g) The sun will come out tomorrow
Yes.
Yes.
No. This is a question.
No. May true or false
No. This is a command.
Yes.
yes
42.1. Propositions and Logical Operation
- Paradox
- A male barber shaves all and only those men
who do not shave themselves -
- Paradox A paradox is a seemingly true
statement or group of statements that lead to a
contradiction or a situation which seems to defy
logic or intuition . Paradox is not a statement.
- Refer to
- http//en.wikipedia.org/wiki/List_of_paradoxe
s -
- Others
- I am lying, this sentence is wrong, so
on and so forth
52.1. Propositions and Logical Operation
- Propositional variables
- In logic, the letters p, q, r denote
propositional variables, which are replaced by
statements - p 12 5
- q It is raining.
- Compound statements
- Propositional variables can be combined by
logical connectives to obtain compound
statements. E.g. - p and q 12 5 and it
is raining.
62.1. Propositions and Logical Operation
- Negation (a unary operation)
- If p is a statement, the negation of p is
the statement not p, denoted by p, meaning it
is not the case that p. - if p is true, then p is false, and if p is
false, then p is true. - Truth Table List the truth value of a
compound statement in terms of its component
parts. -
-
p q
T F
F T
72.1. Propositions and Logical Operation
- Example
- Give the negation of the following statements
- (a) p 23 gt1
- (b) q It is snowing.
- Solution
- (a) p 23 is not greater than 1, namely,
23 lt1 - (b) q It is not the case that it is
snowing. More simply, q It is not snowing. -
-
-
82.1. Propositions and Logical Operation
- Conjunction
- If p and q are statements, the conjunction
of p and q is the compound statement p and q
denoted by p? q . - Truth Table
-
-
-
- Note p? q is T if and only if p is T and q
is T. -
p q p? q
T T T
T F F
F T F
F F F
92.1. Propositions and Logical Operation
- Example
- Form the conjunction of p and q
- p 1gt3 q It
is raining. - Solution
- p? q 1 gt 3 and It is raining.
- Note
- In logic, unlike in everyday English, we may
join two totally unrelated statements by logical
connectives. -
-
102.1. Propositions and Logical Operation
- Disjunction
- If p and q are statements, the disjunction of
p and q is the compound statement p or q,
denoted by p V q -
- Truth Table
- Note p V q is F is and only if q is F and q
is F. -
p q p V q
T T T
T F T
F T T
F F F
112.1. Propositions and Logical Operation
- The connective or
- (a) I left for Spain on Monday or I left for
Spain on Friday. - (b) I passed math or I failed French
-
- Note
- Case (a) Both could not have occurred.
or is an excusive sense in this case. - Case (b) Both could have occurred. or is
an inclusive sense in this case. - In mathematics and computer science. We
agree to use the connective or in inclusive
manner. -
122.1. Propositions and Logical Operation
- Example
- Form the disjunction of p and q
- p 2 is a positive integer q sqrt(2)
is a rational number - Solution
- p V q 2 is a positive integer or sqrt(2)
is a rational number. Since p is true, p V q is
true, even through q is false. -
-
132.1. Propositions and Logical Operation
- Algorithm for making Truth Table
- Step 1 The first n columns of the table are
labeled by the component propositional variables.
Further columns are included for all intermediate
combinations of the variables, culminating in a
column for the full statement. - Step 2 Under each of the first n headings, we
list the 2n possible n-tuples of truth values for
the n component statements. - Step 3 For each of the remaining columns, we
compute, in sequence, the remaining truth values.
-
142.1. Propositions and Logical Operation
- Example 5
- Make a truth table for the statement (p ? q)
V ( p) - Truth Table
-
p q p? q p V
T T T F T
T F F F F
F T F T T
F F F T T
(1)
(2)
(3)
152.1. Propositions and Logical Operation
- Propositional function (predicate)
- An element of a set x P(x) is an object
t for which the statement P(t) is true. P(x) is
called a propositional function (or predicate) ,
because each choice of x produces a proposition
P(x) that is either true or false (well-defined) -
- E.g. Let A x x is an integer less than
8. - Here P(x) is the sentence x is an integer
less than 8 - P(1) denotes the statement 1 is an integer
less than 8 (true) - P(8) denotes the statement 8 is an integer
less than 8 (false)
162.1. Propositions and Logical Operation
- Universal Quantifiers (?)
- The Universal Quantifiers of a predicate
P(x) is the statement for all values of x, P(x)
is true , denoted by ? x P(x) - Example 8
- (a) P(x) -(-x) x is a predicate that
makes sense for all real number x. - then ? x P(x) is true statement.
Since ? x ? R, -(-x) x - (b) Q(x) x1lt4.
- then ? x Q(x) is a false statement,
since Q(5) is false -
172.1. Propositions and Logical Operation
- Existential Quantifiers (?)
- The Existential Quantifiers of a predicate
P(x) is the statement there exists a value of x
for which P(x) is true, denoted by - ? x
P(x) - Example 9
- (a) Let Q(x) x1lt4. then the existential
quantification of Q(x), ? x Q(x), is a true
statement, since Q(1) is a true statement - (b) The statement ? y y2y is false since
there is no value of y for which the
propositional function y2y produces a true
statement. -
182.1. Propositions and Logical Operation
- The order of the Quantifiers ? ?
- The order does not affect the output for the
same quantifiers, while it may produce different
results for different quantifiers. -
- E.g. P(x, y) x y 1
- ? x ? y P(x) is true, ? y ?
x is false. - P(x, y) x y 0
- ? x ? y P(x) is true, ? y ?
x is true too. -
-
192.1. Propositions and Logical Operation
- The negation of Quantifiers ? ?
- (a) let p ?x P(x),
- then p there must be at least one
value of x for which P(x) is false, namely, - ?x P(x) ?x P(x)
- (b) let p ?x P(x),
- then p for all x, P(x) is false,
namely, - ?x P(x) ?x P(x)
202.1. Propositions and Logical Operation
- Homework
- Ex. 2, Ex. 4, Ex. 16, Ex. 28
212.2. Conditional Statements
- Conditional statement
- If p and q are statements, the compound
statement if p then q, denoted pgtq, is called
a conditional statement or implication. - p antecedent or hypothesis q
consequent or conclusion - if then gt
- Truth Table
- Note when p is F, then pgtq is T.
-
-
p q p gt q
T T T
T F F
F T T
F F T
222.2. Conditional Statements
- Example
- Form the implication pgtq for each the
following - (a) p I am hungry. q I will eat.
- (b) p 225 q I am the king
of England. -
- Solution
- (a) If I am hungry, then I will eat
- (b) If 225, then I am the king of English
- Note There is no cause-and effect relationship
between p and q in case (b). And (b) is true,
since 225 is false. -
-
232.2. Conditional Statements
- Converse and Contrapositive
- If pgtq is an implication, then its converse
is the implication q gt p - and its contrapositive is the implication
q gt p -
- E.g. Give the converse and the contrapositive
of the implication If it is raining, then I get
wet - Converse If I get wet, then It is
raining. - Contrapositive If I do not get wet, then It is
not raining. -
242.2. Conditional Statements
- Equivalence (biconditional)
- If p and q are statements, the compound
statement p if and only if q, denoted by p ? q,
is called an equivalence or biconditional. - Truth Table
- Note p ltgt q is T when p and q are both T or
both F.
p q p ? q
T T T
T F F
F T T
F F T
252.2. Conditional Statements
- Example 3
- Is the following equivalence a true
statement? - 3gt2 if and only if 0lt 3
2 - Solution
- Let p 3gt2 and q 0lt 3 2,
- since p and q are both true, we then
conclude that - p ? q is true
statement. -
262.2. Conditional Statements
- Example 4.
- Compute the truth table of the statement
- (pgtq) ? (q gt
p) - Truth Table
p q pgtq q p qgtp (pgtq) ? (q gt p)
T T T F F T T
T F F T F F T
F T T F T T T
F F T T T T T
272.2. Conditional Statements
- Tautology
- A statement that is true for all possible
values of its propositional variables called
Tautology. (e.g. Example 4) - Contradiction (Absurdity)
- A statement that is false for all possible
values of its propositional variables called
Contradiction or Absurdity. (e.g. p ? p) - Contingency
- A statement that can be either true or false,
depending on the truth values of its
propositional variables, is called a contingency.
- (e.g. p gt q)
-
282.2. Conditional Statements
- Logically Equivalent
- If two statements p and q are always either
both true or both false, for any values of the
propositional variables, namely - p ? q is a tautology
- Then we call p and q are logically
equivalent. - Denoted by
- p q
-
292.2. Conditional Statements
- Example 6
- Show that p V q and q V p are logically
equivalent - The truth table of (p V q ) ? (q V p ) are
shown as follows - Truth Table
p q p V q q V p p V q ?q V p
T T T T T
T F T T T
F T T T T
F F F F T
302.2. Conditional Statements
- Two Structures with similar properties
- (Theorem 1)
-
- Structure 1 (logic operations)
- (propositions, ?, V, )
- Structure 2 (sets operations)
- (sets, U, n , - )
-
-
-
-
312.2. Conditional Statements
- Theorem 1
- Commutative properties
- p ? q q ?p, p? q q
? p -
- Associative Properties
- p ? (q ? r) (p ? q) ?r , p? (q ? r)
(p ? q) ? r - Distributive Properties
- p ? (q ? r) (p ? q) ?
(p ? r) - p ? (q ? r) (p ? q) ?
(p ? r) -
-
322.2. Conditional Statements
- Idempotent Properties
- p ? p p , p? p
p - Properties of Negation
- (p) p
- (p ? q) (p) ? (q)
- (p ? q) (p) ? (q) De
Morgans laws
332.2. Conditional Statements
- Theorem 2
- (p ? q) ((p) ? q)
- (p ? q) (q ? p)
- (p ? q) ((p ? q) ? (q ? p))
- (p ? q) (p ? q)
- (p ? q) ((p ? q) ? (q ? p))
-
342.2. Conditional Statements
- Theorem 3
- (?xP(x)) ?xP(x)
- (?xP(x)) ?x(P(x))
- ?x(P(x) ? Q(x)) ?xP(x) ? ?xQ(x)
- ?x(P(x) ? Q(x)) ?xP(x) ? ?xQ(x)
- ?x(P(x) ? Q(x)) ?xP(x) ? ?xQ(x)
- ?xP(x) ? ?xQ(x) ? ?x(P(x) ? Q(x)) is a tautology
- ?x(P(x) ? Q(x)) ? ?xP(x) ? ?xQ(x) is a tautology
352.2. Conditional Statements
- Theorem 4 Each of the following is a tautology
- (p ? q) ? p , (p ? q) ? q
- p ? (p ? q) , q ? (p ? q)
- p ? (p ? q) , (p ? q) ? p
- (p ? (p ? q)) ? q , (p ? (p ? q)) ? q
- (q ? (p ? q)) ? p
- ((p ? q) ? (q ? r)) ? (p ? r)
362.2. Conditional Statements
- Properties of Quantifiers ? and ?
-
- (a) ? x (P(x) V Q(x)) ?x P(x) V ?x (Q(x))
-
- (b) ? x (P(x) ? Q(x)) ?x P(x) ?x Q(x)
-
-
-
-
-
372.2. Conditional Statements
- Homework
- ex.2, ex.7, ex.12, ex.15, ex.21 ex.34
382.3. Method of Proof
- Logically Follow
- If an implication p ? q is a tautology, where
p and q may be compound statements involving any
number of proposition variables, we say that q
logically follows from p. - Suppose that an implication of the form (p1
? p2 ? ? pn) - ? q is a tautology. We say that q logically
follows from p1, p2, - , pn, denoted by
392.3. Method of Proof
- (p1 ? p2 ? ? pn) ? q
- The pis are called the hypotheses or
premises, and q is called the conclusion. - Note we are not trying to show that q is
true, but only that q will be true if all the pi
are true. - ? denotes therefore
402.3. Method of Proof
- Rules of inference
- Arguments based on tautologies represent
universally correct methods of reasoning. Their
validity depends only on the form of the
statements involved and not on the truth values
of the variables they contain. - Example 1
- ((p ? q) ? (q ? r)) ? (p ? r) is tautology,
then the argument - is universally valid, and so is a rule of
inference. -
-
412.3. Method of Proof
- Example 2 Is the following argument valid?
- If you invest in the stock market,
then you will get rich. - If you get rich, then you will be
happy. - ? If you invest in the stock market, then
you will be happy. - let p you invest in the stock market, q
you will get rich - r you will be happy
- The above argument is of the form given in
Example 1, hence the argument is valid! -
-
-
422.3. Method of Proof
- Example 3
- The tautology
- (p ? q) ?((p ? q) ? (q ? p))
- means that the following two arguments are
valid -
p ? q - p ? q
q ? p - ? (p ? q) ? (q ? p) ? p ? q
-
-
432.3. Method of Proof
- Equivalence (p ? q)
- They are usually stated p if and only if q.
We need to prove both pgtq and qgtp by the
tautology mentioned in example 3. - Algorithm
- Step one Assuming p is true, show q must be
true. -
- Step two Assuming q is true, show p must be
true. -
442.3. Method of Proof
- Modus Ponens
- p is true, and pgtq is true, so q is true
- (Theorem 4(g) in Section 2.2.)
-
- p
- pgtq
- ? q
-
-
452.3. Method of Proof
- Example 4 Is the following argument valid?
-
- Smoking is healthy.
- If smoking is healthy, then cigarettes are
prescribed by physicians. - ? Cigarettes are prescribed by physicians
- p Smoking is healthy.
- q cigarettes are prescribed by physicians
- The argument is valid since it is of form modus
ponens.
462.3. Method of Proof
- Example 5 Is the following argument valid?
- If taxes are lowered, then income
rises - Income rises
- ? taxes are lowered
- Solution
- Let p taxes are lowed q income rise
- pgtq
- q
- ? p
- Then argument is not valid , since pgtq and q
can be both true with p being false. -
(or show the truth table of (pgtq) ? q gt p ,
and determine whether or not it is a tautology)
472.3. Method of Proof
- Indirect Method of Proof
- The tautology
- (pgtq ) ? (q) gt (p )
- (An implication is equivalent to its
contrapositive) -
- Note
- To proof pgtq indirectly, we assume q is false
(q) and show that p is then false (p) -
482.3. Method of Proof
- Example 6
- Let n be an integer. Prove that if n2 is odd,
then n is odd. -
- Solution
- Let p n2 is odd , q n is odd.
- To prove that (pgtq)
- We try to prove its contrapositive qgtp
- Assuming that n is even (q), let n2k, k is
an integer, then we have n2 (2k) 2 4k2 is
even (p). - Hence, the given statement has been proved.
-
-
492.3. Method of Proof
- The tautology
- ((p ? q) ? (q)) ? p
- If a statement p implies a false statement q,
then p must be false. - Proof by contradiction
- To prove (p1 ? p2 ? ? pn) ? q ,
- We add (q) into hypothesis p1 ? p2 ? ? pn,
if the enlarged hypothesis p1 ? p2 ? ? pn ? (q)
implies a contradiction, then we can conclude
that q follows from p1 ,p2 , and pn . -
-
-
-
502.3. Method of Proof
- Example 7
- Prove there is no rational number a/b whose
square is 2, namely, sqrt(2) is irrational. - Solution
- Let p a, b are integers and no common
factors, and b is not 0 - q (a/b)2 is not 2
- In order to prove p gt q ,
- We try to find the contradiction from p ? q
- Refer to Example 7 for more details.
-
512.3. Method of Proof
- Example 9
- Prove or disprove the statement that if x and
y are real numbers, (x2y2) ? (xy) - Solution
- Since (1)2(-1)2, but -1 ? 1, the result is
false. Our example is called counterexample, and
any other counterexample would do just as well. - Note
- If a statement claims that a property holds
for all objects of a certain type, then to prove
it, we must use steps that are valid for all
objects of that type and that do not make
references to any particular object. To disprove
such a statement, we need only show one
counterexample.
522.3. Method of Proof
- Homework
- Ex. 6, Ex. 8, Ex. 9
- Ex. 19, Ex. 20
- Ex. 31, Ex. 34
-
532.4 Mathematical Induction
- To Prove
- ?ngtn0 P(n)
- where n0 is some fixed integer
- Two Steps
- 1) Basis Step
- Prove that P(n0) is true
- 2) Induction Step
- Prove that P(k) gt P(k1) is a
tautology - (if P(k) is true, then P(k1) must be true)
-
-
542.4 Mathematical Induction
- Example 1 Prove that for all ngt0
- 123n n(n1)/2
- Solution let P(n)n(n1)/2
- Basis step P(1) 1 12/21 is true.
- Induction Step assuming P(k) is true, then
- P(k1) 12 3 k (k1)
- P(k) (k1)
- k (k1)/2 (k1)
P(k) is true - (k1)(k2)/2
- P(k1)
552.4 Mathematical Induction
- Example 3 Prove any finite, nonempty set is
countable (the elements can be arranged in a
list) - Solution Let P(n) be the predicate that if A is
any set with Angt0, then A is countable - Basis Step P(1) is true ( A x )
- Induction Step assuming P(k) is true
- Let B denotes any finite, nonempty set with
k1 elements. We first pick any element x in B,
then B-x is a set with k elements, and it is
countable (P(k) is true), and listed by x1,x2,xk
. Then we can also list the elements of B as
x1,x2,xk , x (P(k1) is true)
562.4 Mathematical Induction
- Example 5 Consider the following function given
in pseudocode - Function SQ(A)
- 1. C lt- 0
- 2. D lt- 0
- 3. WHILE (D is not A)
- a. C lt- CA
- b. D lt- D1
- 4. RETURN (C)
- End of Function SQ
-
572.4 Mathematical Induction
Weak Induction Strong Induction
Basis Step P(no) is true (or the first several statements are true) P(no) is true (or the first several statements are true)
Induction Step P(k) ? P(k1) is a tautology P(n0) ?P(n1) ? P(k) ? P(k1) is a tautology
582.4 Mathematical Induction
- Example 7 Prove that every positive integer ngt1
can be written uniquely as p1a1p2a2psas, where
pi are primes and p1ltp2lt..Ps - Basis Step P(2) is true, since 2 is prime
and 2 21 (unique ) - Induction Step Assuming P(2), P(3), P(k)
are true - if k1 is prime, then k1 (k1)1
- if k1 is not prime, then we let k1Lm,
where L, m are positive integers less than k1.
Using P(L) and P(m) are true, we have k1Lm
q1b1q2b2qsbs (unique form) - Why? Proof by contradiction
-
592.4 Mathematical Induction
- Homework
- Ex. 2, Ex. 14, Ex. 17, Ex. 22, Ex. 34