Chapter 2. Logic Weiqi Luo ( 骆伟祺 ) School of Software Sun Yat-Sen University Email ： [email protected] Office ： A309 [email protected]

Chapter 2. Logic

Weiqi Luo (骆伟祺 )School of Software

Sun Yat-Sen UniversityEmail ： [email protected] Office ： A309

mailto:[email protected]

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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

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Chapter two: Logic

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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) 2+3=5

(c) Do you speak English?

(d) 3-x=5

(e) Take two aspirins.

(f) The temperature on the surface of the planet Venus is 800 F

(g) The sun will come out tomorrow

2.1. Propositions and Logical Operation

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Yes.

Yes.

No. This is a question.

No. May true or false

No. This is a command.

Yes.

yes

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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_paradoxes

Others

“I am lying”, “this sentence is wrong”, so on and so forth


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Propositional variables

In logic, the letters p, q, r … denote propositional variables, which are replaced by statements

p: 1+2 = 5

q: It is raining.

Compound statements

Propositional variables can be combined by logical connectives to obtain compound statements. E.g.

p and q : 1+2 =5 and it is raining.


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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.


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p ~q

T F

F T

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Example: Give the negation of the following statements

(a) p: 2+3 >1

(b) q: It is snowing.

Solution:

(a) ~p: 2+3 is not greater than 1, namely, 2+3 <=1

(b) ~q: It is not the case that it is snowing. More simply, ~q: It is not snowing.


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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.


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p q p∧ q

T T T

T F F

F T F

F F F

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Example Form the conjunction of p and q

p: 1>3 q: It is raining.

Solution:

p∧ q : 1 > 3 and It is raining.

Note:

In logic, unlike in everyday English, we may join two totally unrelated statements by logical connectives.


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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.


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p q p V q

T T T

T F T

F T T

F F F

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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.


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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.


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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.


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Example 5: Make a truth table for the statement (p ∧ q) V (~ p)

Truth Table:


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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)

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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)


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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): x+1<4.

then ∀ x Q(x) is a false statement, since Q(5) is false


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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): x+1<4. 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 y+2=y is false since there is no value of y ∃for which the propositional function y+2=y produces a true statement.


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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.


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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)


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Homework

Ex. 2, Ex. 4, Ex. 16, Ex. 28


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Conditional statement If p and q are statements, the compound statement “if p then q”, denoted

p=>q, is called a conditional statement or implication.

p : antecedent or hypothesis q : consequent or conclusion

if … then : =>

Truth Table :

Note: when p is F, then p=>q is T.

2.2. Conditional Statements

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p q p => q

T T T

T F F

F T T

F F T

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Example Form the implication p=>q for each the following

(a) p: I am hungry. q: I will eat.

(b) p: 2+2=5 q: I am the king of England.

Solution (a) If I am hungry, then I will eat

(b) If 2+2=5, 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 2+2=5 is false.


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Converse and Contrapositive If p=>q is an implication, then its converse is the implication

q => p

and its contrapositive is the implication ~ q => ~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.


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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 <=> q is T when p and q are both T or both F.


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p q p q

T T T

T F F

F T T

F F T

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Example 3: Is the following equivalence a true statement?

3>2 if and only if 0< 3 – 2

Solution:

Let p: 3>2 and q : 0< 3 – 2,

since p and q are both true, we then conclude that

p q is true statement.


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Example 4. Compute the truth table of the statement

(p=>q) (~q => ~p)

Truth Table :


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p q p=>q ~q ~p ~q=>~p (p=>q) (~q => ~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

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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 => q)


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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


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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 :


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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

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Two Structures with similar properties

(Theorem 1)

Structure 1 (logic operations):

(propositions, ∧ , V, ~)

Structure 2 (sets operations)

(sets, U, ∩ , - )


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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) ∧ ∨ ∧ ∨ ∧


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Idempotent Properties

p p ≡ p , p p ≡ p∨ ∧

Properties of Negation

~(~p) ≡ p

~(p q) ≡ (~p) (~q) ∨ ∧ ~(p q) ≡ (~p) (~q) ∧ ∨ De Morgan’s laws


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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))∧ ∨ ∧


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Theorem 3 ~( xP(x)) ≡ x~P(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∧ ⇒ ∃ ∧ ∃


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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)⇒ ∧ ⇒ ⇒ ⇒


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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)


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Homework

ex.2, ex.7, ex.12, ex.15, ex.21 ex.34


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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

2.3. Method of Proof

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(p1 p∧ 2 … p∧ ∧ n) q⇒

The pi’s 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”


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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.


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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!


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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


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Equivalence (p q) They are usually stated “p if and only if q”. We need to

prove both p=>q and q=>p 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.


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Modus Ponens p is true, and p=>q is true, so q is true

(Theorem 4(g) in Section 2.2.)

p

p=>q

∴ q


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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.


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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

p=>q

q

∴ p

Then argument is not valid , since p=>q and q can be both true with p being false.


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(or show the truth table of (p=>q) q => p , and determine ∧whether or not it is a tautology)

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Indirect Method of Proof The tautology

(p=>q ) (~q) => (~p )

(An implication is equivalent to its contrapositive)

Note:

To proof p=>q indirectly, we assume q is false (~q) and show that p is then false (~p)


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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 (p=>q)

We try to prove its contrapositive ~q=>~p

Assuming that n is even (~q), let n=2k, k is an integer, then we have n2 = (2k) 2 =4k2 is even (~p).

Hence, the given statement has been proved.


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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 p∧ 2 … p∧ ∧ n) q , ⇒

We add (~q) into hypothesis p1 p∧ 2 … p∧ ∧ n, if the enlarged hypothesis p1 p∧ 2 … p∧ ∧ n (~q) implies a contradiction, then ∧we can conclude that q follows from p1 ,p2 , … and pn .


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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 => q ,

We try to find the contradiction from p ~q ∧ Refer to Example 7 for more details.


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Example 9: Prove or disprove the statement that if x and y are real numbers,

(x2=y2) (x=y)

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.


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Homework

Ex. 6, Ex. 8, Ex. 9

Ex. 19, Ex. 20

Ex. 31, Ex. 34


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To Prove

∀n>n0 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) => P(k+1) is a tautology (if P(k) is true, then P(k+1) must be true)

2.4 Mathematical Induction

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Example 1: Prove that for all n>0

1+2+3…+n = n(n+1)/2 Solution: let P(n)=n(n+1)/2

Basis step: P(1) = 1 = 1*2/2=1 is true.

Induction Step: assuming P(k) is true, then

P(k+1) = 1+2 + 3 … + k +(k+1)

= P(k) +(k+1)

= k (k+1)/2 +(k+1) P(k) is true

= (k+1)(k+2)/2

= P(k+1)


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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 |A|=n>0, 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 k+1 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(k+1) is true)


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Example 5: Consider the following function given in pseudocode

Function SQ(A)

1. C < － 0

2. D < － 0

3. WHILE (D is not A)

a. C < － C+A

b. D < － D+1

4. RETURN (C)

End of Function SQ


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Strong Induction


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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(k+1) is a tautology

P(n0) P(n1) … P(k) ∧ ∧ P(k+1) is a tautology

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Example 7: Prove that every positive integer n>1 can be written uniquely as p1

a1p2a2…ps

as, where pi

are primes and p1<p2<..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 k+1 is prime, then k+1= (k+1)1

if k+1 is not prime, then we let k+1=Lm, where L, m are positive integers less than k+1. Using P(L) and P(m) are true, we have k+1=Lm= q1

b1q2b2…qs

bs (unique form)

Why? Proof by contradiction


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Homework

Ex. 2, Ex. 14, Ex. 17, Ex. 22, Ex. 34


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Documents

Chapter 2. Logic Weiqi Luo ( 骆伟祺 ) School of Software Sun Yat-Sen University Email ： [email protected] Office ： A309 [email protected]