Chap. 2 Fundamentals of Logic

Proposition

• Proposition (or statement): an declarative sentence that is either true or false, but not both.

• e.g.– Margret Mitchell wrote Gone with the Wind.– 2+3=6.– What a beautiful evening.– Get up and do your exercise.– Combinatorics is a required course for sophomores.

O

O

O

X

X

Negation of Proposition

• The negation of a proposition p, denoted by ￢ p, is the statement “It is not the case that p”.

• e.g.– p: Combinatorics is a required course for

sophomores.

– ￢ p: It is not the case that combinatorics is a required course for sophomores.

Truth Table of Negation of Proposition

p ￢ p

1

0

1: true

0: false

0

1

Compound Statement

• Compound Statement: a statement that is combined by two or more statements using logic connections, including ⋀ (conjunction), ⋁ (disjunction), ⊻ (exclusive or), → (implication), and ↔ (biconditional).

Conjunction

• The conjunction of statements p and q, denoted by p ⋀ q, is the statement “p and q”.

• e.g.– p: Combinatorics is a required course for

sophomores.– q: Margret Mitchell wrote Gone with the Wind.– p ⋀ q : Combinatorics is a required course for

sophomores and Margret Mitchell wrote Gone with the Wind.

Disjunction

• The disjunction of statements p and q, denoted by p ⋁ q, is the statement “p or q”.

• e.g.– p: Combinatorics is a required course for

sophomores.– q: Margret Mitchell wrote Gone with the Wind.– p ⋁ q : Combinatorics is a required course for

sophomores or Margret Mitchell wrote Gone with the Wind.

Exclusive Or

• The exclusirve or of statements p and q, denoted by p ⊻ q, is the statement “p or q, but not both”.

• e.g.– p: Combinatorics is a required course for

sophomores.– q: Margret Mitchell wrote Gone with the Wind.– p ⊻ q : Combinatorics is a required course for

sophomores or Margret Mitchell wrote Gone with the Wind, but not both.

Implication

• The implication of statements p and q, denoted by p → q, is the statement “if p, then q”.

• e.g.– p: Combinatorics is a required course for

sophomores.– q: Margret Mitchell wrote Gone with the Wind.– p → q : If combinatorics is a required course

for sophomores, then Margret Mitchell wrote Gone with the Wind.

Biconditional

• The biconditional of statements p and q, denoted by p ↔ q, is the statement “p if and only if q”.

• e.g.– p: Combinatorics is a required course for

sophomores.– q: Margret Mitchell wrote Gone with the Wind.– p ↔ q : Combinatorics is a required course for

sophomores if and only if Margret Mitchell wrote Gone with the Wind.

Truth Table of Conjunction, Disjunction, Exclusive Or,

Implication, and Biconditional

p q p ⋀ q p ⋁ q p ⊻ q p → q p ↔ q

0 0

0 1

1 0

1 1

0

1

0

0

1

1

1

0

1

0

1

0

0

1

1

1

0

1

0

1

Truth Table of Implication

p → q p q

1 0 0

1 0 1

1 1 1

p → q means:

If p is true, then q is true.If p is false, then q is true or false.

Thus,

Tautology and Contradiction

• A compound statement is a tautology if it is true for all truth value assignments for its component statements.

• If a compound statement is false for all such assignment, then it is a contradiction.

• e.g.

p ￢p

p ⋀ ￢p

p ⋁ ￢p

0 1

0 1

1 0

1 0

0

0

0

0

1

1

1

1

∴ p ⋀ ￢ p is a p ⋁ ￢ p is a

contradiction.

tautology.

Logical Equivalence

• Two statements p, q are logically equivalent, and we write p ⇔ q, when

p ↔ q is a tautology.

e.g.

p q ￢(p⋀q)

￢ p⋁￢ q

￢ (p⋀q) ↔ ￢ p⋁￢ q

p→q ￢ p⋁q p→q ↔ ￢ p⋁q

0 0

0 1

1 0

1 1

1

1

1

0

1

1

1

0

1

1

1

1

1

1

0

1

1

1

0

1

1

1

1

1

￢ (p⋀q) ⇔ p→q ⇔￢ p⋁ ￢ q ￢ p⋁q

The Laws of Logic

• For any primitive statements p, q, r, any tautology T0, and any contradiction F0.– ￢￢ p ⇔ p (Law of Double Negation)– ￢ (p⋁q) ⇔ ￢ p⋀ ￢ q (DeMorgan’s Law)– p⋁q ⇔ q⋁p (Commutative Law)– p⋁(q⋁r) ⇔(p⋁q)⋁r (Associative Law)– p⋁(q⋀r) ⇔(p⋁q)⋀(p⋁r) (Distributive Law)

The Laws of Logic (2)

• For any primitive statements p, q, r, any tautology T0, and any contradiction F0.– p⋁p ⇔ p (Idempotent Law)– p⋁F0 ⇔ p (Identity Law)– p⋁ ￢ p ⇔ T0 (Inverse Law)– p⋁T0 ⇔ T0 (Domination Laws)– p (⋁ p q)⋀ ⇔ p (Absorption Law)

Dual of Statement

• Let s be a statement. If s contains no logical connectives other than and ⋀ ⋁, then the dual of s, denoted sd, is the statement obtained from s by replacing each occurrence of and by and ⋀ ⋁ ⋁ ⋀, respectively, and each occurrence of T0 and F0 by F0 and T0, respectively.

• e.g. s: (p⋀ ￢ q)⋁(r⋀T0)

sd:(p⋁ ￢ q)⋀(r⋁F0)

Principle of Duality

• The Principle of Duality. Let s and t be statements that contain no logical connectives other than and . If ⋀ ⋁ s ⇔ t, then sd ⇔ td.

• e.g.

￢ (p⋁q) ⇔ ￢ p⋀ ￢ q

￢ (p⋀q) ⇔ ￢ p⋁ ￢ q

(1st DeMorgan’s Law)

(2nd DeMorgan’s Law)

The Laws of Logic (3)

• For any primitive statements p, q, r, any tautology T0, and any contradiction F0.– ￢ (p⋁q) ⇔ ￢ p⋀ ￢ q (1st DeMorgan’s Law)– ￢ (p⋀q) ⇔ ￢ p⋁ ￢ q (2nd DeMorgan’s Law)– p⋁q ⇔ q⋁p (1st Communication Law)– p⋀q ⇔ q⋀p (2nd Communication Law)– p⋁(q⋁r) ⇔ (p⋁q)⋁r (1st Associative Law)– p⋀(q⋀r) ⇔ (p⋀q)⋀r (2nd Associative Law)

The Laws of Logic (4)

• For any primitive statements p, q, r, any tautology T0, and any contradiction F0.– p⋁(q⋀r) ⇔(p⋁q)⋀(p⋁r) (1st Distributive Law)– p⋀(q⋁r) ⇔(p⋀q)⋁(p⋀r) (2nd Distributive Law)– p⋁p ⇔ p (1st Idempotent Law)– p⋀p ⇔ p (2nd Idempotent Law)– p⋁F0 ⇔ p (1st Identity Law)– p⋀T0 ⇔ p (2nd Identity Law)

The Laws of Logic (5)

• For any primitive statements p, q, r, any tautology T0, and any contradiction F0.– p⋁ ￢ p ⇔ T0 (1st Inverse Law)– p⋀ ￢ p ⇔ F0 (2nd Inverse Law)– p⋁T0 ⇔ T0 (1st Domination Laws)– p⋀F0 ⇔ F0 (2nd Domination Laws)– p (⋁ p q)⋀ ⇔ p (1st Absorption Law)– p (⋀ p q)⋁ ⇔ p (2nd Absorption Law)

Substitution Rule 1• Suppose that the compound statement P is a

tautology. If p is a primitive statement that appears in P and we replace each occurrence of p by the same statement q, then the resulting compound statement P1 is also a tautology.

• e.g. P: ￢ (p⋁q)↔( ￢ p⋀ ￢ q) is a tautology

∴ P1: ￢ [(r⋀s)⋁q]↔[ ￢ (r⋀s)⋀ ￢ q] is a tautology

Substitution Rule 1 (2)

• e.g. ￢ ￢ (p∨q) ⇔ (p∨q)

• e.g. (p∨q) (∧ p∨￢ q) ⇔ (p (∨ q∧￢ q))

∵ ￢ ￢ p ⇔ p (Law of Double Negation)

∵ p (∨ q∧r) ⇔ (p∨q) (∧ p∨r)(1st Distributive Law)

Substitution Rule 2• Let P be a compound statement where p

is an arbitrary statement that appears in P, and let q be a statement such that q ⇔ p. Suppose that in P we replace one or more occurrences of p by q. Then this replacement yields the compound statement P1. Under these circumstances P1 ⇔ P.

• e.g. P: (p→q)→r, P1: ( ￢ p∨q)→r

∵ (p→q) ⇔ ￢ p∨q ∴ P1 ⇔ P

Substitution Rule 2 (2)

• e.g. ￢ ￢ (p∨q)∧ ￢ r ⇔ (p∨q)∧ ￢ r

• e.g. (p∨q) (∧ r∧q) ⇔ (p∨q) (∧ q∧r)

∵ ￢ ￢ (p∨q) ⇔ (p∨q) (Law of Double Negation and Substitution Rule 1)

∵ r∧q ⇔ q∧r (Commutative Law)

Example 2.16

• Show that (p∨q)∧ ￢ ( ￢ p∧q) ⇔ p(p∨q)∧ ￢ ( ￢ p∧q) ⇔ (p∨q) (∧ ￢￢ p∨￢ q)

⇔ (p∨q)∧(p∨￢ q)

⇔ p (∨ q∧￢ q)⇔ p∨F0

⇔ p

(2nd DeMorgan’s Law)

(Law of Double Negation)

(1st Distributive Law)

(2nd Inverse Law)

(1st Identity Law)

Example 2.17

• Show that ￢ [ ￢ [(p∨q)∧r]∨ ￢ q] ⇔ q∧r

￢ [ ￢ [(p∨q)∧r]∨ ￢ q] ⇔ ￢￢ [(p∨q)∧r] (∧ ￢￢ q)

⇔ [(p∨q)∧r]∧q

⇔ (p∨q) (∧ r∧q)

⇔ (p∨q) (∧ q∧r)

⇔ [(p∨q)∧q ]∧r)

⇔ q∧r

(1st DeMorgan’s Law)

(Law of Double Negation)

(2nd Associative Law)

(2nd Commutative Law)

(2nd Associative Law)

(2nd Absorption Law)

Logical Implication

• If p, q are arbitrary statements such that p→q is a tautology, then we say that p logically implies q and we write p⇒q to denote this situation.

• e.g. [p (∧ p→q)] ⇒ q

Rule of Inference

• [p (∧ p→q)] ⇒ q

• The actual rule will be written in the tabular form

p

p→q

∴ q

Example 2.31

• Show the following argument is valid

( ￢ p∨￢ q)→(r∧s)

r→t

￢ t

∴ p

Example 2.31 (2)

Steps Reasons

1) r→t

2) ￢ t

3) ￢ r

4) ￢ r∨￢ s

5) ￢ (r∧s)

6) ( ￢ p∨￢ q)→(r∧s)

7) ￢ ( ￢ p∨￢ q)

8) p∧q

9) ∴ p

Premise

Premise

Premise

Steps (1) and (2) and Modus Tollens

Step (3) and Rule of Disjunctive Amplification

Step (4) and 2nd DeMorgan’s Laws

Steps (6) and (5) and Modus Tollens

Step (7), DeMorgan’s Laws, and Law of Double NegationStep (8) and Rule of Conjunctive Simplification

Example 2.32

• Show the following argument is valid by the method of Contradiction

￢ p↔q

q→r

￢ r

∴ p

Example 2.32 (2)

Steps Reasons

1) ￢ p↔q

2) ( ￢ p→q) (∧ q→ ￢ p)

3) ￢ p→q

4) q→r

5) ￢ p→r

6) ￢ p

7) r

8) ￢ r

9) r∧￢ r (⇔ F0)

10) ∴ p

Premise

Step (1) and ￢ p↔q ⇔ ( ￢ p→q) (∧ q→ ￢ p)

Premise (the one assumed)

Step (2) and Rule of Conjunctive Simplification

Premise

Steps (3) and (4) and Law of the Syllogism

Steps (5) and (6) and Rule of Detachment

Premise

Steps (7) and (8) and Rule of Conjunction

Steps (6) and (9) and the method of Proof by Contradiction

Open Statement

• A declarative sentence is an open statement if– 1) it contains one or more variables, and– 2) it is not a statement, but– 3) it becomes a statement when the variables

in it are replaced by certain allowable choices.

• e.g.– p(x): The number x+2 is an even integer.

Open Statement (2)

• e.g.• q(x, y): The numbers y+2, x−y, and x+2y are

even integers.• p(x): x≥0.• q(x): x2≥0.

Quantifier

• Here the universe comprises all real numbers. The open statements p(x), q(x), r(x), and s(x) are given by

p(x): x≥0, r(x): x2−3x−4=0,

q(x): x2≥0, s(x): x2−3>0.

• Then– ∃x [p(x)∧r(x)] is– ∀x [p(x)→q(x)] is– ∀x [q(x)→s(x)] is

(∃: Existential Quantifier)

(∀: Universal Quantifier)

true

true

false

Table 2.21

Statement When Is It True?

When Is It False?

∃x p(x)

∀x p(x)

∃x ￢ p(x)

∀x ￢ p(x)

For some a in the universe, p(a) is true.

For every a in the universe, p(a) is false.

For every a in the universe, p(a) is true.

For some a in the universe, p(a) is false.

For some a in the universe, p(a) is false.

For every a in the universe, p(a) is true.

For every a in the universe, p(a) is false.

For some a in the universe, p(a) is true.

Logical Equivalence of Open Statements

• Let p(x), q(x) be open statements defined for a given universe.

• The open statements p(x) and q(x) are called (logically) equivalent, and we write ∀x [p(x) ⇔ q(x)] when the biconditional p(a) ↔ q(a) is true for each replacement a from the universe (that is, p(a) ⇔ q(a) for each a in the universe).

Logical Equivalence of Open Statements (2)

• For the universe of all triangles in the plane, let p(x), q(x) denote the open statements:

p(x): x is equiangular,

q(x): x is equilateral.

∵ p(a) ↔ q(a) is true for every triangle a in the plane

∴ ∀x [p(x) ⇔ q(x)]

(p(x) and q(x) are logically equivalent)

Logical Implication of Open Statements

• Let p(x), q(x) be open statements defined for a given universe.

• If the implication p(a)→ q(a) is true for each a in the universe (that is, p(a)⇒ q(a) for each a in the universe), then we write ∀x [p(x) ⇒ q(x)] and say that p(x) logically implies q(x).

Logical Implication of Open Statements (2)

• For the universe of all triangles in the plane, let p(x), q(x) denote the open statements:

p(x): x is equiangular,

q(x): x is equilateral.

∵ p(a) → q(a) is true for every triangle a in the plane

∴ ∀x [p(x) ⇒ q(x)]

(p(x) logically implies q(x))

Logical Equivalence and Logical Implication for Qualifier Statement• ∃x [p(x)∧q(x)] [⇒ ∃x p(x)∧∃x q(x)]

• Suppose ∃x [p(x)∧q(x)] is true.

• p(a)∧q(a) is true for some a.

• p(a) is true and q(a) is true for some a.

• ∃x p(x) is true and ∃x q(x) is true.

• [∃x p(x)∧∃x q(x)] is true.

Negating Statement with One Quantifier

• ￢ [∀x p(x)] ⇒ ∃x ￢ p(x)

1.Suppose ￢ [∀x p(x)] is true.2.∀x p(x) is false.

3.p(a) is false for some a.

4.￢ p(a) is true for some a.

5.∃x ￢ p(x) is true.

Negating Statement with One Quantifier (2)

• ∃x ￢ p(x) ⇒ ￢ [∀x p(x)]

• Suppose ∃x ￢ p(x) is true.• ￢ p(a) is true for some a.

• p(a) is false for some a.

• ∀x p(x) is false.

• ￢ [∀x p(x)] is true

See Table 2.21

Example 2.49

• Let p(x, y), q(x, y), and r(x, y) represent three open statements, with replacements for the variables x, y chosen from some prescribed universe(s). What is the negation of the following statement?

∀x∃y [(p(x, y)∧q(x, y))→r(x, y)]

Example 2.49 (2)

• ￢ [∀x∃y [(p(x, y)∧q(x, y))→r(x, y)]]

⇔ ∃x [ ￢ [∃y [(p(x, y)∧q(x, y))→r(x, y)]]]

⇔ ∃x∀y ￢ [(p(x, y)∧q(x, y))→r(x, y)]

⇔ ∃x∀y ￢ [ ￢ [p(x, y)∧q(x, y)]∨r(x, y)]

⇔ ∃x∀y [ ￢￢ [p(x, y)∧q(x, y)]∧ ￢ r(x, y)]

⇔ ∃x∀y [p(x, y)∧q(x, y)∧￢ r(x, y)].

Rule of Universal Specification

• If an open statement becomes true for all replacements by the members in a given universe, then that open statement is true for each specific individual member in that universe.

• (A bit more symbolically—if p(x) is an open statement for a given universe, and if ∀x p(x) is true, then p(a) is true for each a in the universe.)

Example 2.53

• For the universe of all people, consider the open statements

m(x): x is a mathematics professor,

c(x): x has studied calculus.• Now consider the following argument.

– All mathematics professors have studied calculus.

– Leona is a mathematics professor.– Therefore Leona has studied calculus.

∀x [m(x)→c(x)]m(Leona)

∴ c(Leona)

Example 2.53 (2)

• If we let l represent this particular woman (in our universe) named Leona, then we can rewrite this argument in symbolic form as

∀x [m(x)→c(x)]

m(l)

∴ c(l)

Example 2.53 (3)

Steps Reasons

1) ∀x [m(x)→c(x)]

2) m(l)→c(l)

3) m(l)

4) ∴ c(l)

Premise

Step (1) and Rule of Universal Specification

Premise

Steps (2) and (3) and the Rule of Detachment

Rule of Universal Generalization• If an open statement p(x) is proved to be true when x is

replaced by any arbitrarily chosen element c from our universe, then the universally quantified statement ∀x p(x) is true.

• Furthermore, the rule extends beyond a single variable. So if, for example, we have an open statement q(x, y) that is proved to be true when x and y are replaced by arbitrarily chosen elements from the same universe, or their own respective universes, then the universally quantified statement ∀x∀y q(x, y) [or, ∀x,y q(x, y)] is true.

• Similar results hold for the cases of three or more variables.

Example 2.56

• Show the following argument is

∀x [p(x)∨q(x)]

∀x [( ￢ p(x)∧q(x))→r(x)]

∴ ∀x [ ￢ r(x)→ p(x)]

Example 2.56 (2)Steps Reasons

1) ∀x [p(x)∨q(x)]

2) p(c)∨q(c)

3) ∀x [( ￢ p(x)∧q(x))→r(x)]

4) [( ￢ p(c)∧q(c))→r(c)]

5) ￢ r(c)→ ￢ ( ￢ p(c)∧q(c))

6) ￢ r(c)→[p(c)∨ ￢ q(c)]

7) ￢ r(c)

8) p(c)∨ ￢ q(c)

9) [p(c)∨q(c)] [∧ p(c)∨ ￢ q(c)]

10)p(c) [∨ q(c)∧ ￢ q(c)]11)p(c)∨ F0

12)p(c)

13)∀x [ ￢ r(x)→ p(x)]

Premise

Step (1) and Rule of Universal Specification

Step (5), 2nd DeMorgan’s Law, and Law of Double Negation

Premise

Step (3) and Rule of Universal Specification

Step (4) and s →t ⇔ ￢ t → ￢ s

Premise (assumed)

Steps (7) and (6) and Modus Ponens

Steps (2) and (8) and Rule of Conjunction

Step (9) and 1st Distributive Law

Step (10) and 2nd Inverse Law

Steps (7) and (11) and Rule of Universal Generalization

Step (11) and 1st Indentity Law