Propositional (Boolean) Logic

Prof. Elaine Wenderholm

Computer Science Department

SUNY Oswego

Propositions

A (logical) proposition is a statement that is unequivically

either true or false.

Propositions

“13 is an odd number” is ...

Propositions

... true

Propositions

... true

“√

25 = 5” is ...

Propositions

... true

“√

25 = 5” is ...

... false

Propositions

... true

“√

25 = 5” is ...

... false

“calc I is a hard course” is ...

Propositions

... true

“√

25 = 5” is ...

... false

... NOT a proposition!

Propositions

... true

“√

25 = 5” is ...

... false

“x is a prime number” is ...

Propositions

... true

“√

25 = 5” is ...

... false

“x is a prime number” is ...

... not a proposition.

Logical Operators

Logical operators are used to combine existing propositions

to form new propositions.

Logical Operators

Logical operators are used to combine existing propositions

to form new propositions.

Logical operators are defined with truth tables.

AND, OR, NOT Operators

and or not

p q p ∧ q

p q p ∨ q

Venn Diagrams

A Venn Diagram provides a visual depiction of logical operations

in terms of sets .

This is an excellent web reference to learn aboutVenn Diagrams.

http://www.purplemath.com/modules/venndiag.htm

Implication (conditional) and Bicondi-tional Operators

p implies q p iff q

p q p → q

p q p ↔ q

Implication (conditional) and Bicondi-tional Operators

p implies q p iff q

p q p → q

p q p ↔ q

p → q “p implies q” is false only when p (the antecedent) is

true and q (the consequence) is false. Let me explain...

The operators p ↔ q and p ≡ q are logically equivalent.

Logical Equivalence

Two propositions are logically equivalent (or equivalent, or

equal) if their truth tables are identical.

Logical Equivalence

Common equivalences:

Logical Equivalence

p → q ≡ ¬p ∨ q

Logical Equivalence

p → q ≡ ¬p ∨ q

(one of) DeMorgan’s laws: ¬(p ∨ q) ≡ ¬p ∧ ¬q

Logical Equivalence

p → q ≡ ¬p ∨ q

Use truth tables to determine...

Logical Equivalence

p → q ≡ ¬p ∨ q

You answer these

Logical Equivalence

p → q ≡ ¬p ∨ q

You answer these

What is equivalent to ¬(p ∧ q) ? (Hint: the other

DeMorgan’s law)

Logical Equivalence

p → q ≡ ¬p ∨ q

You answer these

DeMorgan’s law)

¬¬p ≡ ?

Logical Equivalence

p → q ≡ ¬p ∨ q

You answer these

DeMorgan’s law)

¬¬p ≡ ?

p ∧ T ≡ ? p ∨ F ≡ ?

Tautologies, Contradictions

A tautology is a proposition that is TRUE under any

assignment of values to propositional variables.

Example: p ∨ ¬p

A contradition is a propostion that is FALSE under any

Example: p ∧ ¬p

Example: p ∨ ¬p

A contradition is a propostion that is FALSE under any

Example: p ∧ ¬p

Other examples? The previous slides lists equivalences –

these are tautologies.

Predicate Logic

Propositions may contain a variable from some domain.

Example: x2 < 4 where x ∈ RIn this case the domain is the set of real numbers.

Predicate Logic

Predicate Logic is propositional logic, and it also

Predicate Logic

Has a domain of discourse (or a Universe)

Predicate Logic

Has variables : free and bound

Predicate Logic

Has variables : free and bound

Has quantifiers :

universal quantifier - “for all” - ∀existential quantifier - “there exists” (for some) - ∃

Free Variables

Example: A proposition involving a free variable over some

domain U .

Free Variables

domain U .

p(x) is some property of x (and evaluates to TRUE or FALSE).

Free Variables

domain U .

Example : x < 4 – x has the property that its value is less than

Free Variables

domain U .

x is a free variable because we are free to substitute any

value form x ∈ U , and evaluate the truth or falsity of the

proposition.

Free Variables

domain U .

x is a free variable because we are free to substitute any

value form x ∈ U , and evaluate the truth or falsity of the

proposition.

p(x) is called a propositional function, an open sentence, or

a condition defined on the domain U .

How do we evaluate a propositionalfunction?

Look at some p(x) defined on domain U .

We can take every element a in U , substitute it into p(x), and

evaluate each one for truth or falsity.

We look at all the elements that make p(x) true and call it the

truth set of p(x).

Examples on the domain N - the Natural numbers. Let’s

define the truth set for each propositinal function.

truth set of p(x).

p(x) is x + 2 > 7. What elements of N make this true?

truth set of p(x).

{x|p(x)} = {6, 7, 8, . . .}

truth set of p(x).

{x|p(x)} = {6, 7, 8, . . .}p(x) is x + 3 < 1 ...

truth set of p(x).

{x|p(x)} = {6, 7, 8, . . .}p(x) is x + 3 < 1 ...

{x|p(x)} = ∅

truth set of p(x).

{x|p(x)} = {6, 7, 8, . . .}p(x) is x + 3 < 1 ...

{x|p(x)} = ∅x < 5 ∧ x%2 = 0

truth set of p(x).

{x|p(x)} = {6, 7, 8, . . .}p(x) is x + 3 < 1 ...

{x|p(x)} = ∅x < 5 ∧ x%2 = 0

{x|p(x)} = {2, 4}

Bound Variables - Quantified Statements

∀x(p(x)) means

for all variables x in the domain, the property p is true for x.

∀x(p(x)) means

x is a bound variable

∀x(p(x)) means

It is bound to the quantifier ∀

∀x(p(x)) means

It is bound to the quantifier ∀Example: domain is N ∀x(x >= 0)

∀x(p(x)) means

This is true for every x ∈ N . In fact it is a tautology.

∀x(p(x)) means

Example: domain is Z - the Integers

∃x(x2 − 4 = 0)

∀x(p(x)) means

∃x(x2 − 4 = 0)

The only values of x to make this true are x = {2,−2}.

∀x(p(x)) means

∃x(x2 − 4 = 0)

The only values of x to make this true are x = {2,−2}.

Only some of the values in Z make the statement true,

but that’sall we need.

Another way to define quantifiers in termsof sets

Given: domain U , and p(x) is a property of x on elements in U .

Universal Quantifier

∀xp(x) is true whenever {x|p(x)} = U , otherwise ∀xp(x) is

false.

Existential Quantifier

false.

Existential Quantifier

∃xp(x) is true whenever {x|p(x)} 6= ∅, otherwise ∃xp(x) is

false.

Negating Quantifiers

¬(∀x(p(x)) ≡ ∃x(¬p(x))

Negating Quantifiers

¬(∀x(p(x)) ≡ ∃x(¬p(x))

¬(∃x(p(x)) ≡ ∀x(¬p(x))

Nested Quantifiers - 1

Domain = Z (the Integers).

How do we evalute ∃x∀y(x + y = 0)? (btw it’s true)

From the inside to the outside:

∃x∀y(x + y = 0)︸︷︷︸

pick any y

∃x∀y(x + y = 0)︸︷︷︸

pick any y

∃x∀y(x + y = 0)︸︷︷︸

pick any y︸︷︷︸

there is some x that makes this true

This is not the same as ∀x∃y(x + y = 0) (btw it’s false)

∀x∃y(x + y = 0)︸︷︷︸

pick some y

∀x∃y(x + y = 0)︸︷︷︸

pick some y

∀x∃y(x + y = 0)︸︷︷︸

pick some y︸︷︷︸

any x makes this true

DeMorgan’s laws in logic and program-ming languages

Logic Software Hardware

¬(p ∨ q) = ¬p ∧ ¬q !(x||y) = !x && !y p · q = p + q

¬(p ∧ q) = ¬p ∨ ¬q !(x&&y) = !x || !y p + q = p · q

¬(p ∨ q) = ¬p ∧ ¬q !(x||y) = !x && !y p · q = p + q

¬(p ∧ q) = ¬p ∨ ¬q !(x&&y) = !x || !y p + q = p · q

In (digital) hardware, the gates are AND gates and OR gates.

(Actually, AND and OR can be made from one gate - NAND

gate.)

¬(p ∨ q) = ¬p ∧ ¬q !(x||y) = !x && !y p · q = p + q

¬(p ∧ q) = ¬p ∨ ¬q !(x&&y) = !x || !y p + q = p · q

In (digital) hardware, the gates are AND gates and OR gates.

(Actually, AND and OR can be made from one gate - NAND

gate.)

We can draw the circuit diagram (a combinatorial circuit) for

any propositional statement in hardware. table.

Algorithm

The proposition is viewed as a function: it has input values (the

logical variables) and an output value (what the proposition

evaluates to)

Write the truth table for the function (How many rows in the

truth table?)

Draw the input variables as vertical lines

For each row in the truth table that evaluates to T do:

Connect the input variables to an AND gate - if the

variable is T , connect it directly; if it is F , put an inverter

between it and the gate

Connect the output of all the AND gates into an OR gate.

The output from the OR gate is the output of the function

Propositional (Boolean) Logic - Oswego

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