Logic Gates

LOGIC

Introduction to logic   

What is logic? Why is it useful? Types of logic  Propositional logic  Predicate logic

Introduction to logic   

What is logic? Why is it useful? Types of logic  Propositional logic  Predicate logic

What is logic?

“Logic is the beginning of wisdom, not the end”

What is logic? 

Logic : The branch of philosophy concerned with analysing the patterns of reasoning by which a conclusion is drawn from a set of premises, without reference to meaning or context

Why study logic? 

Logic is concerned with two key skills, which any computer engineer or scientist should have:  

Abstraction Formalisation

Why is logic important? 

Logic is a formalisation of reasoning.



Logic is a formal language for deducing knowledge from a small number of explicitly stated premises (or hypotheses, axioms, facts) Logic provides a formal framework for representing knowledge Logic differentiates between the structure and content of an argument





What is proposition? Def: A proposition is a statement that is either true or false. or A proposition is a declarative sentence that is either true or false,but not both. e.g. “It is raining in Delhi.” e.g. “The square of 5 is 16”. Some propositions may not be easily verified: e.g. “The universe is infinite.”

Topic #1.0 – Propositional Logic: Operators

The Negation Operator The negation operator “¬” (NOT) transforms a prop. into its logical negation. E.g. If p = “I have brown hair.” then ¬p = “I do not have brown hair.” The truth table for NOT: p ¬p T F T :≡ True; F :≡ False F T “:≡” means “is defined as” Operand column

Result column

Logic •

Notation for propositions: Truth Values – If it’s true, denoted by T; – If it’s false, denoted by F – Used in truth tables: P

P

T

F

F

T

Compound Propositions Composite Composed of subpropositions & various connectives Primitive or not composite E.g. This book is good and cheap

Propositional Variable Symbol representing any proposition real variable (x) not propositon but can be replaced by a

proposition

Basic Logical Operators 1. Conjunction , p∧q (and) 2. Disjunction, p∨q (or) 3. Negation ¬ p(not)

Topic #1.0 – Propositional Logic: Operators

The Conjunction Operator The binary conjunction operator “∧” (AND) combines two propositions to form their logical conjunction. E.g. If p=“I will have salad for lunch.” and q=“I will have soup for dinner.”, then p∧q=“I will have salad for lunch and I will have soup for dinner.”

Logic • •

More with Truth Tables: conjunction If you have propositions p and q, the proposition “p and q” is true when they’re both true, and false otherwise: P

Q

P^Q

T

T

T

T

F

F

F

T

F

F

F

F

Topic #1.0 – Propositional Logic: Operators

The Disjunction Operator The binary disjunction operator “∨” (OR) combines two propositions to form their logical disjunction. p=“My car has a bad engine.” ∨ q=“My car has a bad carburetor.” p∨q=“Either my car has a bad engine, or my car has a bad carburetor.”

After the downwardpointing “axe” of “∨” splits the wood, you can take 1 piece OR the other, or both.

Logic • •

More with Truth Tables: disjunction If you have propositions p and q, the proposition “p or q” is false when they’re both false, and true otherwise: P

Q

PvQ

T

T

T

T

F

T

F

T

T

F

F

F

Propositional calculus. • truth tables for logical connectives

P ~P P Q

P∧Q P∨Q

T F F T

T T T F

T F

T T

F T

F

T

F F

F

F

Example If p represents “ This book is good” and q represents This book is cheap”, write the following sentences in symbolic form: (a) This book is good and cheap. (b) This book is costly but good (c) This book is neither good nor cheap (d) This book is not good but cheap (e) This book is good or cheap (a) p∧q

(b)(¬q) ∧p (c)(¬p) ∧(¬q) (d)(¬ p)∧q (e)p∨q

Topic #1.0 – Propositional Logic: Operators

The Implication (conditional) Operator

The implication p → q states that p implies q. I.e., If p is true, then q is true; but if p is not true, then q could be either true or false. E.g., let p = “You study hard.” q = “You will get a good grade.” p → q = “If you study hard, then you will get a good grade.”

Logic • •

More with Truth Tables: implication p q If you have propositions p and q, the implication p q of p and q is false when p is true and q is false and is true otherwise: p

q

p

q

T

T

T

T

F

F

F

T

T

F

F

T

Logic • •

More with Truth Tables: implication p

q

Other ways to refer to this implication: – q if p if p, q q whenever p – p only if q q is necessary for p – If p, then q p is sufficient for q p implies q p

q

p

q

T

T

T

T

F

F

F

T

T

F

F

T

• •

Logic

More with Truth Tables: implication p q In other words, p is the hypothesis (or antecedent or premise); and q is the conclusion (or consequence)

p

q

p

q

T

T

T

T

F

F

F

T

T

F

F

T

Topic #1.0 – Propositional Logic: Operators

The biconditional operator The biconditional p ↔ q states that p is true if and only if (IFF) q is true. p = “Bush wins the 2005 election.” q = “Bush will be president for all of 2006.” p ↔ q = “If, and only if, Bush wins the 2005 election, Bush will be president for all of 2006.” 2005

2006

I’m still here!

• • •

More with Tables: biconditional p q True when p and q have the same truth values and is false otherwise Other ways to express it: p IFF q; p is necessary and sufficient for q; if p then q, and vice versa p

q

p

q

T

T

T

T

F

F

F

T

F

F

F

T

Proposition Let P(p,q,........) denote an expression constructed from logical variables p,q,......., which take on the value True(T) or False(F), and the logical connectives ∧, ∨, and ¬ E.g. P(p,q) = ¬ (p ∧ ¬ q) p∧¬q ¬ (p ∧ ¬ q) p q ¬q T

T

F

F

T

T

F

T

T

F

F

T

F

F

T

F

F

T

F

T

Well-Formed Formulas(wff) (i) If P is a propositional variable then it is wff. (ii) If x is wff , then ~ x is a wff. (iii) If x and y are wff , then (x∨y), (x∧y), (x⇒y), (x⇔y)are wffs. (iv) A string of symbols is a wff iff it is obtained by finitely many applications of (i)-(iii) A wff is not a proposition , but if we substitute the proposition in place of propositional variable , we get a proposition.

Another method of constructing a truth table p

q

¬

(p

T T F F

T F T F

T F T T 4

step

¬

q)

T T F F

∧ F T F F

F T F T

T F T F

1

3

2

1

Propositional calculus cont. • Truth tables for common sentences • (P⇒Q)=(~Q⇒~P) /contrapositive equivalence P Q ~Q ~P P⇒Q ~Q⇒~P T T

F

F

T

T

T F

T

F

F

F

F T

F

T

T

T

F F

T

T

T

T

Propositional calculus cont. • Truth tables for common sentences • (~P⇒Q)=(P∨Q) and (P ⇒ Q)=(~P ∨ Q) /disjunctive equivalence

P Q ~P ~P⇒ Q P∨ Q P⇒ Q ~P∨ Q T T

F

T

T

T

T

T F

F

T

T

F

F

F T

T

T

T

T

T

F F

T

F

F

T

T

Construct truth table for p∨¬q and (p∨q)

Logic - Equivalences Propositional Equivalences In mathematical arguments, you can replace a statement or proposition with another statement or proposition with the same truth value Tautology: A compound proposition (combination of propositions using logical operators) that is always True, no matter what the truth values of the propositions that are in it Contradiction: a compound proposition that is always false Contingency: proposition that is neither a tautology or a contradiction

Logic - Equivalences •

Propositional Equivalences p

p

pv

p

p^

T

F

T

F

F

T

T

F

Contingency

tautology

p

contradiction

Principle of Substitution Let P(p,q,.......) be a tautology , and let P1(p,q,......),P2(p,q,......),...... be any propositions. Since P(p,q,........) does not depend upon the particular truth values of its variables p,q,..., we can substitute P1 for p , P2 for q, in the tautology P(p,q,.....) and still have tautology.

Theorem- If P(p,q,....) is a tautology, then P(P1,P2,.....) is a tautology for any propositions P1,P2,..........

Logical Equivalence P(p,q,.....) ≡ Q(p,q,........) (if identical truth tables) e.g. ¬ ¬ p ≡ p, p∨p≡p

Show that (p∧q) ∨(p∧ ¬q) ≡ p

PROPOSITIONAL EQUIVALENCES Show ¬( p V q ) using truth tables.

¬ p Λ ¬ q are logically equivalent

and

p

q

pVq

¬(p V q)

T T

T F

T T

F F

F

F

T

T

F

F

F

¬p ¬q

(¬p Λ ¬ q)

F

F T

F F

F

T

F

F

T

T

T

T

Logically equivalent using truth tables

•

Logic - Equivalences Logical Equivalences: compound propositions that have the same truth value in all possible cases words, denotes logical equivalence between p and q, for example. p q pvq T T T

(p v q) p F F

q F

p^ F

T F

T

F

F

T

F

F T

T

F

T

F

F

F F

F

T

T

T

T

q

Truth Table for (p v q) and p ^ q

These are logically equivalent

other tautologies: • commutative law: P∧Q = Q∧ P P∨Q=Q∨P

• associative law: P∧(Q∧R) = (P∧Q)∧R P∨(Q∨R) = (P∨Q)∨R

• distributive law: P∨(Q∧R) = (P∨Q)∧(P∨R) P∧(Q∨R) = (P∧Q)∨(P∧R)

• deMorgan's Law: ∼(P∨Q) = (∼P∧∼Q) ∼(P∧Q) = (∼P∨∼Q)

Logic - Equivalences(Laws of Algebra) Logical Equivalences: (T denotes any proposition that is always true, F denotes one that is always false) p^T p identity laws pvF p pvT T domination laws p^F F pvp p idempotent laws p^p p ( p) p double negation laws pvq qvp commutative laws

Logic - Equivalences(Laws of Algebra) •

Logical Equivalences: (T denotes any proposition that is always true, F denotes one that is always false)

(p v q) v r (p ^ q) ^ r (p v (q ^ r) p ^ (q v r) (p ^ q) (p v q)

p v (q v r) p ^ (q ^ r) (p v q) ^ (p v r) (p ^ q) v (p ^ r) pv q p^ q

Associative laws Distributive laws DeMorgan’s Laws

These laws can be used to prove whether different compound propositions are logically equivalent

Useful Law # 1

p V ¬ p ⇔T

Useful Law # 2

pΛ¬ p⇔ F

Useful Law # 3

p

q⇔¬ pVq

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

This is easy to prove using the truth table. But now we want to prove it using the logical equivalences.

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

Some guidance in proving using logical equivalences. 1. Do implication first Note: How many laws have to do with implies??? When trying to decide which laws to use in a proof, first ask yourself, are there any implications to prove. If there are then use the third useful law

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

2. Do DeMorgan’s second Next ask yourself, are there any negations with and/or operators? If there are, then use DeMorgan’s Law.

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

3. Use Distributative Law Next ask yourself, are there both and & or operators? If there are, then use the Distributive Law.

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

4. Use Double Negation Anytime 5. Use Other Laws as they Apply

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

We are trying to make both sides equivalent. Begin with the left hand side. Try to make it expressed as the right hand side by using your laws.

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

Do we have any implications? no Can we use DeMorgan’s law? Yes ¬ (p V (¬p Λ q))

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

How do we use DeMorgan’s law? DeMorgan’s Law ¬ (p V (¬p Λ q))

¬ ( p V q) ⇔ ⇔

¬p Λ ¬q ¬ p Λ ¬ (¬p Λ q)

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

Now we have ¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬ (¬p Λ q))

DeMorgan’s Law

Do we have any implications? no Can we use DeMorgan’s law? ¬ (¬p Λ q)

yes

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

How do we use DeMorgan’s law? DeMorgan’s Law ¬ (¬p Λ q)

¬( p V q) ⇔ ⇔

¬p

¬ (¬ p)

Λ

V ¬q

¬q

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

Now we have ¬ (p V (¬p Λ q))

⇔ ¬ p Λ ¬ (¬p Λ q)) DeMorgan’s Law ⇔ ¬ p Λ [¬ (¬p) V ¬q)] ) DeMorgan’s Law

Now we can use the double negation. ¬ p Λ [¬ (¬p) V ¬q)]

⇔

¬pΛ (

p V ¬q)

PROPOSITIONAL EQUIVALENCES Now we have ¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬ (¬p Λ q)) DeMorgan’s Law ⇔ ¬ p Λ [¬ (¬p) V ¬q)] ) DeMorgan’s Law ⇔ ¬ p Λ (p V ¬q)

Do we have any implications? Can we use DeMorgan’s law? Can we use the distributive law? ¬ p Λ (p V ¬q)

Double negation

no no Yes

PROPOSITIONAL EQUIVALENCES How do we use the distributive law?

Distributive Law

p Λ (q

V

r) ⇔

(p Λ q) V (p Λ

r)

¬ p Λ (p V ¬q) ⇔ (¬p Λ p) V (¬p Λ ¬q )

PROPOSITIONAL EQUIVALENCES Now we have ¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬ (¬p Λ q)) DeMorgan’s Law ⇔ ¬ p Λ [¬ (¬p) V ¬q)] ) DeMorgan’s Law ⇔ ¬ p Λ (p V ¬q)

Double negation

⇔ (¬ p Λ p) V (¬ p Λ ¬q)

Distributative Law

Do we have any implications? no Can weuse DeMorgan’s law? no Can we use the distributive law? Can we use any of the useful laws? Yes Useful Law # 2

no

pΛ¬ p⇔ F

PROPOSITIONAL EQUIVALENCES

How do we use Useful Law #2?

Useful Law #2p

Λ

⇔

¬ p

(¬ p

Λ F

p)

F

V (¬ p Λ ¬q) V

(¬ p Λ ¬q)

⇔

PROPOSITIONAL EQUIVALENCES Now we have ¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬ (¬p Λ q)) ⇔ ¬ p Λ [¬ (¬p) V ¬q)] )

DeMorgan’s Law DeMorgan’s Law

⇔ ¬ p Λ (p V ¬q)

Double negation

⇔ (¬ p Λ p) V (¬ p Λ ¬q)

Distributative Law

⇔

(F)

V (¬ p Λ ¬q)

Do we have any implications? no Can we use DeMorgan’s law? no Can we use the distributive law? Can we use any of the useful laws? No

no

Now weneed to look at the result and determine how we might get to that answer.

PROPOSITIONAL EQUIVALENCES Now we have ¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬ (¬p Λ q)) ⇔ ¬ p Λ [¬ (¬p) V ¬q)] )

DeMorgan’s Law DeMorgan’s Law

⇔ ¬ p Λ (p V ¬q)

Double negation

⇔ (¬ p Λ p) V (¬ p Λ ¬q)

Distributative Law

⇔

(F)

V (¬ p Λ ¬q)

What do we have?

(F) V (¬ p Λ ¬q)

What are we trying to get?

¬ p Λ ¬q

What do we need to get this result? We need the identity law. How can we get there?

PROPOSITIONAL EQUIVALENCES How do we get to the identity law? Commutative Laws

p V F V

q

⇔

q

V p

(¬ p Λ ¬q) ⇔ (¬ p Λ ¬q) V F

Now apply the identity law? Identity Law

p (¬ p Λ ¬q)

V V

F F

⇔ ⇔

p (¬ p Λ ¬q)

PROPOSITIONAL EQUIVALENCES Prove

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬q,

¬ (p V (¬p Λ q)) ⇔ ¬ p Λ ¬ (¬p Λ q)) DeMorgan’s Law ⇔ ¬ p Λ [¬ (¬p) V ¬q)] ) DeMorgan’s Law ⇔ ¬ p Λ (p V ¬q)

Double negation

⇔ (¬ p Λ p) V (¬ p Λ ¬q)

Distributive Law

⇔

(F)

V (¬ p Λ ¬q)

⇔ (¬ p Λ ¬q) V ⇔ ¬ p Λ ¬q

(F) Commutative Law Identity Law

PROPOSITIONAL EQUIVALENCES Prove

(p Λ q) → ( p V q) is a Tautology.

How do we express this?

Prove

(p Λ q) → ( p V q)

⇔

T

PROPOSITIONAL EQUIVALENCES Prove

(p Λ q) → ( p V q)

⇔

T

(p Λ q) → ( p V q) ⇔ T

Do we have any implications? (p Λ q) → ( p V q) ⇔ T

Yes

PROPOSITIONAL EQUIVALENCES

How do you use Useful Law #3? Useful Law #3

(

p

→

q

)⇔ ¬

p

V

(p Λ q) → ( p V q) ⇔ ¬ ( p Λ q) V

q ( p V q)

PROPOSITIONAL EQUIVALENCES Prove

(p Λ q) → ( p V q) is a Tautology.

Now we have (pΛq)→(pVq)⇔ ⇔ ¬( p Λ q) V ( p V q)

Do we have any implications? Can we use DeMorgan’s law?

no Yes

PROPOSITIONAL EQUIVALENCES

How do we use DeMorgan’s law? DeMorgan’s Law ¬ ( p Λ q)

⇔ ¬ p V ¬q

¬( p Λ q) V ( p V q) ⇔ ( ¬ p V ¬ q) V ( p V q)

PROPOSITIONAL EQUIVALENCES Now we have (p Λ q) → ( p V q) ⇔ T ⇔ ¬( p Λ q) V ( p V q) ⇔ ( ¬ p V ¬ q) V ( p V q)

Do we have any implications? no Can we use DeMorgan’s law? no Can we use the distributive law? Can we use any of the useful laws? No

DeMorgan’s Law

no

PROPOSITIONAL EQUIVALENCES Now we have (p Λ q) → ( p V q) ⇔ T ⇔ ¬( p Λ q) V ( p V q) ⇔ ( ¬ p V ¬ q) V ( p V q)

What do we have? What are we trying to get?

DeMorgan’s Law

( ¬ p V ¬ q) V ( p V q)

T

What do weneed to get this result? We need UL #1 or Identity law or Domination Law.

PROPOSITIONAL EQUIVALENCES

Since we have all V (ors), we will try UL#1. How can we get there? Use the Associative Law

Associative Laws

(p V q) V r ⇔ p V (q V r)

( ¬ p V ¬ q) V ( p V q)

⇔ ¬ p V (¬ q

V p) V

q

PROPOSITIONAL EQUIVALENCES

Then Use the Commulative Law

Commulative Laws

¬ p V (¬ q

pVq⇔ qVp

V p) V q

⇔ ¬ p V (p V ¬ q) V

q

PROPOSITIONAL EQUIVALENCES

Then Again use the Associative Law

Associative Laws

¬ p V (p V ¬ q) V

(p V q) V r ⇔ p V (q V r)

q ⇔ (¬ p V p) V (¬ q V q)

PROPOSITIONAL EQUIVALENCES

Then we can use Useful Law #1.

Useful Law#1

pV¬ p⇔ T

(¬ p V p) V (¬ q V q) ⇔

T

V

T

PROPOSITIONAL EQUIVALENCES

And finally the Domination Law.

pVT⇔T

Domination Laws

T

V

T

⇔

T

PROPOSITIONAL EQUIVALENCES Prove

(p Λ q) → ( p V q) is a Tautology.

(p Λ q) → ( p V q) ⇔ T ⇔ ¬( p Λ q) V ( p V q) ⇔ ( ¬ p V ¬ q) V ( p V q)

DeMorgan’s Law

⇔ ( ¬ p V p) V (¬ q V q)

Associative Law

⇔ ( ¬ p V p) V (¬ q V q)

Commulative Law

⇔ (

Useful Law # 1

T

⇔ T V T

) V (

T

)

Domination Law

Prove ¬(p → q) ⇔ (p Λ ¬q)

Show that ((p∨q) Λ¬(¬p Λ(¬q ∨ ¬r))) ∨(¬p Λ ¬q) ∨(¬p Λ¬r) is a tautology.

Arguments In logical reasoning , a certain number of propositions are assumed to be true and based on the assumption some other proposition is derived(deduced or inferred) premises conclusion Definition - An argument p1,p2,p3,.......,pn q is said to be valid if q is true whenever all premises p1,p2,......,pn are true. valid argument fallacy

Theorem - The argument p1,p2,p3,.......,pn is valid iff the proposition (p1∧p2 ∧........... ∧pn) is a tautology

q q

Inference Rules - General Form • An Inference Rule is – A pattern establishing that if we know that a set of antecedent statements of certain forms are all true, then we can validly deduce that a certain related consequent statement is true.

• antecedent 1 antecedent 2 … ∴ consequent “therefore”

“∴” means

Some Inference Rules •

p ∴ p∨q • p∧q ∴p • p q ∴ p∧q

Rule of Addition Rule of Simplification Rule of Conjunction

Some Inference Rules • p • p→q ∴q •

¬q p→q ∴¬p

(law of detachment) “the mode of affirming” “the mode of denying”

Syllogism Inference Rules •

p→q q→r ∴p→r

• p∨q ¬p ∴q

Rule of hypothetical syllogism

Rule of disjunctive syllogism

Formal Proof Example • Suppose we have the following premises: “It is not sunny and it is cold.” “We will swim only if it is sunny.” “If we do not swim, then we will play.” “If we play, then we will be home early.” • Given these premises, prove the theorem “We will be home early” using inference rules.

Proof Example cont. • Let us adopt the following abbreviations: – p = “It is sunny”; q = “It is cold”; r = “We will swim”; s = “We will play”; t = “We will be home early”.

• Then, the premises can be written as: (1) ¬p∧ q (2) r → p (3) ¬r → s (4) s→ t

Proof Example cont. Step 1. ¬p ∧ q 2. ¬p 3. r→p 4. ¬r 5. ¬r→s 6. s 7. s→t 8. t

Proved by Premise #1. Simplification of 1. Premise #2. rules 2,3. Premise #3. rules 4,5. Premise #4. rules 6,7.

Example Consider the following argument: S1:If a man is a bachelor, he is unhappy S2:If a man is unhappy, he dies young ---------------------------------------------------S:Bachelors die young

Predicate Calculus * Ram is a student * Shyam is a student * x is a student * ‘ is a student ‘ - Predicate e.g. “2x + 3y = 4z” Definition- A part of a declarative sentence describing the properties of an object or relation among objects is called a predicate.

Predicate Calculus * Let A be a given set . A propositional function(or an open sentence or condition) defined on A is an expression p(x) which has property that p(a) is true or false for each a ∈ A. *A - domain of p(x) *T p - all elements of A for which p(a) is true is called the truth set of p(x) *T p= {x:x ∈ A, p(x) is true}or T p = {x:p(x)}

Predicate Calculus e.g.- 1. x is the father of y - P(x,y) 2. 2x+3y = 4z - S(x,y,z) P(x,y) , S(x,y,z) are not propositions but if x=2 , y=0 and z =1 in S(x,y,z) or S(2,0,1) is proposition with truth value T e.g. Find the truth set of each propositional function p(x) defined on the set N of positive integers. (a) Let p(x) be “x+2>7” (b) Let p(x) be “x+5<3” (c) Let p(x) be “x+5>1” (a) {x:x∈N, x+2 >7} = {6,7,8,.......}, (c) {x:x∈N, x+5>1} = N (b) {x:x∈N, x+5<3} = φ

Predicate Calculus Definition - For a declarative sentence involving a predicate , the universe of discourse , or simply the universe, is the set of all possible values which can be assigned to variables. e.g. -1. For p(x): “x is a student “ the universe of discourse is the set of all human names. e.g. - E(n): “n is an even integer”

Logic - Quantifiers •

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Let’s say you have a predicate like P(x) and you want to apply a statement for all possible values of x. You can use quantifiers to do this. The notation x P(x) shows the universal quantification of P(x), with the upside-down A as the universal quantifier. – It says, FOR ALL x P(x) or FOR EVERY x P(x)

Logic - Quantifiers •

An example: Every student in JUIT has studied Maths could be expressed as: –

x (S(x)  P(x)) • Where P(x) denotes that x has studied Maths • And S(x) denotes that x is in JUIT • And the arrow denotes “then”

Logic - Quantifiers – Existential Quantifier of a proposition: there exists an element x in the universe of discourse such that P(x) is true – That is, there is AN x, or at least ONE x, such that P(x) is true – In this case, one would use the backwards E to denote this type quantifier rather than the all inclusive upside down A: • •

x P(x) For example, if P(x) was the statement x > 89, and your data set included test scores of 65, 72, 85, 88, and 95 what would be the existential quantification of P(x)? – TRUE!

Example

“For all x there is a y such that x is greater than y and less than y+1”. In the universe of rational numbers, with the usual interpretation of “+” and “<“, this sentence is true. In the universe of integers, this sentence is false.

Further Examples 1) Similar quantifiers are order independent

2) Different quantifiers are not

3) If P is true of an object, so is Q

4) This is the negation of the above: for some object, P is true but Q is false.

Relations between negation, universal and existential quantifiers ~∃X p(X) = ∀X ~p(X) ~ ∀X p(X) = ∃X ~ p(X) ∃X p(X) = ∃Y p(Y) ∀X q(X) = ∀Y q(Y) ∀X (p(X) ∧ q(X)) = ∀X p(X) ∧ ∀Y q(Y) ∃X (p(X) ∨ q(X)) = ∃X p(X) ∨ ∃Y q(Y)

Negation of Quantified Statements “All math majors are male” “It is not the case that all math majors are male” or “There exists at least one math major who is a female” M- the set of math majors ¬(∀ x ∈M)(x is a male) ≡ (∃x ∈M)(x is not male) ¬(∀ x ∈M)p(x) ≡ (∃x ∈M) ¬p(x) or ¬ ∀ x p(x) ≡ ∃x ¬p

De Morgans Law ¬ (∀x∈A)p(x) ≡ ( ∃ x∈A) ¬ p(x) ¬ ( ∃ x∈A) p(x) ≡ (∀x∈A) ¬ p(x) e.g. “For all positive integers n we have n+2 >8” “There exists a positive integer n such that n+2not>8”

Negating Quantified Statements with More than One Variable e.g. ¬[(∀x ∃y ∃z, p(x,y,z)] ≡ ∃x ¬[∃y ∃z, p(x,y,z)] ≡ ∃x ∀y [¬ ∃z, p(x,y,z)] ≡ ∃x ∀y ∀ z, ¬ p(x,y,z)] e.g. L is the limit of a sequence a1 ,.............. follows ∀ε > 0, ∃n0 ∈ N , ∀n > n0 , a n − L < ε

L is not the limit of the sequence a1 , a 2 ,............ when ∃ε > 0, ∀n0 ∈ N , ∃n > n0 , a n − L ≥ ε