Logic gates

LOGICLOGIC

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

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.”

What is proposition?

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 F T

T :≡ True; F :≡ False“:≡” means “is defined as”

Operandcolumn

Resultcolumn

Topic #1.0 – Propositional Logic: Operators

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 , pq (and)

2. Disjunction, pq (or)

3. Negation p(not)

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 pq=“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 F

F T

F F

T

F

F

F

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.”

pq=“Either my car has a bad engine, or my car has a bad carburetor.”

After the downward-pointing “axe” of “””splits the wood, yousplits the wood, youcan take 1 piece OR the can take 1 piece OR the other, or both.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 P v Q

T T

T F

F T

F F

T

T

T

F

Propositional calculus.

• truth tables for logical connectives

P ~P P Q P Q P Q

T F T T T TF T T F F T

F T F T

F F F F

ExampleIf 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) pq (b)(q) p (c)(p) (q)

(d)( p)q (e)pq

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 F

F T

F F

T

F

T

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 F

F T

F F

T

F

T

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 F

F T

F F

T

F

T

T

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 I’m stillhere!

2006

http://www.rtvbih.ba/2001/vijesti/juli/04/george%20bush.jpg

http://images.google.com/imgres?imgurl=www.uncwil.edu/sga/election.gif&imgrefurl=http://www.uncwil.edu/sga/&h=178&w=178&prev=/images%3Fq%3Delection%26start%3D80%26svnum%3D10%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26sa%3DN

• 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 F

F T

F F

T

F

F

T

Let P(p,q,........) denote an expression constructed fromlogical 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 q p q (p q)

T

T

T

T

T

T

T

T

T

T

F

F

F

F

F

F

F

F

F

F

Proposition

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 (xy), (xy), (xy),(xy)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 geta proposition.

p q

TTFF

TFTF

(p q)

step

TTFF

1

TFTF

12

FTFT

3

FTFF

4

TFTT

Another method of constructing a truth table

Propositional calculus cont.

• Truth tables for common sentences• (PQ)=(~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• (~PQ)=(PQ) 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 pq and (pq)

Propositional Equivalences In mathematical arguments, you can replace a statement or proposition with another statement or proposition with the same truth valueTautology: 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 itContradiction: a compound proposition that is always falseContingency: proposition that is neither a tautology or a contradiction

Logic - Equivalences

Logic - Equivalences• Propositional Equivalences

p p p v p p ^ p

T F T F

F T T F

Contingency tautology contradiction

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,..........

Principle of Substitution

Logical Equivalence

P(p,q,.....) Q(p,q,........) (if identical truth tables)

e.g. p p, p p p

Show that (pq) (p q) p

Show ¬( p V q ) and ¬ p ¬ q are logically equivalentusing truth tables.

PROPOSITIONAL EQUIVALENCES

Logically equivalent using truth tables

F

p q p V q ¬ p ¬ q

T T

F T

T T

T

T T

T

T

T T

T F

F F F

F

F F

F

F

F

F

F

F

¬(p V q) (¬p ¬ q)

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

T T T F F F F

T F T F F T F

F T T F T F F

F F F T T T T

These are logically equivalent

Truth Table for (p v q) and 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

p v F p

p v T T domination laws

p ^ F F

p v p p idempotent laws

p ^ p p

( p) p double negation laws

p v q q v p commutative laws

p ^ q q ^ p

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 v (q v r) Associative laws

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

(p v (q ^ r) (p v q) ^ (p v r) Distributive laws

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

(p ^ q) p v q DeMorgan’s Laws

(p v q) p ^ q

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

Useful Law # 1 p V ¬ p T

Useful Law # 2 p ¬ p F

Useful Law # 3 p q ¬ p V q


Prove ¬ (p V (¬p q)) ¬ p ¬q,

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


Prove ¬ (p V (¬p q)) ¬ p ¬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


Prove ¬ (p V (¬p q)) ¬ p ¬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.


Prove ¬ (p V (¬p q)) ¬ p ¬q,

3. Use Distributative Law

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


Prove ¬ (p V (¬p q)) ¬ p ¬q,

4. Use Double Negation Anytime

5. Use Other Laws as they Apply

Prove ¬ (p V (¬p q)) ¬ p ¬q,


Prove ¬ (p V (¬p q)) ¬ p ¬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.

Prove ¬ (p V (¬p q)) ¬ p ¬q,


Prove ¬ (p V (¬p q)) ¬ p ¬q,

Do we have any implications?no

Can we use DeMorgan’s law?Yes

¬ (p V (¬p q))

Prove ¬ (p V (¬p q)) ¬ p ¬q,


Prove ¬ (p V (¬p q)) ¬ p ¬q,

How do we use DeMorgan’s law?

¬ (p V (¬p q)) ¬ p ¬ (¬p q)

DeMorgan’s Law ( p V q) ¬ p ¬ q

Prove ¬ (p V (¬p q)) ¬ p ¬q,


Prove ¬ (p V (¬p q)) ¬ p ¬q,

Now we have

¬ (p V (¬p q)) ¬ p ¬ (¬p q)) DeMorgan’s Law

Do we have any implications?no

Can we use DeMorgan’s law?yes

¬ (¬p q)


¬ (¬p q) ¬ (¬ p) V ¬ q

DeMorgan’s Law ¬( p V q) ¬ p ¬ q

Prove ¬ (p V (¬p q)) ¬ p ¬q,


Prove ¬ (p V (¬p q)) ¬ p ¬q,

Prove ¬ (p V (¬p q)) ¬ p ¬q,


Prove ¬ (p V (¬p q)) ¬ p ¬q,

Now we have

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

Now we can use the double negation.

¬ p [¬ (¬p) V ¬q)] ¬ p ( p V ¬q)


Now we have


¬ p [¬ (¬p) V ¬q)] ) DeMorgan’s Law

¬ p (p V ¬q) Double negation

Do we have any implications? no Can we use DeMorgan’s law? no Can we use the distributive law? Yes

¬ p (p V ¬q)

How do we use the distributive law?

¬ p (p V ¬q) (¬p p) V (¬p ¬q )

Distributive Law p (q V r) (p q) V (p r)



Now we have




(¬ p p) V (¬ p ¬q) Distributative Law

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

Useful Law # 2 p ¬ p F

How do we use Useful Law #2?

(¬ p p) V (¬ p ¬q)

F V (¬ p ¬q)


Useful Law #2 p ¬ p F

PROPOSITIONAL EQUIVALENCESNow we have




(F) V (¬ p ¬q)

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

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

PROPOSITIONAL EQUIVALENCESNow we have




(F) V (¬ p ¬q)

What do we have? (F) V (¬ p ¬q)

What are we trying to get? ¬ p ¬q

What do we need to get this result?

We need the identity law. How can we get there?

How do we get to the identity law?

F V (¬ p ¬q) (¬ p ¬q) V F

Now apply the identity law?

(¬ p ¬q) V F (¬ p ¬q)


Commutative Laws p V q q V p

Identity Law p V F p


Prove ¬ (p V (¬p q)) ¬ p ¬q,Prove ¬ (p V (¬p q)) ¬ p ¬q,




(¬ p p) V (¬ p ¬q) Distributive Law

(F) V (¬ p ¬q)

(¬ p ¬q) V (F) Commutative Law

¬ p ¬q Identity Law


Prove (p q) ( p V q) is a Tautology.

How do we express this?

Prove (p q) ( p V q) T


(p q) ( p V q) T

Prove (p q) ( p V q) T

Do we have any implications? Yes

(p q) ( p V q) T

How do you use Useful Law #3?

(p q) ( p V q) ¬ ( p q) V ( p V q)


Useful Law #3 ( p q ) ¬ p V q



Now we have

(pq)(pVq)

¬( p q) V ( p V q)

Do we have any implications? no Can we use DeMorgan’s law? Yes


¬( p q) V ( p V q) ( ¬ p V ¬ q) V ( p V q)

DeMorgan’s Law ¬ ( p q) ¬ p V ¬q



Now we have

(p q) ( p V q) T

¬( p q) V ( p V q)

( ¬ p V ¬ q) V ( p V q) DeMorgan’s Law

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


What do we have? ( ¬ p V ¬ q) V ( p V q)

What are we trying to get? T

What do weneed to get this result?

We need UL #1 or Identity law or Domination Law.

Now we have

(p q) ( p V q) T

¬( p q) V ( p V q)


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

Use the Associative Law

( ¬ p V ¬ q) V ( p V q) ¬ p V (¬ q V p) V q


Associative Laws (p V q) V r p V (q V r)

Then Use the Commulative Law

¬ p V (¬ q V p) V q ¬ p V (p V ¬ q) V q


Commulative Laws p V q q V p

Then Again use the Associative Law

¬ p V (p V ¬ q) V q (¬ p V p) V (¬ q V q)


Associative Laws (p V q) V r p V (q V r)

Then we can use Useful Law #1.

(¬ p V p) V (¬ q V q) T V T


Useful Law#1 p V ¬ p T

And finally the Domination Law.

T V T T


Domination Laws p V T T



(p q) ( p V q) T

¬( p q) V ( p V q)


( ¬ p V p) V (¬ q V q) Associative Law

( ¬ p V p) V (¬ q V q) Commulative Law

( T ) V ( T ) Useful Law # 1

T V T Domination Law

Prove

¬(p q) (p ¬q)

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

Arguments

In logical reasoning , a certain number of propositionsare assumed to be true and based on the assumption someother proposition is derived(deduced or inferred)

premises

conclusion

valid argument

qDefinition - An argument p1,p2,p3,.......,pn is said to be valid if q is true whenever all premises

p1,p2,......,pn are true.

fallacy

Theorem - The argument p1,p2,p3,.......,pn q

is valid iff the proposition (p1p2 ........... pn) q

is a tautology

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 “” means “therefore”

Some Inference Rules

• p Rule of Addition pq

• pq Rule of Simplification p

• p Rule of Conjunction q pq

Some Inference Rules

• p

• pq (law of detachment)q

• qpq p

“the mode of affirming”

“the mode of denying”

Syllogism Inference Rules

• pq Rule of hypothetical qr syllogismpr

• p q Rule of disjunctive p syllogism q

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 Proved by1. p q Premise #1.2. p Simplification of 1.3. rp Premise #2.4. r rules 2,3.5. rs Premise #3.6. s rules 4,5.7. st Premise #4.8. t rules 6,7.

Example

Consider the following argument:S1:If a man is a bachelor, he is unhappyS2: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 describingthe properties of an object or relation among objects iscalled 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 - all elements of A for which p(a) is true is called the truth set of p(x)

*T = {x:x A, p(x) is true}or T = {x:p(x)}

p

p p

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:xN, x+2 >7} = {6,7,8,.......},

(b) {x:xN, x+5<3} = (c) {x:xN, x+5>1} = N

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 discourseis the set of all human names.

e.g. - E(n): “n is an even integer”

Logic - Quantifiers• 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 Examples1) 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

(xA)p(x) ( xA) p(x)

( xA) p(x) (xA) 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

LannNnfollowsa n,,,0 .....,......... 001

L is not the limit of the sequence LannNnaa n,,,0 when ...,........., 0021

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