1

Propositional Equivalences

From Aaron Bloomfield..Used by Dr. Kotamarti

2

Tautology and Contradiction

A tautology is a statement that is always true p ¬p will always be true (Negation Law)

A contradiction is a statement that is always false p ¬p will always be false (Negation Law)

p p ¬p p ¬pT T FF T F

3

Logical Equivalence

A logical equivalence means that the two sides always have the same truth values Symbol is ≡or (we’ll use ≡)

4

p T ≡ p Identity law

p F ≡ F Domination law

Logical Equivalences of And

p T pTT T TF T F

p F pFT F FF F F

5

p p ≡ p Idempotent law

p q ≡ q p Commutative law

Logical Equivalences of And

p p ppT T TF F F

p q pq qpT T T TT F F FF T F FF F F F

6

(p q) r ≡ p (q r) Associative law

Logical Equivalences of And

p q r pq (pq)r qr p(qr)T T T T T T TT T F T F F FT F T F F F FT F F F F F FF T T F F T FF T F F F F FF F T F F F FF F F F F F F

7

p T ≡ T Identity lawp F ≡ p Domination lawp p ≡ p Idempotent lawp q ≡ q p Commutative law(p q) r ≡ p (q r) Associative law

Logical Equivalences of Or

8

Corollary of the Associative Law

(p q) r ≡ p q r(p q) r ≡ p q rSimilar to (3+4)+5 = 3+4+5Only works if ALL the operators are the same!

9

¬(¬p) ≡ p Double negation lawp ¬p ≡ T Negation lawp ¬p ≡ F Negation law

Logical Equivalences of Not

10

Sidewalk chalk guy

Source: http://www.gprime.net/images/sidewalkchalkguy/

11

DeMorgan’s Law

Probably the most important logical equivalenceTo negate pq (or pq), you “flip” the sign, and negate BOTH p and q

Thus, ¬(p q) ≡ ¬p ¬q Thus, ¬(p q) ≡ ¬p ¬q

p q p q pq (pq) pq pq (pq) pqT T F F T F F T F FT F F T F T T T F FF T T F F T T T F FF F T T F T T F T T

12

Yet more equivalences

Distributive:p (q r) ≡ (p q) (p r)p (q r) ≡ (p q) (p r)

Absorptionp (p q) ≡ pp (p q) ≡ p

13

How to prove two propositions are equivalent?

Two methods: Using truth tables

Not good for long formulaIn this course, only allowed if specifically stated!

Using the logical equivalencesThe preferred method

Example: Rosen question 23, page 35 Show that: rqprqrp )()()(

14

Using Truth Tables

p q r p→r q →r (p→r)(q →r) pq (pq) →r

T T T T T T T TT T F F F F T F

T F T T T T F T

T F F F T T F T

F T T T T T F T

F T F T F T F T

F F T T T T F T

F F F T T T F T

rqprqrp )()()(

(pq) →rpq(p→r)(q →r)q →rp→rrqp

15

Idempotent Law

Associativity of Or

rqprqrp )()()( Definition of implication

Using Logical Equivalences

rqprqrp )()()(

rqprqp rqprrqp

rqprqrp )()()(

rqprqrp

Re-arranging

Original statement

DeMorgan’s Law

qpqp

qpqp )(

rqrprqrp )()(

rrr

16

Quick survey I understood the logical

equivalences on the last slidea) Very wellb) Okayc) Not reallyd) Not at all

18

Logical ThinkingAt a trial:

Bill says: “Sue is guilty and Fred is innocent.” Sue says: “If Bill is guilty, then so is Fred.” Fred says: “I am innocent, but at least one of the

others is guilty.”Let b = Bill is innocent, f = Fred is innocent, and s = Sue is innocentStatements are:

¬s f ¬b → ¬f f (¬b ¬s)

19

Can all of their statements be true?Show: (¬s f) (¬b → ¬f) (f (¬b ¬s))

b f s ¬b ¬f ¬s ¬sf ¬b→¬f ¬b¬s f(¬b¬s)

T T T F F F F T F FT T F F F T T T T T

T F T F T F F T F F

T F F F T T F T T F

F T T T F F F F T T

F T F T F T T F T T

F F T T T F F T T F

F F F T T T F T T F

¬b ¬f ¬s ¬b→¬f f(¬b¬s)¬b¬s¬sfsfb

20

Are all of their statements true?Show values for s, b, and f such

that the equation is trueOriginal statementDefinition of implicationAssociativity of ANDRe-arrangingIdempotent lawRe-arrangingAbsorption lawRe-arrangingDistributive lawNegation lawDomination lawAssociativity of ANDTsbf

Tsbffbfs ))(()()(Tsbffbfs ))(()()(Tsbffbfs )()(Tsbfbffs )()(Tsbfbfs )()(Tbssfbf )()(Tsfbf )(

Tsfbf )( Tsffbf )()(

TsFbf )(Tsbf )(

21

What if it weren’t possible to assign such values to s, b, and f?

Original statementDefinition of implication... (same as previous slide)Domination lawRe-arrangingNegation lawDomination lawDomination lawContradiction!

Tssbf

Tssbffbfs ))(()()(Tssbffbfs ))(()()(

Tssbf )(

TFbf TFf TF

22

Quick survey I feel I can prove a logical

equivalence myselfa) Absolutelyb) With a bit more practicec) Not reallyd) Not at all

23

Logic PuzzlesRosen, page 23, questions 19-23Knights always tell the truth, knaves always lieA says “At least one of us is a knave” and B says nothingA says “The two of us are both knights” and B says “A is a knave”A says “I am a knave or B is a knight” and B says nothingBoth A and B say “I am a knight”A says “We are both knaves” and B says nothing

24

Sand Castles

25

Quick survey I felt I understood the material in this

slide set…a) Very wellb) With some review, I’ll be goodc) Not reallyd) Not at all

26

Quick survey The pace of the lecture for this

slide set was…a) Fastb) About rightc) A little slowd) Too slow