propositional equivalences

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Propositional Equivalences

From Aaron Bloomfield..Used by Dr. Kotamarti

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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

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Logical Equivalence

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

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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

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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

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(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

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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

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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!

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

Logical Equivalences of Not

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Sidewalk chalk guy

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

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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

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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

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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 )()()(

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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

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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

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Quick survey I understood the logical

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

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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)

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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

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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 )(

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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

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Quick survey I feel I can prove a logical

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

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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

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Sand Castles

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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

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Quick survey The pace of the lecture for this

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