Propositional Equivalences
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Transcript of 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