Lecture v Propositional Equivalencies

8/7/2019 Lecture v Propositional Equivalencies

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Lecture 5: PropositionalEquivalences

Andrew Katumba2011



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Announcements

CAT I is on Thursday 24th March :

8am 10 am



3

Agenda

Tautologies

Logical Equivalences



Readings

Sections 4.5, 4.6, 4.7 and 4.8 of Schaums Outline of Discrete

Mathematics (pg 77)

Solved problems are on page 82

4



5

Tautologies, contradictions,contingenciesDEF: A compound proposition is called a tautology if

no matter what truth values its atomic propositionshave, its own truth value is T.

EG: p � ¬p (Law of excluded middle)

The opp osite to a tautology, is a comp ound p rop ositionthat ¶s always false ± a c ontradi c tion .

EG: p � ¬p

On the other hand, a comp ound p rop osition whose

truth value isn¶t constant is called a

c

ontingen c

y.EG: p p ¬p



6

Tautologies and contradictions

The easiest way to see if a compoundproposition is a tautology/contradiction

is to use a truth table.

T

F

F

T

�pp

T

T

p ��p

T

F

F

T

�pp

F

F

p ��p



7

Tautology example Part 1

Demonstrate that

[¬p �(p �q )]pq

is a tautology in two ways:

1. Using a truth table ± show that [¬p �(p �q )]pq is always true

2. Using a p roof (will get to this later).



8

Tautology by truth table

p q ¬p p �q ¬p �(p �q ) [¬p �(p �q )]pq

T T

T F

F T

F F



9


p q ¬p p �q ¬p �(p �q ) [¬p �(p �q )]pq

T T F

T F F

F T T

F F T



10


p q ¬p p �q ¬p �(p �q ) [¬p �(p �q )]pq

T T F T

T F F T

F T T T

F F T F



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p q ¬p p �q ¬p �(p �q ) [¬p �(p �q )]pq

T T F T F

T F F T F

F T T T T

F F T F F



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p q ¬p p �q ¬p �(p �q ) [¬p �(p �q )]pq

T T F T F T

T F F T F T

F T T T T T

F F T F F T



13

Tautologies, contradictionsand programming

Tautologies and contradictions in yourcode usually correspond to poor

programming design. EG: while(x <= 3 || x > 3)

x++;

if(x > y)

if(x == y)

return ³never got here´;



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DEF: Two compound propositions p, q arelogi c ally equivalent if their biconditional

joining p q is a tautology. Logicaleq uivalence is denoted by p � q .

EG: The c ontrapositive of a logical implicationis the reversal of the implication, while

negating both components. I.e. thecontrapositive of p pq is ¬q p¬p . As we¶llsee next: p pq � ¬q p¬p



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Logical Equivalence of Conditional and Contrapositive

The easiest way to check for logical equivalenceis to see if the truth tables of both variantshave identical last columns:

p pqqp

Q: why does this work given definition of � ?

¬q p¬p p ¬p q ¬q



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T

F

T

T

T

F

T

F

T

T

F

F

p pq q p


¬q p¬p p ¬p q ¬q



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T

F

T

T

T

F

T

F

T

T

F

F

p pq q p


T

T

F

F

¬q p¬p p ¬p

T

F

T

F

q ¬q



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T

F

T

T

T

F

T

F

T

T

F

F

p pq q p


T

T

F

F

¬q p¬p p ¬p

T

F

T

F

q

F

T

F

T

¬q



L3 19



T

F

T

T

T

F

T

F

T

T

F

F

p pq q p


T

T

F

F

¬q p¬p p

F

F

T

T

¬p

T

F

T

F

q

F

T

F

T

¬q



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T

F

T

T

T

F

T

F

T

T

F

F

p pq q p


T

F

T

T

T

T

F

F

¬q p¬p p

F

F

T

T

¬p

T

F

T

F

q

F

T

F

T

¬q



21


A: p �q by definition means that p m qis a tautology. Furthermore, the

biconditional is true exactly when thetruth values of p and of q are identical.So if the last column of truth tables of p

and of q is identical, the biconditional join of both is a tautology.



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Logical Non-Equivalence of Conditional and Converse

The c onverse of a logical implication is thereversal of the implication. I.e. the converseof ppq is q pp .

EG: The converse of ³If Donald is a duck thenDonald is a bird.´ is ³If Donald is a bird thenDonald is a duck.´

As we¶ll see next: p pq and q pp are not logically eq uivalent.



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p q p pq q pp (p pq) m (q pp )



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T

TF

F

T

FT

F



25



T

TF

F

T

FT

F

T

FT

T



26



T

TF

F

T

FT

F

T

FT

T

T

TF

T



27



T

TF

F

T

FT

F

T

FT

T

T

TF

T

T

FF

T



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Derivational Proof Techniques

When compound propositions involve more andmore atomic components, the size of thetruth table for the compound propositionsincreases

Q1: How many rows are required to construct the truth-table of:( (qm(ppr )) � (�(s�r)��t) ) p (�qpr )

Q2: How many rows are required to construct the truth-table of a proposition involving natomic components?



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A1: 32 rows, each additional variable doublesthe number of rows

A2: In general, 2n rowsTherefore, as compound propositions grow in

complexity, truth tables become more andmore unwieldy. Checking for

tautologies/logical equivalences of complexpropositions can become a chore, especially if the problem is obvious.



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EG: consider the compound proposition

(p pp ) � (�(s�r)��t) ) � (�qpr )

Q: Why is this a tautology?



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A: Part of it is a tautology (p pp ) andthe disjunction of True with any other

comp ound p rop osition is still True:(p pp ) � (�(s�r)��t )) � (�qpr )

� T � (�(s�r)��t )) � (�qpr )

�TDerivational techniques formalize the

intuition of this examp le.



L3 32

Tables of Logical Equival ences

Ident i ty la ws

Li ke a dding 0

Domina t ion la ws

Li ke mul t i pl ying by 0

Idempo ten t la wsDel ete redun danci es

Dou bl e n ega t ion

³I don¶t like you, not ́

Commutativity

Like ³x+y = y+x´

Associativity

Like ³(x+y)+z = y+(x+z)´

Distributivity

Like ³(x+y)z = xz+yz´

De Morgan



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Ta bl es of Logical Equival enc es

Exclu ded mi ddl e

Nega t ing c rea tes o pposi te

Defini t ion of i mplica t ion in terms of No t an d Or



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DeMorgan Identities

DeMorgan¶s identities allow for simplification of negations of complex expressions

Conjunctional negation:�(p 1�p 2�«�p n ) � (�p 1��p 2�«��p n )

³It ¶s n ot the case that all are true iff on e is false.´

Disjun ction al n egation :

�(p 1�p 2�«�p n ) � (�p 1��p 2�«��p n )

³It ¶s n ot the case that on e is true iff all are false.´



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Tautology example Part 2

Demonstrate that

[¬p �(p �q )]pq

is a tautology in two ways:1. Using a truth table (did above)

2. Using a p roof relying on Tables 5 and

6 of Rosen, section 1.2 to derive Truethrough a series of logicalequivalences



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Tautology by proof [¬p �(p �q )]pq



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� [(¬p �p )�(¬p �q )]pq Distributive



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� [ F � (¬p �q )]pq ULE



39



� [ F � (¬p �q )]pq ULE

� [¬p �q ]pq Identity



40



� [ F � (¬p �q )]pq ULE


�¬

[¬

p �q ] � q ULE



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� [ F � (¬p �q )]pq ULE


�¬

[¬

p �q ] � q ULE� [¬(¬p )� ¬q ] � q DeMorgan



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� [ F � (¬p �q )]pq ULE


�¬

[¬


� [p � ¬q ] � q Double Negation



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� [ F � (¬p �q )]pq ULE


�¬

[¬



� p � [¬q �q ] Associative



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� [ F � (¬p �q )]pq ULE


� ¬

[¬

p �

q ] � q ULE� [¬(¬p )� ¬q ] � q DeMorgan



� p � [q �¬q ] Commutative



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� [ F � (¬p �q )]pq ULE


� ¬

[¬

p �

q ]�

q ULE� [¬(¬p )� ¬q ] � q DeMorgan



� p � [q �¬q ] Commutative� p � T ULE



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� [ F � (¬p �q )]pq ULE


� ¬

[¬

p �

q ]�

q ULE� [¬(¬p )� ¬q ] � q DeMorgan



� p � [q �¬q ] Commutative� p � T ULE

� T Domination



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Exercise

1. ³I don¶t drink and drive´ is logicallyequivalent to ³If I drink, then I don¶t

drive´

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Lecture v Propositional Equivalencies

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