Propositional Equivalences

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Propositional Equivalences

• CS/APMA 202, Spring 2005
• Rosen, section 1.2
• Aaron Bloomfield

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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 ? p
T T F
F 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 ? (well use )

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Logical Equivalences of And

• p ? T p Identity law
• p ? F F Domination law

p T p?T
T T T
F T F
p F p?F
T F F
F F F
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Logical Equivalences of And

• p ? p p Idempotent law
• p ? q q ? p Commutative law

p p p?p
T T T
F F F
p q p?q q?p
T T T T
T F F F
F T F F
F F F F
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Logical Equivalences of And

• (p ? q) ? r p ? (q ? r) Associative law

p q r p?q (p?q)?r q?r p?(q?r)
T T T T T T T
T T F T F F F
T F T F F F F
T F F F F F F
F T T F F T F
F T F F F F F
F F T F F F F
F F F F F F F
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Logical Equivalences of Or

• p ? T T Identity law
• p ? F p Domination law
• p ? p p Idempotent law
• p ? q q ? p Commutative law
• (p ? q) ? r p ? (q ? r) Associative law

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Corollary of the Associative Law

• (p ? q) ? r p ? q ? r
• (p ? q) ? r p ? q ? r
• Similar to (34)5 345
• Only works if ALL the operators are the same!

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Logical Equivalences of Not

• (p) p Double negation law
• p ? p T Negation law
• p ? p F Negation law

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Sidewalk chalk guy

• Source http//www.gprime.net/images/sidewalkchalk
  guy/

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DeMorgans Law

• Probably the most important logical equivalence
• To negate p?q (or p?q), you flip the sign, and
  negate BOTH p and q
• Thus, (p ? q) p ? q
• Thus, (p ? q) p ? q

p q ?p ?q p?q ?(p?q) ?p??q p?q ?(p?q) ?p??q
T T F F T F F T F F
T F F T F T T T F F
F T T F F T T T F F
F 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)
• Absorption
• p ? (p ? q) p
• p ? (p ? q) p

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How to prove two propositions are equivalent?

• Two methods
• Using truth tables
• Not good for long formula
• In this course, only allowed if specifically
  stated!
• Using the logical equivalences
• The preferred method
• Example Rosen question 23, page 27
• Show that

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Using Truth Tables
p q r p?r q ?r (p?r)?(q ?r) p?q (p?q) ?r
T T T T T T T T
T 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
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Using Logical Equivalences
Original statement
Definition of implication
DeMorgans Law
Associativity of Or
Re-arranging
Idempotent Law
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Quick survey

• I understood the logical equivalences on the last
  slide
• Very well
• Okay
• Not really
• Not at all

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End of lecture on 25 January 2005
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Logical Thinking

• At 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 innocent
• Statements 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 s?f b?f b?s f?(b?s)
T T T F F F F T F F
T 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
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Are all of their statements true?Show values for
s, b, and f such that the equation is true
Original statement Definition of
implication Associativity of AND Re-arranging Idem
potent law Re-arranging Absorption
law Re-arranging Distributive law Negation
law Domination law Associativity of AND
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What if it werent possible to assign such values
to s, b, and f?
Original statement Definition of implication ...
(same as previous slide) Domination
law Re-arranging Negation law Domination
law Domination law Contradiction!
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Quick survey

• I feel I can prove a logical equivalence myself
• Absolutely
• With a bit more practice
• Not really
• Not at all

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Logic Puzzles

• Rosen, page 20, questions 51-55
• Knights always tell the truth, knaves always lie
• A says At least one of us is a knave and B says
  nothing
• A 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
  nothing
• Both 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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Functional completeness

• Functional completeness is discussed on page 27
  (questions 37-39) of the text
• All the extended operators have equivalences
  using only the 3 basic operators (and, or, not)
• The extended operators nand, nor, xor,
  conditional, bi-conditional
• Given a limited set of operators, can you write
  an equivalence of the 3 basic operators?
• If so, then that group of operators is
  functionally complete

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Rosen, 1.2 question 46

• Show that (NAND) is functionally complete
• Equivalence of NOT
• p p ?p
• ?(p ? p) ?p Equivalence of NAND
• ?(p) ?p Idempotent law

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Rosen, 1.2 question 46

• Equivalence of AND
• p ? q ?(p q) Definition of nand
• p p How to do a not using nands
• (p q) (p q) Negation of (p q)
• Equivalence of OR
• p ? q ?(?p ? ?q) DeMorgans equivalence of OR
• As we can do AND and OR with NANDs, we can thus
  do ORs with NANDs
• Thus, NAND is functionally complete

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Quick survey

• I felt I understood the material in this slide
  set
• Very well
• With some review, Ill be good
• Not really
• Not at all

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Quick survey

• The pace of the lecture for this slide set was
• Fast
• About right
• A little slow
• Too slow