Discrete Mathematics: Lecture 2 - Propositional Logic II.pdf

Discrete Math

1. Math211
Discrete Mathematics
Propositional Logic II
SAHAR SELIM

2. Agenda
1.2 Applications of Propositional Logic
Translating English to logical expressions
1.3 Propositional Equivalences
Discrete Mathematics and Its Applications
Kenneth H. Rosen
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3. 1.2 Applications of Propositional Logic
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4. Translating English to logical
expressions
Why?
Englishisoftenambiguousandtranslatingsentencesinto
compoundpropositionsremovestheambiguity
Usinglogicalexpressions,wecananalyzethemanddetermine
theirtruthvalues
Wecanuserulesofinferencestoreasonaboutthem
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5. Example 1
Let p = “It is hot”, and q = “It is sunny”
It is not hot.
It is hot and sunny.
It is hot or sunny.
It is not hot but sunny.
It is neither hot nor sunny.
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¬ p
p ^ q
p ˅q
¬ p ^ q
¬ p ^ ¬ q

6. Example 2
“You can access the internet from campus
if you are a computer science major or you are not a freshman. ”
p : “You can access the internet from campus”
q : “You are a computer science major”
r : “You are freshmen”
( q v ┐r ) p
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Hypothesis
Conclusion
H C

7. System specification
Translating sentences in natural language into logical
expressions is an essential part of specifying both hardware and
software systems.
Consistency of system specification.
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8. “The automated reply cannotbe sent whenthe file system is full”
1.Let p denote “The automated reply can be sent”
2.Let q denote “The file system is full”
The logical expression is
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Example: System specification

9. Determine whether these system specifications are consistent:
1.The diagnostic message is stored in the buffer or it is retransmitted.
2.The diagnostic message is not stored in the buffer.
3.If the diagnostic message is stored in the buffer, then it is
retransmitted.
Let p denote “The diagnostic message is stored in the buffer”
Let q denote “The diagnostic message is retransmitted”
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Example: System specification

10. Compound Propositions
Example:
p ∨q∧r : Could be interpreted as(p ∨q)∧r orp ∨(q ∧r)
Precedence order: ¬∧∨⊕ → ↔(IMP!) (Overruled by brackets)
We use this order to compute truth values of compound
propositions.
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11. 1.3 Propositional Equivalences
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12. Propositional Equivalence
Two syntactically(i.e., textually) different compound propositions
may be the semantically identical (i.e., have the same
meaning). We call them equivalent .
Learn:
Various equivalence rules orlaws.
How to proveequivalences using symbolic derivations.
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13. Proofs of Equivalence
Show that A ↔ B and ¬(A ⊕B) are equivalent
≡≡
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•Truthtablescanbeusedto
showtheequivalenceoftwo
logicalexpressions
•Thelasttwocolumnshavethe
same values--sothe
expressionsareequivalent

14. Some Terminology
P ¬P P ˅¬P P ^ ¬P
T F T F
F T T F
Result is always
true, no matter
what P is
Result is always
false, no matter
what P is
Tautology Contradiction
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Result is neither
tautology nor
contradiction
Contingency

15. A tautology is a proposition which is always true
Classic Example: P ∨¬P
A contradiction is a proposition which is always false
Classic Example: P ∧¬P
A contingency is a proposition which neither a tautology nor a
contradiction.
Example:(P ∨Q) →¬R
Some Terminology
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16. Logical Equivalence
Two propositions P and Q are logically equivalent
if P ↔Q is a tautology.
We write: P ⇔ Q orP ≡Q
i.e. p and q contain the same truth values as each other in all rows of their truth
tables.
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17. Proving Equivalence via Truth Tables
p→q⇔¬p ∨q
p q (p→q) ¬p ¬p ∨q
F F T T T
F T T T T
T F F F F
T T T F T
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Important for
removing
implication

18. Prove that p ∨q⇔¬(¬p ∧¬q).
Proving Equivalence via Truth Tables
pqpp∨∨qq¬¬pp¬¬qq¬¬pp ∧∧ ¬¬qq¬¬((¬¬pp ∧∧ ¬¬qq))
FF
FT
TF
TT
F
T
T
T
T
T
T
T
T
T
F
F
F
F
F
F
F
F
T
T
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19. Equivalence Laws
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order

20. Equivalence Laws
Domination: p∨ T⇔T p∧F⇔F
Idempotent: p∨ p⇔p p∧ p⇔p
Double negation: ¬¬ p ⇔p
Associative: ( p∨q)∨r⇔p∨(q∨r)
(p∧q)∧r⇔p∧(q∧r)
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Grouping

21. More Equivalence Laws
Distributive: p∨(q∧r) ⇔(p∨q)∧(p∨r)
p∧(q∨r) ⇔(p∧q)∨(p∧r)
De Morgan’s:
¬(p∧q) ⇔¬p ∨¬q
¬(p∨q) ⇔¬p ∧¬q
Trivial tautology/contradiction:
p∨¬p⇔T p∧¬p⇔F
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22. Defining Operators via Equivalences
Using equivalences, we can define operators in terms of
other operators.
(a^b^c) ˅ (a^¬b^c) ˅ (a^¬b^c)
≡
(a^c) ˅ (a^b)
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23. Defining Operators via Equivalences
Exclusive Or
p⊕q⇔(p∨q)∧¬(p∧q)
p⊕q⇔(p∧¬q)∨(q∧¬p)
Implies
p→q ⇔¬p ∨q
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24. Equivalences with Conditionals and
Biconditionals
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Important for
removing
implication

25. Converse, Inverse, Contrapositive
Some terminology, for an implication p →q:
Its converseis: q →p
Its inverseis: ¬p→¬q
Its contrapositive: ¬q →¬p
One of these three has the same meaning (same truth table) as
p→q. Can you figure out which?
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26. Sahar Selim MATH211 Lecture 2 | Propositional Logic II
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27. Truth Table
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28. Statement ≡Contrapositive
•The Statement If x=2, then x
2
=4 is true.
•Its Contrapositive If x
2
ǂ4, then x ǂ 2 so is true.
Converse ≡inverse
•Its Converse If x
2
=4, then x=2 is not true (x could be equal to - 2).
•Its inverse If x ǂ2, then x
2
ǂ4 so is not true (x could be equal to - 2).
p →q q →p ¬p →¬q ¬q →¬p
statement converse inverse contrapositive
Example
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29. Ex.2
•Statement : If x is positive and x
2
=4, then x=2.
•Contrapositive : If x ǂ2 , then x is not positive or x
2
ǂ4.
Ex.3
•Statement: If the sun is shining then I go to store.
•Contrapositive: If I don’t go to store then the sun is not shining.
•Converse: If I go to store then the sun is shining.
•Inverse: If the sun is not shining then I don’t go to store
p →q q →p ¬p →¬q ¬q →¬p
statement converse inverse contrapositive
Example
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30. Manipulating Propositions
Compound propositions can be simplified by using
simple rules.
Some are obvious, e.g. Identity, Domination,
Idempotence, double negation, commutativity,
associativity
Less obvious: Distributive, De Morgan’s laws
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31. Example 1
p→q
¬p ∨q removing implication
¬p ∨¬(¬q) ¬¬x = x (double negation of q)
¬(¬q) ∨¬p commutative law
(¬q) →(¬p) Implication x→y⇔¬x ∨y
Show that p→q⇔¬q→¬pby logic algebra
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p↔q⇔(p→q)∧(q→p)
p→q⇔¬p ∨q

32. Example 2
Show that (p ∧q) → (p ∨q) is a tautology using logical algebra
De Morgan law
Associative and commutative
law for disjunction
Commutative law for disjunction
Domination law
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33. Example 3
Show that p↔q⇔(p ^ q) ∨(¬p ^¬q)by logic algebra
p↔q= (p→q)∧(q→p)
p↔q⇔(p→q)∧(q→p)
p→q⇔¬p ∨q
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35. Problem 1
Show that ￢(p ∨(￢p ∧q)) and ￢p ∧￢q are logically
equivalent by developing a series of logical equivalences.
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36. Problem 2
Show p ˅(q ^ r) ≡(p ˅q) ^(p ˅r) using Truth Table
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37. Problem 3
Check using a symbolic derivation whether
(p ∧¬q) →(p⊕r)⇔¬p ∨q∨¬r
(p ∧¬q) →(p⊕r)⇔
[Expand definition of →] ¬(p ∧¬q) ∨(p⊕r)
[Defn. of ⊕] ⇔ ¬(p ∧¬q) ∨((p∨r) ∧¬(p∧r))
[DeMorgan’sLaw]
⇔(¬p∨q)∨((p∨r) ∧¬(p∧r))
⇔[associative law] cont.
p→q⇔¬p ∨q
p⊕q⇔(p∨q)∧¬(p∧q)
p⊕q⇔(p∧¬q)∨(q∧¬p)
¬(p∧q) ⇔¬p ∨¬q
¬(p∨q) ⇔¬p ∧¬q
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38. Problem 3 Continued...
(¬p ∨q) ∨((p∨r) ∧¬(p∧r))⇔[∨commutes]
⇔(q∨¬p)∨((p∨r) ∧¬(p∧r))[∨associative]
⇔q∨(¬p∨((p∨r) ∧¬(p∧r))) [distrib. ∨over ∧]
⇔q∨(((¬p∨(p∨r)) ∧(¬p∨¬(p∧r)))
⇔q∨(((¬p∨p) ∨r) ∧(¬p∨¬(p∧r))) [assoc.]
[trivial taut.] ⇔q∨((T∨r) ∧(¬p∨¬(p∧r)))
[domination]⇔q∨(T∧(¬p∨¬(p∧r)))
[identity] ⇔q∨(¬p∨¬(p∧r))⇔cont.
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39. Problem 3 Continued...
q∨(¬p∨¬(p∧r))
[DeMorgan’s] ⇔q∨(¬p∨(¬p∨¬r))
[Assoc.] ⇔ q∨((¬p∨¬p) ∨¬r)
[Idempotent] ⇔ q∨(¬p∨¬r)
[Assoc.] ⇔ (q∨¬p) ∨¬r
[Commut.] ⇔ ¬p ∨q∨¬r
(Which was to be shown.)
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40. Problem 4: Biconditionaloperator
p↔q⇔(p→q)∧(q→p)
p q p ↔ q (p→q) (q→p)(p→q)∧(q→p)
F F T T T T
F T F T F F
T F F F T F
T T T T T T
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Try p↔q ⇔¬(p⊕q)

41. Summary of Lecture: Propositional Logic
Atomic propositions: p, q, r, …
Boolean operators:¬∧∨⊕→↔
Compound propositions: s:≡(p∧¬q) ∨r
Equivalences:p∧¬q ⇔¬(p →q)
Proving equivalences using:
Truth tables
Symbolic derivations. p⇔q ⇔r …
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42. Supplementary Material
Kimberly BrehmChannel
Discrete Math 1.2.1 -Translating Propositional Logic Statements
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