Logic
1
Outline
•  Basics
•  TruthTablesofOperators
•  LogicFormulae
–  Evalua:on;TruthTables
•  Equivalence
–  Laws;Proving
•  Inferences
CSE 215: Logic
2
Logic
•  AllaboutwhatisTRUEandwhatisFALSE.
•  Proposi:on/Statements:
–  “Sentences”thatareeitherTorF,andneverboth.
–  E.g.,(a)TodayisSunday.(b)IftodayisSunday,
thenitwillraintoday.
–  NOTproposi:ons:(a)“x+y=5”sincexandyare
unknown,(b)“Howfaristhenexttown?”
CSE 215: Logic
3
LogicVariablesandOperators
LogicVariables:
•  Ofunknownvalue(buts:llTorF).
•  p,q,r,tarecommonlyusedaslogicvariables.
LogicOperators
1.  NOT(~) E.g.,~p
2.  AND(∧) E.g.,p∧q
3.  OR (∨) E.g.,p∨q
4.  IMPLIES(→) E.g.,p→q
CSE 215: Logic
4
CompoundStatements
Formedofvariablesandlogicoperators.
E.g.,
•  ~p
•  ~(pV~q)
CSE 215: Logic
5
Outline
•  Basics
•  TruthTablesofOperators
•  LogicFormulae
–  Evalua:on;TruthTables
•  Equivalence
–  Laws;Proving
•  Inferences
CSE 215: Logic
6
TruthTablefor~
Whenis(~p)TRUE?
WhenpisFALSE.
CSE 215: Logic
7
TruthTablefor∧
Whenis(p∧q)TRUE?
Whenbothpandqaretrue.
CSE 215: Logic
8
TruthTablefor∨
Whenis(p∨q)true?
Wheneitheroneistrue.
CSE 215: Logic
9
TruthTablefor“→”
Whenis(p→q)true?
= “ifpthenq”
Always,unlesspistrueand
qisfalse.
CSE 215: Logic
10
p→q
•  Whyis(p→q)alwaystruewhenpisF.
•  Letp=“Skyisblue”
•  Letq=“itsday:me”.
p→qis“If(skyisblue)then(itsday:me).
•  Consideranight.Here,pisFandqisF.
•  Or,consideracloudyday.Here,pisF,qisT.
•  Shouldp→qhold,atnightsorcloudydays?
•  Yes,p→qisavalidclaim(evenatnightor
cloudydays).
CSE 215: Logic
11
Outline
•  Basics
•  TruthTablesofOperators
•  LogicFormulae
–  Evalua:on;TruthTables
•  Equivalence
–  Laws;Proving
•  Inferences
CSE 215: Logic
12
LogicFormulae
(~(~p ∧ ~q)) ∧ (~(p ∧ q))
•  When is the above true?
•  Can the above be simplified?
Construct its truth table.
Parentheses used to avoid ambiguity.
CSE 215: Logic
13
TruthTableforLogicFormulae
(~(~p ∧ ~q)) ∧ (~(p ∧ q))
•  Evaluate the above for all combination of values of
variables (in this case, p and q).
•  Thus,
p
q
(~(~p ∧ ~q)) ∧ (~(p ∧ q))
T
T
?
T
F
?
F
T
?
F
F
?
CSE 215: Logic
14
(1)Enumera:ngallcombina:ons
•  ConsideraformulacomposedofNvariables.
•  Howdoyouenumerateallcombina:onof
values?
–  Binarymethod:
–  Forloop:
–  Recursive:
Thereare2Ncombina:ons.
CSE 215: Logic
15
(2)Evalua:ngaFormula
(~(~p ∧ ~q)) ∧ (~(p ∧ q))
•  What is its value, when p is TRUE and q is FALSE?
è(~(~T ∧ ~F)) ∧ (~(T ∧ F))
= (~(F ∧T)) ∧ (~F)
= (~F) ∧ (T)
= T.
CSE 215: Logic
16
TruthTableforFormula
(~(~p ∧ ~q)) ∧ (~(p ∧ q))
p
q
(~(~p ∧ ~q)) ∧ (~(p ∧ q))
T
T
F
T
F
T
F
T
T
F
F
F
CSE 215: Logic
17
Outline
•  Basics
•  TruthTablesofOperators
•  LogicFormulae
–  Evalua:on;TruthTables
•  Equivalence
–  Laws;Proving
•  Inferences
CSE 215: Logic
18
Equivalence
•  Twoformulaeareequivalent,iftheyhavethe
sametruthtable.
•  E.g.,(p→q)and(~p∨q)areequivalent.
•  Nota:on:X≡Y,whereXandYareformulae. CSE 215: Logic
19
EquivalenceLaws–I
1.  ∧ and ∨ are commutative.
§  p ∧ q ≡ q ∧ p
§  p ∨ q ≡ q ∨ p
2.  ∧ and ∨ are associative.
§  p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r
§  p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r
3.  ∧ and ∨ are distributive.
§  p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
§  p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
CSE 215: Logic
20
EquivalenceLaws–II
1.  De Morgan’s Laws.
§  ~(p ∧ q) ≡ ~p ∨ ~q
§  ~(p ∨ q) ≡ ~p ∧ ~q
CSE 215: Logic
21
ProvingEquivalence
1.  Showthetruthtablesaresame.(Obvious)
2.  Proveusinglaws.E.g.,(p ∨ ~q) ∧ q ≡ p ∧ q
Proof:
(p ∨ ~q) ∧ q ≡ q ∧ (p ∨ ~q)
≡ (q ∧ p) ∨ (q ∧ ~q)
≡ (q ∧ p) ∨ F
≡ (q ∧ p)
≡p∧q
CSE 215: Logic
22
MoreFactsabout(p→q)
(p→q)isequivalenttoallofthebelow:
•  (~q→~p)
•  (~pVq)
•  ~(p∧~q)
Easytoconfirmusingtruthtables.
Toseeintui:vely,try:
p=Alexisafather
q=Alexisamale
Here,(p→q)holds.Dotheothersmeanthesame?
CSE 215: Logic
23
OtherConcepts
•  Tautology:Aformulathatisalwaystrue.
–  E.g.,(p∨~p),(p∧q)∨(~p∨~q).
•  Contradic:on:Aformulathatisalwaysfalse.
•  Bi-condi:onal(p↔q)
–  Equivalentto:(p→q)∧(q→p).
–  (p↔q)istrueifandonlyifpandqhavethesame
values(check).
CSE 215: Logic
24
NecessaryandSufficient
1.  “pissufficientforq”means(p→q).
2.  “pisnecessaryforq”means(~p→~q).Right?
abovealsoimpliesthat:
“pisnecessaryforq”means(q→p).Why?
Thus,p↔qmeanspisnecessaryandsufficientforq.
CSE 215: Logic
25
Ifandonlyif
•  (p→q)means“ifpthenq”OR“qistrueifpis
true”
•  Whatdoes“qistrueonlyifpistrue”mean?
–  pisnecessaryforq.
–  (~pà~q)
–  (qàp)
CSE 215: Logic
26
Interesting Implication Example
•  Prosecutor:
–  "If the defendant is guilty, then he had an accomplice."
•  Defense Attorney:
–  "That's not true!!"
•  What did the defense attorney just claim??
~(p → q) ≡ ~(~p ∨ q) ≡ ~~(p∧~q) ≡ p ∧ ~q
CSE 215: Logic
27
Outline
•  Basics
•  TruthTablesofOperators
•  LogicFormulae
–  Evalua:on;TruthTables
•  Equivalence
–  Laws;Proving
•  Inferences
CSE 215: Logic
28
Arguments;MakingInferences
• 
• 
• 
• 
Wedon’tknowthevaluesofpandq.
But,weknow(p∨q)and~paretrue.
Canwesayanythingaboutp,q?
Yes.
–  pisfalse.
–  qistrue.
•  How?
CSE 215: Logic
29
Inferences
Given(astrue)formulaeXandY.Canweinfer
(deduceastrue)Z?Wrilenas:
X,Y
Z
When:
•  Wecanshowthat(X∧Y)→Zisalwaystrue
(i.e.,isaTautology).How?Truthtable!
CSE 215: Logic
30
InferenceExample
Given(p→(qV~r))and(q→(p∧r))
Canweinfer(p→r)?
CSE 215: Logic
31
CommonInferenceRules
CSE 215: Logic
32
InferenceExample–I
1.  IfIwasreadinginthekitchen,thenmyglasseswouldbeonthe
kitchentable.
2.  Ifglassesareonthekitchentable,thenIsawthematbreakfast.
3.  Ididnotseemyglassesatbreakfast.
4.  Iwasreadingthenewspaperinthelivingroomorthekitchen.
5.  IfIwasreadingthenewspaperinthelivingroom,thenmyglasses
areonthecoffeetable.
Wherearetheglasses?
6.  Glassesnotonkitchentable [From2,3,ModusTollens].
7.  Iwasnotreadinginthekitchen [From1,6,ModusTollens]
8.  Readinginthelivingroom [From4,7,Elimina:on]
9.  Glassesonthecoffeetable. [From5,8,ModusPonens]
33
InferenceExample–II
Knights and Knaves: knights always tell the truth and knaves
always lie
A says: B is a knight.
B says: A and I are of opposite type.
Suppose A is a knight.
∴ What A says is true.
by definition of knight
∴ B is also a knight.
That’s what A said.
∴ What B says is true.
by definition of knight
∴ A and B are of opposite types. That’s what B said.
∴ We have arrived at the following contradiction: A and B are
both knights and A and B are of opposite type.
∴ The supposition is false.
by the contradiction rule
∴ A is not a knight.
negation of supposition
∴ A is a knave.
since A is not a knight.
∴ What A says is false.
by definition of knave
∴ B is not a knight.
~(what A said).
∴ B is also a knave.
by elimination
34
Useofcontradic:onRule
• 
• 
• 
• 
• 
• 
• 
• 
Givensomefacts.
P(supposi:on)
Q(inferred/givenfromabove).
~Q(inferred/givenfromabove).
False(fromQ∧~Q).
Thus,wehaveshown(Facts∧P)àFalse.
Thus,~(Facts∧P).
Thus,~p.
CSE 215: Logic
35
UseofTransi:vity
• 
• 
• 
• 
• 
• 
• 
• 
GivenXandY,weinferZ.
Now,fromYandZ,weinferW.
CanwesaythatWcanbeinferredfromXandY?
(X∧Y)àZ.
(X∧Y)àY. (trivial)
Thus,(X∧Y)à(Y∧Z)
Also,(Y∧Z)àW.
Thus,(X∧Y)àW(bytransi:vity)
CSE 215: Logic
36