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CMPUT 272
Formal Systems
Logic in CS
I. E. Leonard
University of Alberta
http://www.cs.ualberta.ca/~isaac/cmput272/f03
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Today
Sets (5.1, 5.2, 5.3)
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Sets (chapter 5)
QuickTi me™ and a TIFF (U ncompressed) decompressor are needed to see this picture.
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Sneak-preview : Sets
What is a set?
A collection of elements:
Order is irrelevant
No repetitions
Can be infinite
Can be empty
Examples:
{Angela, Belinda, Jean}
{0,1,2,3,…}
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Operations on sets
S is a set
Membership:
xS
x is an element of S
Angela{Angela, Belinda, Jean}
Subset:
S1  S
Set S1 is a subset of set S
All elements of S1 are elements of S
{Angela,Belinda}  {Angela, Belinda, Jean}
Proper subset S1S
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Operations on sets
S, S1 are sets
Equality:
S = S1
iff they have the same elements
Difference:
S \ S1
is a set of all elements that belong to S but
NOT to S1
{Angela, Belinda, Jean} \ {Angela,Dana} =
{Belinda, Jean}
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Operations on sets
S, S1 are sets
Intersection:
S  S1
is a set of all elements that belong to both
{Angela, Belinda, Jean}  {Angela,Dana} =
{Angela}
Union:
S  S1
is a set of all elements that belong to either
{Angela, Belinda, Jean}  {Angela,Dana} =
{Angela,Belinda,Jean,Dana}
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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More notation
In mathematics sets are often specified
with a predicate and an enveloping set
as follows:
S={xA | P(x)}
S is the set of all elements of A that satisfy
predicate P
Example:
Q={xR | a,bZ b0 & x=a/b}
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Set Equality
Two sets are equal iff they have the
same elements
Theorem: for any sets A and B, A=B iff
AB & BA
Proof
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Subset and Membership
Book example 5.1.5
2{1,2,3}
{2}{1,2,3}
2{1,2,3}
{2}{1,2,3}
{2}{{1},{2}}
{2}{{1},{2}}
?
?
?
?
?
?
How about set A such that {2} is a
subset of it and A is an element of it?
A={1,2,{1},{2}}
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Universal Set
If we are dealing with sets which are all
subsets of a larger set U then we call it
a universal set U
All of your sets will be subsets of U
When does such a U exist?
Always, for we can set U to the union of all
sets involved
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Complement
So if I am dealing with set A which is a
subset of the universal set U then:
I can define complement of A:
AC=U\A
That is the set of all elements (of U)
that are not in A
Often “of U” is dropped and people say
that AC is the set of everything that is
not in A
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Quick questions
What is the complement of U?
UC = Ø
What set has U as its complement?
ØC=U
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Sets & Predicate Logic
All of the set operations and relations
above can be defined in terms of
Boolean connectives:
AB={x | xA v xB}
AB={x | xA & xB}
A\B={x | xA & not xB}
AC={x | not xA}
A=B iff x (xA  xB)
AB iff x (xA  xB)
AB iff (x (xA  xB)) & not (A=B)
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Questions
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Symmetric Difference
Set C is the symmetric difference of sets
A and B iff every element of C belongs
to A or B but not both
ABC [C=A  B  a (aC 
(aA xor aB))]
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Examples
A={1,2}, B={2,3}
A  B={1,3}
A={Clinton,Reagan}, B={Gorbachov,Bregnev}
A  B={Clinton,Reagan,Gorbachov,Bregnev}
A={CMPUT272 students}, B={CMPUT272 students}
A  B = {}
In general: A  A = {}
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Exercise 2
Intersection of two sets is contained in their union:
AB [ (A  B)  (A  B) ]
Proof:
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Exercise 3
Union is commutative
AB [ A  B = B  A ]
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Exercise 4
Intersection is commutative
AB [ A  B = B  A ]
Proof: very similar to the one we just
did. Try it yourself.
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Exercise 5
Intersection distributes over union:
ABC [ A  (B  C) =(A  B)  (A  C) ]
There is an analogy between logical operations v and &
and arithmetic operations:
v feels like +
& feels like *
So A & (B v C) = A & B v A & C
[just like A*(B+C) = A*B + A*C]
How about A+(B*C) --- is it (A+B)*(A+C)?
NO
So what about A v (B & C) = (A v B) & (A v C)?
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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The Analogy
The analogy is incomplete:
Arithmetic:
Logic:
A+(B*C)  (A+B)*(A+C)
A v B&C = (A v B) & (A v C)
Proof of the latter:
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Exercise 6
In Exercise #5 we proved:
ABC [ A  (B  C) =(A  B)  (A  C) ]
using the fact that A&(B v C)=A&B v A&C
Given the statement just proved
A v B&C = (A v B) & (A v C)
what can we now prove in terms of sets?
Union distributes over intersection:
ABC [ A  (B  C) =(A  B)  (A  C) ]
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Proof of Exercise 6
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Questions
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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More Identities
See Theorem 5.2.2 in the book
The proofs can often be done using:
the logical definitions of set operations
logical identities we have proven before
I recommend doing some/all of them as
an exercise
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Boolean Algebra
Are these similarities between set
identities and logical identities
incidental?
It turns out that both systems are
examples of a more general construct
called Boolean algebra
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Boolean Algebra
Boolean algebra is a set S together with two operations: +
and * defined on S such that the following identities hold: For
all a,b,c in S
a+b is in S and a*b is in S (closure laws)
a+b=b+a and a*b = b*a (commutative laws)
(a+b)+c=a+(b+c) and (a*b)*c = a*(b*c) (associative laws)
a*(b+c)=a*b+a*c and a+(b*c) = (a+b)*(a+c) (distributive laws)
There exist distinct 0,1 in S:
a+0=a (additive identity)
a*1=a (multiplicative identity)
For each a from S there exists a complement a’ such that:
a+a’=1 (complement laws)
a*a’=0
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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Boolean Algebra - Logic - Sets
S
+
*
a+b=b+a
a*b=b*a
(a+b)+c=a+(b+c)
(a*b)*c=a*(b*c)
a*(b+c)=(a*b)+(a*c)
a+(b*c)=(a+b)*(a+c)
0
1
a+0=a
a*1=a
complement (a’)
a+a’=1
a*a’=0
Oct 30, 2003
{true,false}
v
&
avb=bva
a&b=b&a
(avb)vc=av(bvc)
(a&b)&c=a&(b&c)
a&(bvc)=(a&b)v(a&c)
av(b&c)=(avb) & (avc)
false
true
a v false = a
a & true = a
~a
a v ~a = true
a & ~a = false
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
P(U) (i.e., all sets)


ab=ba
ab=ba
(ab)c=a(bc)
(ab)c=a(bc)
a(bc)=(ab)(ac)
a(bc)=(ab)(ac)
Ø
U
aØ=a
aU=a
aC
a  aC = U
a  aC = Ø
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Notes
How about numbers with addition and
multiplication? Do they form a Boolean
algebra?
Not quite
Why are these 5 groups of identities
(page 266 in the book) important?
Other standard identities of any Boolean
algebra (including Boolean logic and sets
can be derived from them)
Oct 30, 2003
© Vadim Bulitko : CMPUT 272, Fall 2003, UofA
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