IST 4
Information and Logic
Claude Shannon April 30, 1916 - 2001
mon tue wed thr fri
28 M1 1
4 M1
11 1 2 M2
18 2
25 M2
2 3
9 3 4
16 4 5
23
30 5
x= hw#x out x= hw#x due
Mx= MQx out Mx= MQx due
midterms
oh oh
oh oh
oh oh
oh
oh
oh oh
oh
oh oh oh
oh
oh oh
oh = office hours oh
T
= today
T oh
Boolean Proofs: Axioms + Fun
A1-A4
L1 T1
T2 T3
L2
T4 T0
Proof
Dependency
Syllogism to Algebra George Boole, 1847
George Boole 1815 –1864
George Boole Early Days
George Boole 1815-1864
Born in Lincoln, England an industrial town
His father was a shoemaker with a passion for mathematics and science
When George was 8 he surpassed his father’s
knowledge in mathematics By age 14 he was fluent in Latin, German, French, Italian and English… and algebra…
When his was 15 he had to go to work to support his family, he became a math teacher in the Wesleyan Methodist academy in Doncaster (40 miles…)
Lost his job after two years….
Lost two more teaching jobs…
When he was 20 he opened his own school in his hometown - Lincoln
source: wikipedia
George Boole 1815-1864
Born in Lincoln, England an industrial town
In 1841 (26) he published three papers in the newly established Cambridge Mathematical Journal (edited by DG).
In 1844 (29) he published “On a General Method of Analysis”; he considered it to be his best paper. This paper won the first (newly established) Gold medal for Mathematics awarded by Royal Society
In 1846 (31) he applied for a professor position in the newly established Queen’s College – 3 campuses in Ireland
In 1847 (32) ‘while waiting to hear from Ireland’, he published “The Mathematical Analysis of Logic”
George Boole Early Career
George Boole 1815-1864
Born in Lincoln, England an industrial town
In 1847 (32) ‘while waiting to hear from Ireland’, he published “The Mathematical Analysis of Logic”
In 1849 (34), his was offered a position of the first professor of mathematics at Queen’s college at Cork
He married Mary Everest (1832- 1916) in 1855 (23,40) and they had five daughters
Niece of George Everest (Mt. Everest…) led the expedition to map the Himalayas
8848 m 29,029 ft
George Boole Ireland
source: wikipedia
In 1864, died of pneumonia (49)
Born in Lincoln, England an industrial town
1849-1864, taught at
at Queen’s college in Cork
George Boole Geography
source: wikipedia
Cork, Ireland Source: www.flickr.com
1849-1855:
lived here until got married
Boole was never a student at a university…
only a professor…
Leibniz never held an academic position…
George Boole 1815-1864 Gottfried Leibniz
1646-1716
It is all about
Algorizmi
780-850AD Gottfried Leibniz
1646-1716
George Boole 1815 –1864
???
Binary
Boolean
Boolean Algebra
Boolean is
notBinary...
Two-valued Boolean Algebra
Boolean Algebra: set of elements B={0,1},
two binary operations OR and AND xy OR(x,y)
00 01 10 11
0 1 1 1
xy AND(x,y)
00 01 10 11
0 0 0 1
0 iff both x and y are 0 1 iff both x and y are 1 We proved it is a Boolean algebra
Four-valued Boolean Algebra
Two-valued Boolean Algebra:
set of elements B={0,1},
two binary operations OR and AND Four-valued Boolean Algebra:
set of elements
??
two binary operations
??
and??
Elements:
0-1 vectors:
(10)
(00) (01) (11) operations?
+
Elements are: (00), (11), (10), (01)
00 01 10 11 00
01 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
+
Elements are: (00), (11), (10), (01)
00 01 10 11 00
01 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
Bitwise OR
+
Elements are: (00), (11), (10), (01)
00 01 10 11 00
01 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
Bitwise OR
00 01 xy OR(x,y)10 11
0 1 1 1
Elements are: (00), (11), (10), (01)
00 01 10 11 00
00 00 00
00 01 00
01
00 00 00 01
10 10 10 11 00
01 10 11
Elements are: (00), (11), (10), (01)
00 01 10 11 00
00 00 00
00 01 00
01
00 00 00 01
10 10 10 11 00
01 10 11
Bitwise AND
Elements are: (00), (11), (10), (01)
00 01 10 11 00
00 00 00
00 01 00
01
00 00 00 01
10 10 10 11 00
01 10 11
Bitwise AND
00 01 xy AND(x,y)10 11
0 0 0 1
Elements:
Operations:
(10)
(00) (01) (11)
Bitwise AND Bitwise OR
Is it a Boolean algebra?
0 1
Bitwise Complement
+
00 01 10 11 0001 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
00 01 10 11 00
00 00 00
00 01 00
01
00 00 00 01
10 10 10 11 00
01 10 11
+
00 01 10 11 0001 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
00 01 10 11 00
00 00 00
00 01 00
01
00 00 00 01
10 10 10 11 00
01 10 11
+
00 01 10 11 0001 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
00 01 10 11 00
00 00 00
00 01 00
01
00 00 00 01
10 10 10 11 00
01 10 11
+
00 01 10 11 0001 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
00 01 10 11 00
00 00 00
00 01 00
01
00 00 00 01
10 10 10 11 00
01 10 11
We can prove it!
Elements:
Operations:
(10)
(00) (01) (11)
Bitwise AND Bitwise OR
It is a Boolean algebra!
0 1
Bitwise Complement
In general: 0-1 vectors a Boolean algebra?
Elements:
Operations:
(001)
(000) (010) (111)
Bitwise AND Bitwise OR
0 1
Bitwise Complement
True for a two-valued Boolean algebra True for any Boolean algebra with
0-1 vectors and bitwise OR and AND??
0-1 vectors Boolean algebra
(011) (100) (101) (110)
n=3
n=1
arbitrary finite n
|s|
Idea 1:
Any identity that is true for 0-1 is true for 0-1 vectors with bitwise OR/AND
Idea 2:
Axioms are identities
True for 0-1 à true for 0-1 vectors
The 0-1 Theorem
0-1 Theorem:
An identity is true for 0-1 vectors with bitwise OR and AND if and only if
it is true for two valued (0-1) with bitwise OR and AND
Proof: The easy direction
Assume an identity true for 0-1 vectors True for 0-1
The 0-1 Theorem
0-1 Theorem:
Proof: The non-obvious direction
Assume an identity true for 0-1
Need to prove true for any 0-1 vectors
An identity is true for 0-1 vectors with bitwise OR and AND if and only if
it is true for two valued (0-1) with bitwise OR and AND
Example:
Theorem 2:
Proof:
0 + 0 ⋅ 0 = 0 0 + 0 ⋅1 = 0 1 + 1⋅ 0 = 1 1 + 1⋅1 = 1
The identity is true for 0-1
Need to prove it for 0-1 vectors
Theorem 2:
Proof (for 0-1 vectors):
If an identity is not true in general;
then there is an assignment of
elements that violates the equality
Hence, there must be a position in the binary vector that is violated
There exists a 0-1 assignment to the identity that violates the equality, CONTRADICTION!!
By contradiction Assume true for all 0-1 assignments
and not true for some other assignment
Example:
Recap: The 0-1 Theorem
An identity is true for 0-1 vectors with bitwise OR/AND
if and only if it is true for two valued (0-1) with OR/AND Proof: The easy direction
• Assume an identity true for 0-1 vectors
• 0-1 is a special case: True for 0-1 The non-obvious direction Q
• Assume an identity true for 0-1
• Need to prove true for 0-1 vectors
• Assume there exists a general ‘identity violating’ assignment
• Show that there is a 0-1 ‘identity violating’ assignment CONTRADICTION!!
Operations:
Bitwise AND
Bitwise OR Bitwise Complement
It is a Boolean algebra for any finite n
0-1 vectors of length n, there are 2
nvectors Elements:
Application of the the 0-1 Theorem
It is a Boolean algebra with for n=1
By the 0-1 Theorem:
Boolean Algebra
Two-value? Or not?
Prove or Disprove
At least one of the following is true:
Prove or Disprove
At least one of the following is true:
xy OR(x,y)
00 01 10 11
0 1 1 1
True
for two-valued Boolean algebra
Is it true for 0-1 vectors Boolean algebras?
+
00 01 10 11 0001 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
Prove or Disprove
At least one of the following is true:
Is it true for 0-1 vectors Boolean algebras?
NO
If
Then or
Claim is NOT TRUE in general!
The 0-1 Theorem:
True only for identities!!!
+
00 01 10 11 0001 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
Examples ‘other’ Boolean Algebras
• 0-1 vectors
• Arithmetic Boolean algebras
• Algebra of subsets union / intersection
next next
Boolean algebra
Boolean integers
|s|Euclid, 300BC
405 = 3x3x3x3x5
297 = 3x3x3x11 27 = 3x3x3
884 = 2x2x13x17
68 = 2 x 2 x 17 612 = 2x2x3x3x17
Greatest Common Divisor
Application: Simplifying fractions
Euclid, 300BC
405 = 3x3x3x3x5
297 = 3x3x3x11 4455 = 3x3x3x3x5x11
884 = 2x2x13x17
7956 = 2x2x3x3x13x17 612 = 2x2x3x3x17
Least Common Multiple
Application: Adding fractions
Euclid, 300BC
297x405 = 120,285
gcd(297,405) = 27 lcm(297,405) = 4455
Least Common Multiple Greatest Common Divisor
27x4455 = 120,285
gcd(a,b) x lcm(a,b) = a x b
405 = 3x3x3x3x5 297 = 3x3x3x11 405 = 3x3x3x3x5
297 = 3x3x3x11
Arithmetic Boolean Algebra
lcm = lowest common multiple gcd = greatest common divisor
The set of elements: {1,2,3,6}
The operations: lcm and gcd
1 is Boolean 0 6 is Boolean 1
George Boole 1815 –1864
What is the complement?
1 2
3 6
The set of elements: {1,2,3,6}
The operations: lcm and gcd
1 is Boolean 0 6 is Boolean 1
The set of elements: {1,2,3,6}
The operations: lcm and gcd
1 is Boolean 0 6 is Boolean 1
For which n does it work?
Elements:
Set of all the divisors of an integer n
complement? The
The set of elements: {1,2,4,8}
The operations: lcm and gcd
1 is Boolean 0 8 is Boolean 1
1 2
4 8
The set of elements: {1,2,3,6}
The operations: lcm and gcd
1 is Boolean 0 6 is Boolean 1
For which n does it work?
Elements:
Set of all the divisors of an integer n
Prime factors appear at most once in n
Bunitskiy Algebra 1899
Elements:
The set of divisors of a Boolean integer {1,2,3,5,6,10,15,30}
The operations: lcm and gcd
The 0 and 1 elements: 1 and 30
Boolean Integers 2 x 3 x 5 = 30
2 x 3 x 7 = 42
Every prime in the prime factorization is a power of one (square-free integer)
Euclid, 300BC
George Boole 1815 –1864
Is Bunitskiy Algebra a Boolean Algebra?
6 = 3x2 3 = 3
2 = 2 1 = 1
11
00 10 01
LCM GCD
11 01
Bitwise OR Bitwise AND
YES
+
00 01 10 11 0001 10 11
01 01 11 11
10 11 11 11 10 11 11 11 00
01 10 11
00 01 10 11 00
00 00 00
00 01 00
01
00 00 00 01
10 10 10 11 00
01 10 11
1 2 3 6
1
1 2 3 6
1 2 3 6
2 2 6 6
3 6 3 6
6 6 6 6 1 2 3 6
1 1 3 3
1 2 1 2
1 1 1 1
1 2 3 6
2 3 6
lcm gcd
Bunitskiy algebra is isomorphic to the algebra of 0-1 vectors,
it is a Boolean algebra!
syntax the same – different semantics
Boolean algebra
subsets of a set
|s|Algebra of Subsets
S is the set of all points
Elements: all possible subsets of a set S
+ is union of sets:
is intersection of sets
How many elements? 2|s|
Example:
Elements:
S =
Operations: union and intersection
Complement:
Elements:
Corresponding 0-1 vectors:
(10)
(00) (11) (01)
S =
Algebra of subsets is isomorphic to the algebra of 0-1 vectors,
it is a Boolean algebra!
Is the algebra of subsets
a Boolean Algebra? YES
syntax the same – different semantics
01 11 00 10
00
10
01
11
Elements are: (00), (11), (10), (01)
01 11 00 10
00
10
01
11
Elements are: (00), (11), (10), (01)
00 10 01 11
10
11 01
10 11 11
11
11 11 11
11 01
Union
Bitwise OR
Elements are: (00), (11), (10), (01)
01 11 00 10
00
10
01
11
00 00 00 00
00
00
00
10 00 10
00
10 01 11
01 01
Intersection
Bitwise AND
Boolean algebra
an amazing theorem
Examples of Boolean Algebras
• 0-1 (two valued) Boolean algebra OR / AND
• 0-1 vectors
bitwise OR / bitwise AND
• Arithmetic Boolean algebras lcm / gcd
• Algebra of subsets union / intersection
Size 2
kThey are isomorphic!
Representation Theorem (Stone 1936):
Every finite Boolean algebra is isomorphic to a Boolean algebra of 0-1 vectors
Algebra 1 Algebra 2
elements elements
operations operations
Representation Theorem (Stone 1936):
Every finite Boolean algebra is isomorphic to a Boolean algebra with elements being bit vectors of finite length with bitwise operations OR and AND
Two Boolean algebras with m elements are isomorphic
Provides intuition beyond the axioms:
We can ‘naturally’ reason about results in Boolean algebra Every Boolean algebra has 2 elements k
Marshall Stone
Marshall Stone 1903-1989
Proved in 1936
90AB = years After Boole
The Boolean Syntax invented in 1847 has a unique representative semantic!!!
Harlan Fiske Stone
12th Chief Justice of the US 1941-1946
Marshall entered Harvard in 1919 intending to continue his studies at Harvard law school; fell in love with Mathematics, and the rest is history…
Marshall had a passion for travel. He began traveling when he was young and was
on a trip to India when he died....
Quiz #3 – 10min
What is the complement of
Use the following DeMorgan Laws:
And the axioms:
Show
your work andjustify
every step!Solution in this form:
What is the complement of
DeMorgan Laws:
And the axioms:
DeMorgan
DeMorgan
A4 A4 A3
A2 A3
A1
One week!
2 symbol adder c
s
d1 d2
c
parity
majority
Need to provide a complete proof No duality arguments
For (c): Sum of products (no need to expand to DNF) See the solution to Quiz #3
1 2
3 4
Prove 1 Prove 2
multiply