Boolean Algebras

(ブール代数)

Discrete Mathematics I

12th lecture, December 12, 2025

https://www.sw.it.aoyama.ac.jp/2025/Math1/lecture12.html

Martin J. Dürst

© 2005-25 Martin J. Dürst Aoyama Gakuin University

Today's Schedule

• Remaining schedule, final exam
• Summary and homework for last lecture
• Algebraic structures related to groups
• Boolean algebra
• Examples of Boolean algebras
• Bitstrings and Bitwise Operations
• Structure of Boolean algebras

Remaining Schedule

• December 12 (today): this lecture
• December 19: 13th lecture (Modular Arithmetic)
• January 9: 14th lecture (Mathematical Induction and Other Proof Methods)
• January 16: 15th lecture (Remainder, Review)
  
• [January 23: makeup lectures]
• January 30, 11:10-12:35 (85 min): Term Final Exam

Final Exam・期末試験

Coverage:

Complete contents of lecture and handouts

Past exams:

2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017, 2018, 2019, (2020 online) 2021, 2022, 2023, 2024

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Important Points about Final Exam

• Most problems are in Japanese and English
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• Be able to do calculations (base conversions, truth tables,...) speedily
• Combine and apply knowledge from different lectures
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Questions about Final Exam

(for additional questions, use the Q&A Forum)

Summary of Last Lecture

• Algebraic structures are defined as sets with operations and axioms
• Groups are the best known example of an algebraic structure, with a rich theory
• Groups have one associtive binary operation, an identity element, and inverse elements
• Examples of groups are (ℤ, +), (ℝ-{0}, ·), (ℝ⁺, ·), and symmetric groups (permutations of n elements with composition)
• Finite groups are represented by Cayley tables, and compared using the concept of isomorphism (one-to-one mapping conserving structure)

Homework Due November 6, Problem 1

Prove/check the following laws using truth tables (important: always add a sentence stating the conclusion):

1. Reductio ad absurdum: A→¬A = ¬A
   

   A	¬A	A→¬A
   F	T	T
   T	F	F

   Because the columns for ¬A and A→¬A are the same, this law is correct.

2. The associative law for conjunction: (A∧B) ∧ C = A ∧ (B∧C)
   

   A	B	C	A∧B	(A∧B) ∧ C	B∧C	A ∧ (B∧C)
   F	F	F	F	F	F	F
   F	F	T	F	T	T	T
   F	T	F	T	T	T	T
   F	T	T	T	T	T	T
   T	F	F	T	T	F	T
   T	F	T	T	T	T	T
   T	T	F	T	T	T	T
   T	T	T	T	T	T	T

   Because the columns for (A∧B) ∧ C and A ∧ (B∧C) are the same, this law is correct.

3. Contraposition: A→B = ¬B→¬A
   

   A	B	A→B	¬B	¬A	¬B→¬A	(A→B) ↔(¬B→¬A)
   F	F	T	T	T	T	T
   F	T	T	F	T	T	T
   T	F	F	T	F	F	T
   T	T	T	F	F	T	T

   Because the column for (A→B) ↔ (¬B→¬A) is all Ts, this law is correct.

4. One of De Morgan's laws: ¬(A∧B) = ¬A ∨ ¬B
   

   A	B	A∧B	¬(A∧B)	¬A	¬B	¬A ∨ ¬B
   F	F	F	T	T	T	T
   F	T	F	T	T	F	T
   T	F	F	T	F	T	T
   T	T	T	F	F	F	F

   Because the columns for ¬(A∧B) and ¬A ∨ ¬B are the same, this law is correct.

Homework Due November 6, Problem 2

Prove transitivity of implication (((A→B) ∧ (B→C)) ⇒ (A→C)) by formula manipulation.

For each simplification step, give the name of the law you used.

Hint: Show that ((A→B) ∧ (B→C)) → (A→C) is a tautology by simplifying it to T.

Details: Multiple parallel applications of the same law can be done in one step. Multiple sequential applications of the same law must be separated into different steps. Application of commutativity and associativity laws can be combined in one step.

((A→B) ∧ (B→C)) → (A→C) [removing implication]
= ¬((A→B) ∧ (B→C))∨ (A→C) [removing implication, 3 times]
= ¬((¬A∨B) ∧ (¬B∨C)) ∨ (¬A∨C) [DeMorgan]
= (¬(¬A∨B) ∨ ¬(¬B∨C)) ∨ (¬A∨C) [DeMorgan, twice]
= ((¬¬A∧¬B) ∨ (¬¬B∧¬C)) ∨ (¬A∨C) [double negation, twice]
= ((A∧¬B) ∨ (B∧¬C)) ∨ (¬A∨C) [associative/commutative laws]
= (A∧¬B) ∨ ¬A ∨ (B∧¬C) ∨ C [distributive law, twice]
= (A∨¬A)∧(¬B∨¬A) ∨ (B∨C) ∧ (¬C∨C) [law of excluded middle, twice]
= T ∧ (¬B∨¬A) ∨ (B∨C) ∧ T [property of true, twice]
= (¬B∨¬A) ∨ (B∨C) [associative/commutative laws]
= ¬A ∨ B∨¬B ∨ C [law of excluded middle]
= ¬A ∨ T ∨ C [property of true, twice]
= T ∴

By reducing it to a tautology, we showed that implication is transitive.

Homework Comment

This homework is a good example of a general strategy to simplify formulæ:

• Remove equivalences and implications
• Push negation inside using DeMorgan's laws
• Remove double negations
• Regroup formula using associative/commutative/distributive laws
• Reduce opposites to T/F with law of excluded middle/law of noncontradiction
• Remove T/F or variable terms with properties of T/F

Last Lecture's Homework 1:
Symmetric Group S₃

Create a Cayley table of the symmetric group S₃ (3!=6 elements). Use the notation introduced in this lecture. Use lexical order for the permutations. Do not use any abbreviations.

Last Lecture's Homerwork 2:
Non-Isomorphic Groups of Size 4

If we define isomorphic groups as being "the same", there are two different groups of size 4. Give an example of each group as a Cayley table. Hint: Check all the conditions (axioms) for a group. There will be a deduction if you use the same elements of the group as another student.

Solution 1 (cyclic group of order 4)

Interesting examples: Addition modulo 4 (below), 10/20/30/40 and addition modulo 10,..., rotation by multiples of 90°, {1, 2, 3, 4} and multiplication modulo 5 (see next lecture)

Solution 2 (Klein four-group):

Interesting examples: bitstrings of length 2 with XOR (below), reflections and rotations by 180°

Algebraic Structures Related to Groups

• Abelian group: Group where the operation is also commutative
  Examples:
  • All groups of size 3 or 4
  • (ℤ; +): Integers with addition
  • (ℝ-{0}; ·) Reals (excluding 0) with multiplication
• Semigroup: Group without unit element and without inverse elements

More Algebraic Structures

• Ring: Two operations, 'addition' and 'multiplication'; forms an Abelian group under addition and a semigroup under multiplication, and addition distributes over multiplication.
  Examples: ℝ, ring of polynomials
• Field: Same as ring, but commutative, and 'multiplication' has to have an inverse for all non-zero elements.
• Lattice: Two operations; both operations are associative and commutative, and respect absorption laws
• Boolean Algebra

Boolean Algebra

• A Boolean algebra is an algebraic structure
• A Boolean algebra has one set (B), two special elements (0, 1), and three operations (⫬, △, ▽)
• B = (B, 0, 1, ⫬, △, ▽) is a Boolean algebra if and only if it meets the following conditions:
  • 0 ∈ B, 1 ∈ B (1 is called unit element, 0 is called zero element)
  • ⫬ is an unary operation on B, △ and ▽ are binary operations on B
  • △ (called join) and ▽(called meet) are associative and commutative, and distribute over each other
  • ∀a∈B: a▽0 = a, a△1 = a (0 is the neutral element for ▽, 1 is the neutral element for △)
  • ∀a∈B: a▽⫬a = 1, a△⫬a = 0
  • Axioms (Huntington axioms (1904), see lecture 5):
    a▽0 = a [+D]; a▽b=b▽a [+D];
    a▽(b△c)=(a▽b)△(a▽c) [+D]; a▽⫬a=1 [+D]
• This definition (operations, axioms) can be simplified
• Boolean algebras define a (partial) order, with 1 at the top and 0 at the bottom (∀a∈B: 0≤a≤1)

Boolean Algebra Example 1:
Basic Logic

• B is the set {false, true}
• 0 is false (0)  
• 1 is true (1)  
• ⫬ is negation (logical not, ¬)
• △ is conjunction (logical and, ∧)
• ▽ is disjunction (logical or, ∨)
• The (partial) order relation uses false<true  , or a≤b ⇔ a→b
• Hasse diagram: BAbasic_logic.svg

Comments on Boolean Algebra

• Symbols 0, 1 (bold), ⫬, △, and ▽ are abstract and do not have any concrete meaning
• A Boolean algebra is a (partial) order and a lattice
• Boolean algebra is a general concept covering logical operations, set operations, some (partial) orders, ... (see examples)
• Because a Boolean algebra is a partial order, it can be shown with a Hasse diagram

Boolean Algebra Example 2:
A Powerset with Set Operations

• B is the powerset of a (finite) set U (B = P(U))
• 0 is the empty set ({})
• 1 is U (the universal set)
• ⫬ is the complementary set operation (^c)
• △ is set intersection (∩)
• ▽ is set union (∪)
• The (partial) order relation is ⊂ (subset)
• Hasse diagram for U={a, b, c}: BApowerset.svg

Bitwise Operations

• Inside a computer, information is represented as strings of bits (e.g. a=00111010, b=10011100).
• Bitwise operations are bitwise logic operations.
• Bitwise not: ¬ is applied to each bit (~ in C^*, e.g. ¬_bitwisea=11000101)
• Bitwise and: ∧ is applied to bits in the same position (& in C^*, e.g. a∧_bitwiseb=00011000)
• Bitwise or: ∨ is applied to bits in the same position (| in C^*, e.g. a∨_bitwiseb=10111110)
• Bitwise xor: XOR (⊕) is applied to bits in the same position (^ in C^*, e.g. a⊕_bitwiseb=10100110)

^*) and many other programming languages

Boolean Algebra Example 3:
Bitstrings and Bitwise Operations

• B is the set of all the bit strings of length n (e.g. for n=2, B={00, 01, 10, 11})
• 0 is the bit string of length n with all bits set to 0 (e.g. 00)    
• 1 is the bit string of length n with all bits set to 1 (e.g. 11)    
• ⫬ is bitwise not    
• △ is bitwise and    
• ▽ is bitwise or    
• The (partial) order relation is the conjunction of the order relation at each bit position
  (a≤b ↔ ∀i∈{1,...,n}: a_i≤b_i; a has 0 in all those bit positions where b has 0; in C: ~(b|~a) == 0).
• Hasse diagram for bit strings of length 4: BAbitstrings.svg

Boolean Algebra Example 4:
Integers and Divisibility

• B is formed by taking a set of n pairwise coprime integers^* (e.g. {2, 3, 5}), and all the multiples that include or exclude each of the integers (including 1) (e.g. B = {1, 2, 3, 5, 6, 10, 15, 30})
• 0 is 1    
• 1 is the product of all the original n integers (e.g. 30)    
• ⫬a is 1/a    
• △ is the greatest common divisor (GCD) of the operands    
• ▽ is the least common multiple (LCM) of the operands    
• The (partial) order relation is divisibility  
• Hasse diagram for {1, 2, 3, 5, 6, 10, 15, 30}: BAdivisibility.svg

^*) This restriction can be slightly relaxed.

The Structure of Boolean Algebras

The following can be proven from the axioms of Boolean algebras:

• The size of a finite boolean algebra is always |B| = 2ⁿ
• All Boolean algebras of the same size have the same structure (they are isomorphic)
• The structure of a finite boolean algebra is an n-dimensional cube
  (animation of a 4-dimensional cube, even higher dimensions (code))
• The Hasse diagram consists of the edges of the n-dimensional cube, when placing 0 at the bottom and 1 at the top
• The result of △ is the highest vertex equal or lower than both operands (infimum, greatest lower bound)
• The result of ▽ is the lowest vertex equal or higher than both operands (supremum, least upper bound)
• The result of ⫬ is the vertex diagonally oposite of its operand
• For all boolean algebras, the same laws apply (they can be proven from the axioms)

Isomorphisms for Examples

• Basic logic is isomorphic to bitstrings of length 1 (0=false, 1=true)
• The subsets of a set of size n can be expressed as bit strings of size n:
  Each bit indicates presence (1) or absence (0) of the respective element from the subset
• Set operations can be defined by logical operations
  Examples: A ∪ B = {x| x∈A ∨ x∈B}; A ∩ B = {x| x∈A ∧ x∈B}; A^c = {x| x∈U∧¬(x∈A)}
• Examples with divisibility are constructed to include/exclude each factor; this can be mapped to bitstrings or to set operations

Axioms for Boolean Algebras

The axioms for Boolean algebras are the same as the axioms for basic logic (standard/Huntington/Robbins/Sheffer/Wolfram).

There is a choice between compactness and obviousness.

We obtained the axioms by starting with basic logic and trying to find axiomatizations.

We obtain Boolean algebras by trying to find combinations of sets and operations that conform to these axioms.

Basic logic ⇒ Axioms of basic logic/boolean algebras ⇒ Boolean algebras

The Magic Garden of George B.

(The Magic Garden of George B. And Other Logic Puzzles, Raymond Smullyan, Polimetrica, 2007)

• A very special garden: Flowers can change colors every day
  (set of flowers F, set of days D, color of flower a on day d: color(a, d))
• Each flower is either blue or red for a whole day
  (∀f∈F: ∀d∈D: color(f,d)=red⊕color(f,d)=blue)
• For any flowers a and b, there is a flower c that is red on all and only those days on which a and b are both blue.
  (∀a, b∈F: ∃c∈F: ∀d∈D: color(a,d)=color(b,d)=blue ↔ color(c,d)=red)
• Any two different flowers a and b have different colors on at least one day.
  (∀a, b∈F: ∃d∈D: a≠b→color(a,d)≠color(b,d))
• We know that the garden has more than 200 but less than 500 flowers. (200<|F|<500)
• Question 1: How many flowers does the garden have?
• Question 2: What is the minimum of days that the flowers bloom?
• Question 3: Who is George B.?

How to Solve the Magic Garden Puzzle

• Let red stand for 1, blue for 0
• Let each day be a bit position
• Let each flower be a bit combination
• "there is a flower c that is red on all and only those days on which a and b are both blue" means that bitwise NOR is always defined
• Because any (bitwise) logical operation can be defined based on bitwise NOR, all bitwise logical operations are defined
• The flowers therefore form a Boolean algebra

The Magic Garden Boolean Algebra

• B is F (set of flowers)  
• 0 is a flower that is always^(*) blue   
• 1 is a flower that is always^(*) red (*assuming |F|=2^|D|)    
• ⫬a is the flower that is red on those days where a is blue and blue on those days where a is red  
• △ is the flower that is red on those days where both          of its operands are red, otherwise blue
• ▽ is the flower that is red on those days where one or more    of its operands are red, otherwise blue
• The (partial) order relation is: a flower a is smaller or equal than a flower b if on all days where b is blue, a is also blue    
• Hasse diagram for 32 flowers: BAgarden.svg

Answers to the Magic Garden Quiz

• The size of a finite Boolean algebra is 2ⁿ, hence 200<|F|<500 → |F|=256=2⁸; min(|D|)=8
• Question 1: How many flowers does the garden have? 256  
• Question 2: What is the minimum of days that the flowers bloom? 8 days (dimension of this Boolean algebra)    
• Question 3: Who is George B.? George Boole, after which Boolean algebra is named    

Summary

• Boolean algebra is (an example of) an algebraic structure
• Boolean algebras have one set (B), two special elements (0 and 1), and three operations (⫬, △, and ▽)
• The axioms of boolean algebra are the same as the axioms of basic logic; there are various choices with tradeoffs between obviousness and compactness
• Boolean algebras are (partial) orders and lattices
• There are many examples of Boolean algebras: basic logic (with logic operations), bit strings (with bitwise logic operations), powersets (with set operations), certain sets of integers (with divisibility), gardens with flowers,...
• All finite Boolean algebras are of size 2ⁿ, and are isomorphic to an n-dimensional cube

Homework

Deadline: December 18, 2025 (Thursday), 18:30.

Format: A4 single page (NO cover page), easily readable handwriting (NO printouts), name (kanji and kana) and student number at the top right

Where to submit: Box in front of room O-529 (building O, 5th floor)

Draw the Hasse diagram of a Boolean algebra of dimension 4 (16 elements). There will be a deduction if you use the same elements as another student.

Additional Homework (no need to submit): Which Boolean function(s) make(s) a group out of any Boolean algebra?

Glossary

coverage

(試験) 範囲

cyclic group

巡回群

Klein four-group

クラインの四元群

Abelian group

アベル群、可換群

semigroup

半群

ring

環 (かん)

polynomial

多項式

field

体 (たい)

lattice

束 (そく)

Boolean algebra

ブール代数

zero element

零元

unary operation

単項演算

binary operation

二項演算

bitwise operation

ビット毎演算

bitwise not

ビット毎否定

bitwise and

ビット毎論理積

bitwise or

ビット毎論理和

bitwise xor (exclusive or)

ビット毎排他的又は

coprime

互いに素

greatest common divisor

最大公約数

least common multiple

最小公倍数

coprime

互いに素

pairwise coprime

対ごとに素、どの二つも互いに素

n-dimensional

n 次元 (の)

cube

立方体

supremum (least upper bound)

上限、最小上界

infimum (greatest lower bound)

下限、最大下界