Propositional Logic, Normal Forms

(命題論理、標準形)

Discrete Mathematics I

4th lecture, October 10, 2025

https://www.sw.it.aoyama.ac.jp/2025/Math1/lecture4.html

Martin J. Dürst

© 2005-25 Martin J. Dürst Aoyama Gakuin University

Today's Schedule

• Summary of last lecture
• Last lecture's homework
• Laws for logical operations
• From truth table to logical formula
  • Normal forms and their simplification
  • Karnaugh map
• This week's homework

Summary of Last Lecture

• Propositions: sentences which are objectively either correct (true) or wrong (false)
• Basic logical (Boolean) operations: Conjunction (and, ∧), disjunction (or, ∨), negation (not, ¬)
• Precedence: high: ¬, medium: ∧, low: ∨
• Representation of formulæ as trees
• Use of truth tables:
  • Definition of logical operations
  • Evaluation of a Boolean formula for all values of its variables
  • Proof of the equivalence of two Boolean formulæ (next lecture)

Overview of Basic Logical Operations

Last Week's Homework

1. Solve the quiz Homework: Propositions: True or False
2. Solve the quiz Homework: Truth Table 1
3. Solve the quiz Homework: Truth Table 2
4. Use highschool texts or the Web to find out about other laws for logical operations.
5. Bring highlighter pens to the next lecture (ideally green and blue)

List of Laws for Logical Operations

1. Idempotent laws: A∧A = A,   A∨A = A
2. Commutative laws: A∧B = B∧ A,   A∨B = B∨A
3. Associative laws: (A∧B) ∧ C = A ∧ (B∧C),   (A∨B) ∨ C = A ∨  (B∨C)
4. Distributive laws: A ∧ (B∨C) =  A∧B ∨ A∧C,
   A ∨ B∧C =  (A∨B) ∧ (A∨C)
5. Absorption laws: A ∧ (A∨B) = A,   A ∨ A∧B = A
6. Double negative: ¬¬A = A
7. Law of excluded middle: A∨¬A = T
8. Law of (non)contradiction: A∧¬A = F
9. Properties of true and false: T∧A = A,   T∨A = T,   F∧A = F,   F∨A = A
10. De Morgan's laws: ¬(A∧B) = ¬A∨¬B,
    ¬(A∨B) = ¬A∧¬B

Questions about the Laws

• Which laws look familiar from other areas of Mathematics?

  commutative laws (+, ×), associative laws (+, ×), distributive law (× distributes over +, but not + over ×!)  

• Which laws are specific to two-valued logic?

  law of excluded middle, law of noncontradiction  

• Which laws look (almost) obvious?

  idempotent laws, double negative, law of (non)contradiction, properties of true and false  

• Which laws look difficult?

  absorption laws, De Morgan's laws  

Use of Laws: Rewriting Logical Formulæ (Simplification)

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

¬(A ∨ ¬B) = ¬A ∧ ¬¬B   = ¬A ∧ B  

Application: Proof of absorption law from other laws

A ∧ (A ∨ B) = (A∨F) ∧ (A∨B)   = A ∨ F∧B   = A ∨ F   = A  

Pairs of Laws

When looking at the laws of logical operations, we see many laws that come in pairs:

Examples:

• T∨A = T   and   F∧A = F
• T∧A = A   and    F∨A = A
• A∨¬A = T   and   A∧¬A = F
• Idempotent laws, commutative laws, associative laws, distributive laws, De Morgan's laws, absorption laws

The only exception is the double negative

This suggests some general principle

The Duality Principle for Logical Operations

When looking at the laws of logical operations, we see the following:

If we start with a true formula and exchange all instances of ∧/∨, and T/F, we obtain another true formula.
(Caution: We may have to adjust parentheses; e.g. for the distributive law)

The new formula is called the dual of the old formula.

This is true in general. If a formula is true, its dual is also true. This can be proved using the truth tables for ∧ and ∨.

This is called the duality principle.

It is very useful for remembering the laws of logical operations.

Application example:

The dual of (¬A∨B) ∧ C = C∧¬A ∨ B∧C is: ¬A∧B ∨ C = (C∨¬A) ∧ (B∨C)  

The dual of (A∧T) ∨ (A∧B) = A is (A∨F) ∧ (A∨B) = A

From a Truth Table to a Logical Formula

Assume we are given a Boolean function as a truth table, for example:

Can you find a logical formula for this truth table?

Is there a way to find a logical formula for every Boolean function (truth table)?

Two Normal Forms

The easiest way to create a logical formula for a Boolean function is to use a normal form.

There are two normal forms:

Disjunctive normal form (DNF):

Disjunction of conjunctions (of negations) of variables

Example: A∧¬B ∨ ¬A∧B

Conjunctive normal form (CNF):

Conjunction of disjunctions (of negations) of variables

Example: (A∨B) ∧ (¬A∨¬B)

Construction of Normal Forms

For disjunctive normal form [conjunctive normal form is given in [], based on the duality principle]

1. Look only at the rows in the truth table where the result is T [F]
2. If for all rows, the result is F [T], the overall result is F [T]
3. For each of these rows, construct the conjunction [disjunction] of all the variables (A, B, C,...)
4. If the variable's value in a row is F [T], then add a negation to this variable in this row
5. Construct the disjunction [conjunction] of all the formulæ created

Example of Normal Forms

DNF: ¬A∧¬B∧¬C ∨ ¬A∧¬B∧C ∨ A∧¬B∧C ∨ A∧B∧C  

CNF: (A∨¬B∨C) ∧ (A∨¬B∨¬C) ∧ (¬A∨B∨C) ∧ (¬A∨¬B∨C)  

Example of Normal Forms

DNF: ¬A∧¬B∧¬C ∨ ¬A∧¬B∧C ∨ A∧¬B∧C ∨ A∧B∧C

CNF: (A∨¬B∨C) ∧ (A∨¬B∨¬C) ∧ (¬A∨B∨C) ∧ (¬A∨¬B∨C)

Reason for Correctness

The constructed normal form is correct because:

• Each of the terms (rows) is a conjunction [disjunction]. Therefore, the term is only T [F] if all variables match (with or without negation). All other terms are F [T].
• The overall formula is a disjunction [conjunction]. Therefore, if any of the terms is T [F], the overall result is T [F]. Otherwise, it is F [T].

Properties of Normal Forms

• Easy to construct
• Possible to construct for any Boolean function (truth table)
• Low depth (→fast logical circuit (see lecture 5))
• Possibly long formula (→circuits may need lots of space)

Simplification of Normal Forms

Normal forms can be very long. It helps to simplify them. There are two methods:

• Rewrite (transform) the normal form
• Karnaugh map

Both methods do the same, but with different tools (formulæ vs. a diagram).

The Karnaugh map keeps the structure of the normal form
(for disjunctive normal form: disjunction of conjunctions (of negations)).

Using a different structure may allow an even shorter formula.
(example: (A∧B∨C)∧D)

Simplification by Rewriting

• Try to use any laws/properties to simplify the normal form.
• Most frequent simplification step:
  • Example: A∧B∧C ∨ A∧¬B∧C ⇒ A∧C∧ (B ∨ ¬B) ⇒ A∧C∧ T ⇒ A∧C
  • Look for two terms where only the presence/absence of negation for one variable differs.
  • Use a distributive law (backwards), the law of the excluded middle or of (non)contradiction, and a property of true and false to eliminate the variable.
    (Commutative laws and associative laws are usually also needed. But their use is not made explicit.)
  • This corresponds to the graphical grouping in the Karnaugh map

Example for three-variable normal form: A∧B∧C ∨ A∧¬B∧C ∨ ¬A∧¬B∧C ∨ ¬A∧¬B∧¬C ⇒ A∧C ∨ ¬A∧¬B  

Attention: There may be more than one solution to simplification. Different simplification paths with different steps may lead to different (but equivalent) results.

Karnaugh Map Example

Final result: ¬B∧¬C∧¬D ∨ A∧C ∨ ¬A∧B∧¬C  

Karnaugh Map Construction

Creating a simplification of a disjunctive normal form [conjunctive normal form]:

1. Create a two-dimensional truth table. Each dimension uses (roughly) half of the variables.
   (3 variables: 4×2; 4 variables: 4×4; 5 variables: 8×4;...)
2. Think about this table as a torus:
   • The rightmost field in each row is the left neigbor of the leftmost field.
   • The bottommost field in each column is the top neighbor of the topmost field.
3. Arrange the truth values in each direction so that only one variable's value differs from row to row and from column to column.
4. Concentrate on the fields with a T [F] (fields with F [T] can be left blank)
5. Surround any two neighboring (according to step 2) fields with a line.
6. Combine any neighboring groups of two fields from step 5 by surrounding them with a differently colored line. Extend this to groups of eight fields, and so on.
7. Select a minimal number of groups so that all Ts [Fs] are included. The groups can overlap. There may be several equally minimal solutions. If there are uncovered Ts [Fs], surround them to make them groups of 1.
8. Each surrounded group corresponds to a term of a simplified formula. Construct this formula as follows: Eliminating variables that are both T and F for fields in the group. Keep the variables that have an uniform value, adding a negation if that value is F [T].

This Week's Homework

Submit the solutions to the following two problems to Moodle as assignment Boolean Formulæ and Normal Forms.

Deadline: Thursday October 16, 2025, 22:00.

Format: Start with template, use plain text only; use a plain text editor (such as Microsoft Notepad)

Keep all text, formatting, and character encoding (UTF-8)

Replace all the question marks; do not replace/translate anything else, do not remove spaces or empty lines

For operators,... use the symbols already used in the template

Caution: NO Microsoft Word, Microsoft Excel,...

File name: solution4.txt

Homework Problem One: All Boolean Functions of Two Variables

• List all the possible Boolean functions of two variables A and B in a table.
  Use one row for each Boolean function.
• For each function, find the/a simplest formula (using ¬, ∧, and ∨).
• Example answer (incomplete):

  A = F B = F	A = F B = T	A = T B = F	A = T B = T	simplest formula
  F	F	F	F	F
  F	F	F	T	A ∧ B
  F	F	T	F	?
  ?	?	?	?	?

• Explanation: An independent truth table for the Boolean function in the third row of the above table looks as below. Parts with the same color represent the same information. Overall there are 16 Boolean functions of two variables.

  A	B	A ∧ B
  F	F	F
  F	T	F
  T	F	F
  T	T	T

Homework Problem Two: Normal Forms

• Create a truth table for a Boolean function with four variables (A, B, C, D).
• Decide on the result (truth value, T or F) for each row of the truth table with a random function.
• As a random function, use e.g. a coin toss.
• Decide which side of the coin corresponds to which truth value
  (e.g. Japanese 500-yen coin: 500 side → true; flower side → false)
• Toss the coin as many times as necessary (16   times).
• Your Boolean function will be different from the Boolean function of all other students.
• If your Boolean function is the same as that of another student, there will be a deduction.
• Calculate the two normal forms and a simplified formula for your Boolean function.

宿題: 論理関数と標準形

提出: 10月16日 (木)、22:00 (厳守)、Moodle にて。形式は雛形に厳守 (プレーンテキスト, ? のところだけ埋めて、残りを全部そのまま残す)。ファイル名は solution4.txt。

Glossary

idempotent law

べき等律 (冪等律)

commutative law

交換律

absorption law

吸収律

double negative

二重否定

simplification

簡略化

law of excluded middle

排中律

law of (non)contradiction

矛盾律

properties of true and false

真偽の性質

De Morgan's law

ド・モルガンの法則

simplification

単純化

duality principle

双対原理

dual

双対

normal form

標準形

disjunctive normal form

加法標準形 (選言標準形、変数の (否定の) 積の和)

conjunctive normal form

乗法標準形 (連言標準形、変数の (否定の) 和の積)

property

性質

low depth

深さが浅い

(logical, electronic) circuit

回路

term

項

manipulate

操作する

transform

変換

Karnaugh map

カルノー図表

torus

トーラス (ドーナツ型)

format

形式

template

雛形

plain text

プレーンテキスト

notepad (Windows application)

メモ帳

deduction

減点