Propositional Logic

(命題論理)

Discrete Mathematics I

4th lecture, Oct. 10, 2014

http://www.sw.it.aoyama.ac.jp/2014/Math1/lecture4.html

Martin J. Dürst

© 2005-14 Martin J. Dürst Aoyama Gakuin University

Today's Schedule

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

Homework: Quizes

• Propositions: True or False: not tried: 10; not full points: 1
  
• Truth Table 1: not tried: 17; not full points: 1
• Truth Table 2: not tried: 14; not full points: 2

Summary of Last Lecture

• Prepositions are sentences which are objectively either correct (true) or wrong (false)
• Conjunction (and, ∧), disjunction (or, ∨), and negation (not, ¬) are three basic logical (Boolean) operations
• ¬ has higher precedence than ∧, which has higher precedence than ∨
• Truth tables can be used for
  • Evaluating a Boolean formula for all values of its variables
  • Prove the equivalence of two Boolean formulæ

Overview of Logical Operations

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∧C ∨ B∧C,
   A∧B ∨ C =  (A∨C) ∧ (B∨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 law is specific to two-valued logic?

  law of excluded middle

• 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

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

The Duality Principle for Logical Operations

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

If we exchange all instances of ∧ and ∨, and T and F, we get another law.

Examples:
T ∧ A = A; dual: F ∨ A = A
(¬A∨B) ∧ C = C∧¬A ∨ B∧C; dual: ¬A∧B ∨ C = (C∨¬A) ∧ (B∨C)

This is true in general. It 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.

From a Truth Table to a Logical Formula

Assume we are given a truth table (boolean function) such as the following:

Can you find a logical formula for this truth table?

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

Normal Forms

Disjunctive normal form: Disjunction of conjunction (of negation) of variables

Conjunctive normal form: Conjunction of disjunction (of negation) of variables

Properties of Normal Forms

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

Construction of Normal Forms

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

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

Reason for Correctness

The constructed normal form is correct because:

• Each of the terms (rows) is a disjunction [conjunction]. 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 conjunction [disjunction]. Therefore, if any of the terms is T [F], the overall result is T [F]. Otherwise, it is F [T].

Example of Normal Form

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)

Simplification of Normal Forms

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

• Manipulate (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
(conjunction of disjunction (of negation) for CNF).

Using a different structure may allow a shorter formula.

Manipulation of normal form

• Try to use any laws/properties to simplify the normal form.
• Most frequent simplification step:
  • Look for two terms where only the presence/absence of negation for one variable differs.
    
  • Use a distributive law (backwards), an idempotent law, 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.)
  • Example: A∧B∧C ∨ A∧¬B∧C ⇒ A∧C∧ (B ∨ ¬B) ⇒ A∧C
  • 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 is no single solution to simplification. Different simplification paths with different steps may lead to different results.

Karnaugh Map Construction

Creating a simplification of a 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 (fields with F 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 are included. The groups can overlap. There may be several equally minimal solutions.
8. Each surrounded group corresponds to a term 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 false.

The same procedure can be used to create a disjunctive normal form (based on the duality principle).

Karnaugh Map Example

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, 2014, 22:00.

Format: Start with example, use plain text only; use Windows Notepad or another plain text editor (NO Microsoft Word, Microsoft Excel,...)

File name:

solution4.txt

Homework Problem One

• 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.
• 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

• 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.)
• 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 some 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: 来週の木曜日 (10月23日)、22:00 (厳守)、Moodle にて。形式は例に準拠 (プレーンテキスト, メモ帳など)。ファイル名は solution4.txt。

Glossary

idempotent law

べき等律 (冪等律)

commutative law

交換律

distributive law

分配律

distribute

分配する

absorption law

吸収律

double negative

二重否定

simplification

簡略化

double negative

二重否定

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

plain text

プレーンテキスト

notepad (Windows application)

メモ帳

deduction

減点