Propositional Logic, Normal Forms
(命題論理、標準形)
Discrete Mathematics I
4th lecture, October 12, 2018
http://www.sw.it.aoyama.ac.jp/2018/Math1/lecture4.html
Martin J. Dürst
© 2005-18 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
About Moodle
If you have not yet registered for "Discrete Mathematics I" at https://moo.sw.it.aoyama.ac.jp, you
must come to the front immediately after this lecture.
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: highest: ¬, lower: ∧, lowest: ∨
- 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
Overview of Logical Operations
|
disjunction |
conjunction |
negation |
or |
and |
not |
precedence |
low |
middle |
high |
A |
B |
A ∨ B |
A ∧ B |
¬B |
F |
F |
F |
F |
T |
F |
T |
T |
F |
F |
T |
F |
T |
F |
|
T |
T |
T |
T |
Last Week's Homework
- Solve the quiz Propositions:
True or False
- Solve the quiz Truth Table
1
- Solve the quiz Truth Table
2
- Use highschool texts or the Web to research about laws for logical
operations.
Laws for Logical Operations
- Idempotent laws: A∧A = A,
A∨A = A
- Commutative laws: A∧B = B∧
A,
A∨B = B∨A
- Associative laws:
(A∧B) ∧ C = A ∧ (B∧C),
(A∨B) ∨ C = A ∨
(B∨C)
- Distributive laws: (A∨B) ∧ C =
A∧C ∨ B∧C,
A∧B ∨ C =
(A∨C) ∧ (B∨C)
- Absorption laws:
A ∧ (A∨B) = A,
A ∨ A∧B = A
- Double negative: ¬¬A = A
- Law of excluded middle: A∨¬A = T
- Law of (non)contradiction: A∧¬A = F
- Properties of true and false: T∧A = A,
T∨A = T, F∧A = F,
F∨A = A
- 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.
(We may have to adjust parentheses.)
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 the duality principle.
It is very useful for remembering the laws of logical operations.
Number of Boolean Functions for n Variables
- Number of rows of a truth table with n variables: 2n
- Each row can have either T or F as a result
- Number of Boolean functions with n variables: 2(2n)
- Examples:
Number of variables (n) |
1 |
2 |
3 |
4 |
Number of rows |
2 |
4 |
8 |
16 |
Number of Boolean functions |
4 |
16 |
256 |
65536 |
Question: Can we find a logic formula for any such function?
From a Truth Table to a Logical Formula
Assume we are given a truth table (boolean function) such as the
following:
A |
B |
C |
? |
F |
F |
F |
F |
F |
F |
T |
T |
F |
T |
F |
F |
F |
T |
T |
T |
T |
F |
F |
T |
T |
F |
T |
F |
T |
T |
F |
T |
T |
T |
T |
F |
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]
- Look only at the rows in the truth table where the result is T [F]
- For each of these rows, construct the conjunction [disjunction] of
all the variables (A, B, C,...)
- If the variable's value in a row is F [T], then add a negation to this
variable in this row
- Construct the disjunction
[conjunction] of all the
formulæ created
Example of Normal Forms
A |
B |
C |
? |
Disjunctive Normal
Form |
Conjunctive Normal
Form |
F |
F |
F |
T |
¬A ∧ ¬B ∧
¬C |
|
F |
F |
T |
T |
¬A ∧ ¬B ∧
C |
|
F |
T |
F |
F |
- |
A ∨ ¬B ∨
C |
F |
T |
T |
F |
- |
A ∨ ¬B ∨
¬C |
T |
F |
F |
F |
- |
¬A ∨ B ∨
C |
T |
F |
T |
T |
A ∧ ¬B ∧
C |
|
T |
T |
F |
F |
- |
¬A ∨ ¬B ∨
C |
T |
T |
T |
T |
A ∧ B ∧
C |
|
Disjunctive NF: ¬A∧¬B∧¬C ∨
¬A∧¬B∧C ∨
A∧¬B∧C ∨
A∧B∧C
Conjunctive NF: (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].
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
(disjunction of conjunction (of negation) for Disjunctive Normal Form).
Using a different structure may allow a shorter formula.
Simplification by Formula Manipulation
- 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∧ T ⇒ 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 may be more than one solution to simplification. Different
simplification paths with different steps may lead to different results.
Karnaugh Map Example
|
A=F
B=F |
A=T
B=F |
A=T
B=T |
A=F
B=T |
C=F
D=F |
T |
T |
F |
T |
C=T
D=F |
F |
T |
T |
F |
C=T
D=T |
F |
T |
T |
F |
C=F
D=T |
F |
F |
F |
T |
Karnaugh Map Construction
Creating a simplification of a conjunctive normal form:
- 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;...)
- 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.
- Arrange the truth values in each direction so that only one variable's
value differs from row to row and from column to column.
- Concentrate on the fields with a T (fields with F can be left blank)
- Surround any two neighboring (according to step 2) fields with a
line.
- 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.
- Select a minimal number of groups so that all Ts are included. The groups
can overlap. There may be several equally minimal solutions.
- 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).
This Week's Homework
Submit the solutions to the following two problems to Moodle as assignment
Boolean
Formulæ and Normal Forms.
Deadline: Thursday October 18, 2018, 22:00.
Format: Start with example, use plain text only;
use Microsoft
Notepad or another plain text editor
Keep the format and character encoding (UTF-8), only replace the question
marks
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.
提出: 10月18日 (木)、22:00 (厳守)、Moodle にて。形式は例に厳守 (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
- 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