Section1-1

. Preface

目標這門課應該教導同學如何用邏輯與數學
來思考 (how to think logically and
mathematically)。
內容包括五個部份：mathematical
reasoning(數學推理)、combinatorial
analysis(組合分析)、discrete
structure(離散結構)、algorithmic
thinking(演算法的思考)、applications
and modeling(應用與模型)

. . . . . .

洪春男 1. The Foundations: Logic and Proofs March 1, 2011 1 / 41

. 教材
課本 Discrete Mathematics and its
Applications(sixth edition), Kenneth H.
Rosen
參考書籍離散數學 (Discrete Mathematics and its
Applications 中譯本) sixth edition, 謝良
瑜陳志賢譯

. . . . . .


. 自我介紹

姓名洪春男
email spring@mail.dyu.edu.tw
電話 04-8511888 轉 2410
辦公室工學院 H311
Homepage http://www.dyu.edu.tw/ spring

. . . . . .


. 評分標準

期中考 20%
期末考 30%
平常分數 50%(點名與隨堂測驗、作業、平常考
大約各佔 1/3)

. . . . . .


. Contents
1. The Foundations: Logic and Proofs(1.1-1.7)
.
2 Basic Structures: Sets, Functions, Sequences, and

Sums(2.1-2.4)
3. The Fundamentals: Algorithms, the Integers, and
Matrices(3.1-3.5, 3.8)
4. Induction and Recursion(4.1-4.3)
.
5 Counting(5.1-5.3)

6. Discrete Probability(6.1)
.
7 Advanced Counting Techniques(7.1, 7.5)

.
8 Relations(8.1, 8.3, 8.5)

9. Graphs(9.1-9.5)
.
10 Trees(10.1)

. Boolean Algebra
11

.
12 Modeling Computation
. . . . . .


. 1. The Foundations: Logic and Proofs

Logic is the basis of all mathematical reasoning,
and of all automated reasoning. 邏輯是所有數
學推理與自動推理的基礎。
To understand mathematics, we must
understand what makes up a correct
mathematical argument, that is, a proof. 要了
解數學，必須了解建構正確的數學論證，也
就是證明。

. . . . . .


. 1.1 Propositional Logic 命題邏輯
A proposition is a declarative sentence(that is,
a sentence that declares a fact) that is either
true or false, but not both. 命題是一個述句
(宣告事實的句子)，它可能是真、也可能是
假，但不能旣真又假。
.
Example 1
.
1. Washington, D.C., is the capital of the United
States of America. 華盛頓特區是美國首都。
2. Toronto is the capital of Canada. 多倫多是加
拿大首都。
3. 1 + 1 = 2.
4. 2 + 2 = 3.
. 洪春男 . .

1. The Foundations: Logic and Proofs
. . .

March 1, 2011
.

7 / 41


.
Example 2
.
下列是錯誤的 propositions
1. What time is it? 現在幾點？
2. Read this carefully. 小心閱讀。
3. x + 1 = 2.
.
. x + y = z.
4

. . . . . .


propositional variables 命題變數, p, q, r, s, · · ·
truth value: T(真)、F(假)
The area of logic that deals with propositions is
called the propositional calculus or
propositional logic. 專門處理命題的邏輯稱
為命題演算或命題邏輯，亞里斯多德
(Aristotle) 最早開始使用。
New propositions, called compound
propositions, are formed from existing
propositions using logical operators. 由已存在
的命題加上邏輯運算子形成新的命題，稱為
複合命題。 . . . . . .



.
Deﬁnition 1
.
Let p be a proposition. The negation of p,
denoted by ¬p(also denoted by p), is the statement
“It is not the case that p.”
The proposition ¬p is read “not p”. The truth value
of the negation of p, ¬p, is the opposite of the
truth value of p.
令 p 為一命題， p 的否定句為「p 不成立」，以
¬p 表示 (有時也用 p 表示)。 ¬p 讀作「非 p」，
其真假值與 p 的真假值剛好相反。
.

. . . . . .



.
Example 3
.
Find the negation of the proposition
“Today is Friday.”
and express this in simple English.
找出「今天是星期五」的否定命題，且用簡單的
英文表示。
“It is not the case that today is Friday.”
“Today is not Friday.”
“It is not Friday today.”
「今天是星期五不成立」或「今天不是星期五」
.

. . . . . .


.
Example 4
.
Find the negation of the proposition
“At least 10 inches of rain fell today in Miami.”
and express this in simple English.
找出「邁阿密今天至少下 10 英吋的雨」的否定
命題，且用簡單的英文表示。
“It is not the case that at least 10 inches of rain fell
today in Miami.”
“Less than 10 inches of rain fell today in Miami.”
「邁阿密今天至少下 10 英吋的雨不成立」或
「邁阿密今天下不到 10 英吋的雨」
. 。
. . . . . .



.
Deﬁnition 2
.
Let p and q be propositions. The conjunction of p
and q, denoted by p ∧ q, is the proposition “p and
q.”. The conjunction p ∧ q is true when both p and
q are true and is false otherwise.
令 p 與 q 都是命題， p 與 q 同時發生為「p 和
q」，記成 p ∧ q，當 p 與 q 都是真時 p ∧ q 為真，
否則為假。
.

. . . . . .



.
Example 5
.
Find the conjunction of the propositions p and q
where p is the proposition “Today is Friday” and q
is the proposition “It is raining today”.
令命題 p 為「今天是星期五」，命題 q 為「今天
下雨」，請找出 p 與 q 的 conjunction。
“Today is Friday and it is raining today.”
「今天是星期五且下雨」
.

. . . . . .



.
Deﬁnition 3
.
Let p and q be propositions. The disjunction of p
and q, denoted by p ∨ q, is the proposition “p or q.”.
The disjunction p ∨ q is false when both p and q are
false and is true otherwise.
令 p 與 q 都是命題， p 與 q 的分裂為「p 或
q」，記成 p ∨ q，當 p 與 q 都是假時 p ∨ q 為假，
否則為真。
.
or 有 inclusive 與 exclusive 的分別， ∨ 是
inclusive。
. . . . . .



.
Example 6
.
What is the disjunction of the propositions p and q
where p and q are the same propositions as in
Example 5.
令命題 p 為「今天是星期五」，命題 q 為「今天
下雨」，請問 p 與 q 的 disjunction。
“Today is Friday or it is raining today.”
「今天是星期五或今天下雨」
.

. . . . . .



.
Deﬁnition 4
.
Let p and q be propositions. The exclusive or of p
and q, denoted by p ⊕ q, is the proposition that is
true when exactly one of p and q is true and is false
otherwise.
令 p 與 q 都是命題， p 與 q 的互斥或，記成
p ⊕ q，當 p 與 q 恰為一真一假時 p ⊕ q 為真，否
則為假。
.

. . . . . .



The truth table 真值表
p q ¬p p ∧ q p ∨ q p ⊕ q
T T F T T F
T F F F T T
F T T F T T
F F T F F F

. . . . . .


.
Deﬁnition 5
.
Let p and q be propositions. The conditional
statement p → q is the proposition “if p, then q.”
The conditional statement p → q is false when p is
true and q is false, and true otherwise. In the
conditional statement p → q, p is called the
hypothesis(or antecedent or premise) and q is
called the conclusion(or consequence).
令 p 與 q 都是命題，條件句 p → q 代表「若 p
則 q」的命題。當 p 真 q 假時，條件句 p → q 為
假，否則為真。其中 p 稱為假設 (或前提)、而 q
稱為結論。
.
. . . . . .


A conditional statement is also called an
implication. 條件句有時也稱為隱涵。
下列都是「若 p 則 q」的寫法： p, then 「if
q」「p implies q」「if p, q」, 「p only if q」,
、、
「p is sufficient for q」, 「a sufficient condition
for q is p」, 「q if p」, 「q whenever p」, 「q
when p」「q is necessary for p」「a necessary
、、
condition for p is q」「q follows from p」「q
、、
unless ¬p」
“If I am elected, then I will lower taxes.” 若我當
選就減稅。
“If you get 100% on the final, then you will get
an A.” 期末考 100 分就得 A。 . . . . . .


.
Example 7
.
Let p be the statement “Maria learns discrete
mathematics” and q the statement “Maria will find
a good job.” Express the statement p → q as a
statement in English.
令 p 是「瑪麗亞學離散數學」 q 為「瑪麗亞將，
找到好工作」，請用英文表達 p → q。
“If Maria learns discrete mathematics, then she will
find a good job.”，「若瑪麗亞學離散數學，她將
找到好工作」，“Maria will find a good job when
she learns discrete mathematics.”，“Maria will find
a good job unless she does not learn discrete
mathemathics.”
. . . . . . .



“If it is sunny today, then we will go to the
beach.” 若今天出太陽，我們將去海邊玩。
“If today is Friday, then 2 + 3 = 5.” 若今天是
星期五，則 2 + 3 = 5。
“If today is Friday, then 2 + 3 = 5.” 若今天是
星期五，則 2 + 3 = 6。
前題與結果未必需要有因果關係。
.
Example 8
.
if 2 + 2 = 4 then x := x + 1.
若在這個 statement 之前 x = 0 的話，執行之後 x
的值是多少？
.
. . . . . .



The proposition q → p is called the converse
(相反) of p → q.
The contrapositive (對換) of p → q is the
proposition ¬q → ¬p.
The proposition ¬p → ¬q is called the inverse
(相反) of p → q.
When two compound propositions always have
the same truth value we call them equivalent.

. . . . . .


.
Example 9
.
What are the contrapositive, the converse, and the
inverse of the conditional statement
“The home team wins whenever it is raining.”? 每
當下雨時地主隊獲勝。
contrapositive “If the home team doesn’t win, then
it is not raining.” 若地主隊沒贏就沒有下雨。
converse “If the home team wins, then it is raining.”
若地主隊贏就下雨
inverse “If it is not raining, then the home team
doesn’t win.” 若沒下雨地主隊就沒贏。
.
. . . . . .


.
Deﬁnition 6
.
Let p and q be propositions. The biconditional
statement p ↔ q is the proposition “p if and only
if q.” The biconditional statement p ↔ q is true
when p and q have the same truth values, and is
false otherwise. Biconditional statements are also
called bi-implications.
令 p 與 q 都是命題，雙條件句 p ↔ q 代表「p 若
且唯若 q」的命題。當 p 與 q 有相同真假值時，
雙條件句 p ↔ q 為真，否則為假。雙條件句又稱
為雙蘊涵。
.
. . . . . .



當 p → q 與 q → p 都是 true 時 p ↔ q 才為
true。
「p is necessary and sufficient for q」「if p
、
then q, and conversely」「p iff q」都是 p ↔ q
、
的意思。iff 是 if and only if 的縮寫。
.
Example 10
.
Let p be the statement “You can take the flight”
and let q be the statement “You buy a ticket.”
Then p ↔ q is the statement
“You can take the flight if and only if you buy a
ticket.”
.
. . . . . .



The truth table 真值表
p q p → q q → p ¬q → ¬p ¬p → ¬q p ↔ q
T T T T T T T
T F F T F T F
F T T F T F F
F F T T T T T

. . . . . .



.
Example 11
.
Construct the truth table of the compound
proposition
(p ∨ ¬q) → (p ∧ q).
.
p q ¬q p ∨ ¬q p ∧ q (p ∨ ¬q) → (p ∧ q)
T T F T T T
T F T T F F
F T F F F T
F F T T F F

. . . . . .



Precedence of Logical Operators
邏輯運算子的優先順序
Operator Precedence
¬ 1
∧ 2
∨ 3
→ 4
↔ 5

. . . . . .


.
Example 12
.
How can this English sentence be translated into a
logical expression?
“You can access the Internet from campus only if
you are a computer science major or you are not a
freshman.”
只有當你主修電腦或不是新鮮人，才能在校園中
使用網路
a 代表 “You can access the Internet from campus.”
c 代表 “You are a computer science major.”
f 代表 “You are a freshman.”
前面的句字可翻譯為 a → (c ∨ ¬f)
.
. . . . . .


.
Example 13
.
How can this English sentence be translated into a
logical expression?
“You cannot ride the roller coaster if you are under
4 feet tall unless you are older than 16 years old.”
若你不到 4 英呎高就不能坐雲霄飛車，除非你超
過 16 歲。
q 代表 “You can ride the roller coaster.”
r 代表 “You are under 4 feet tall.”
s 代表 “You are older than 16 years old.”
前面的句字可翻譯為 (r ∧ ¬s) → ¬q
.
. . . . . .



.
Example 14
.
Express the specification “The automated reply
cannot be sent when the file system is full” using
logical connectives. 使用邏輯連詞表達下列規
定：「當檔案系統滿了，自動回覆功能不能被送
出」。
p 代表 “The automated reply can be sent.”
q 代表 “The file system is full.”
前面的句字可翻譯為 q → ¬p
.

. . . . . .



.
Example 15
.
Determine whether these system specifications are consistent:
“The diagnostic message is stored in the buffer or it is
retransmitted.”
“The diagnostic message is not stored in the buffer.”
“If the diagnostic message is stored in the buffer, then it is
retransmitted.”
p 代表 “The diagnostic message is stored in the buffer.”
q 代表 “The diagnostic message is retransmitted.”
前面三個句字為 p ∨ q, ¬p, p → q，當 p 為 F 而 q 為 T
時，三個句子都成立，因此 consistent。
.

. . . . . .



.
Example 16
.
Do the system specifications in Example 15 remain
consistent if the specification “The diagnostic
message is not retransmitted” is added?
p 代表 “The diagnostic message is stored in the
buffer.”
q 代表 “The diagnostic message is retransmitted.”
四個句字為 p ∨ q, ¬p, p → q, ¬q，顯然無法使四
個句子都為 true，因此不 consistent。
.

. . . . . .


. 1.1 Propositional Logic 命題邏輯 - boolean searches

可在搜尋引擎中輸入下列關鍵字，看看結果如何
.
Example 17
.
New and Mexico and University「New Mexico
University」
(New and Mexico or Arizona) and University
(Mexico and Universities) not New 「Mexico
Universities -New」
大葉 -高島屋
大葉資訊 -資管 -會計
.

. . . . . .


. 1.1 Propositional Logic 命題邏輯 - Logic Puzzles

.
Example 18
.
An island that has two kinds of inhabitants(居民),
knights(騎士),who always tell the truth, and their
opposites, knaves(無賴), who always lie. You
encounter two people A and B. What are A and B
if A says “B is a knight” and B says “The two of us
are opposite types”?
.

. . . . . .


. 1.1 Propositional Logic 命題邏輯 - Logic Puzzles

答案：Both A and B are knaves.
.
Example 19
.
兩個兄妹在後院玩，兩人的前額都沾了泥巴，父
親說：「你們兩人中至少一人前額有泥巴。」父親
問：「你知道你自己的前額有沒有沾泥巴呢？」
父親問兩次，請問兩個小朋友會如何回答？假設
小朋友可看到另一人的前額、看不到自己的前
額，且都說實話，兩人只能回答 Yes / No。
.

. . . . . .


1.1 Propositional Logic 命題邏輯 - Logic and Bit
. Operations

答案：第一次都回答 NO，第二次都是 Yes。
A bit is a symbol with two possible values,
namely, 0(zero) and 1(one). 可代表 true(1) 與
false(0)。
A variable is called a Boolean variable if its
value is either true or false.
.
Deﬁnition 7
.
A bit string is a sequence of zero or more bits. The
length of this string is the number of bits in the
string.
.
. . . . . .


. Operations

.
Example 20
.
101010011 is a bit string of length nine.
.
We deﬁne the bitwise OR, bitwise AND, and
bitwise XOR of two strings of the same length to
be the strings that have as their bits the OR, AND,
and XOR of the corresponding bits in the two
strings, respectively.

. . . . . .


. Operations

.
Example 21
.
Find the bitwise OR, bitwise AND, and bitwise XOR
of the bit strings 0110110110 and 1100011101.
0110110110
1100011101
bitwise OR 1110111111
bitwise AND 0100010100
. bitwise XOR 1010101011

. . . . . .


謝謝大家的聆聽！

. . . . . .


Section1-1

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Section1-1

Similar to Section1-1 (20)

Recently uploaded

Recently uploaded (20)

Section1-1