英中UF-1英文先修專欄-方問
WAH RIO YAT PD
蟹
clauses
Lesson 13
Kinds of CTaúses! (3)))
立蒸0
,足爲就讀者理想
籬,教新
4【現料,他容界
生、士、榔| 英語:黃兆滿(工 家碩士、擔任經濟及 英語,王師,牛津大 文京及
·英語,應國宏(大
日十月八年六六九一磨公年五十五國民国中
悲敦交来考科 預博
|敎華
鳥海海學
育僑
只有下列两可能
( 1 ) P. ( p. q. r. - - - ) Q ( p q r
--)不是邏輯蕴涵Q(P84
X (2) PCP Jr.
[定理邏輯蕴涵是一個關係(Relation)它符合下列
两定律:
WIRIE PC4, q, r.TM*~~ ) ⇒ P ( p, q, r, - - - - ) (2)可移律:如果PCRE
- ) →→ Q ( p, q, r- ~ 2}
→R (4 q A
記者附註邏輯蕴涵”点“时稍性”此马典選輯等價
最大之不同處
香港教師會與私校協進會合辦
喬運筆記 現代數學講座 杜安講師主講
伍岳峰 All H (Algebra of Proposition) 恒真命題與邏輯等償之關係
8. =png (邏輯等價)
~~
g→~ong. 是恒真命題!
(証你真值表
p q
TIT
F
F F
T
T
証明從略
F
AF
F. F
T
T
TIF
I
步驟3
2
1.
2 J
4
We have known that an adjective clauza in
Conglow Sentence 18 4 subordinate clause which does the
of un adjective.
(例) IL quaTITIAS & Youn or pronoun in the Principod Cause.(我們已經知道在一個模的中的形容
瀛
Tere`ta me moy who got a prize.]
In the above sentenca." the clause "who got a
prize is an Argective Clause which quarries the Boun
1979 or Alivotave Clayse?
培
l'udinctive Clause may qualify"
[一個形容詞子句可以形容:
she Subject of tio Principal clause: (羊系列中的主詞).
Cxamples
The woman yno VTSTted you was my aunt
2. The mouse that he lived in belongs to John.
The pen which you gave me was lost.
4. The man who won the prize is his friends 5. The time when he left was unknown.i
(b) the Ch fect of the Principal Clanse: (主要子句中的電訊了
1.a hao-ayriend who was a famous "writer
2.xie net a boy whom you know.
3.She had lost the book which you gave ner
4.*I like the song which she gang.
5. Jine can answer the question that you asked.
(e) some other Ndun of Tronoun in the Primeipat Clause.
Examples,
Tom was sitting av the door wich was painted
gress.
2.7 The window of the house thap belongs.
was broken.
3.kud you know the road leading to the templa
which was built in 1901.
af without the help which he ra
finish the work.
(could)
wo lieing up to the deem which
Locked®®
Irang
Exercise 38
Underline the Adverb Clause In each of the following sentences, and say what kind of Adverb Clause it is (0)} Te) mouse ran away because it saw the cat.
比較上表第三行及第八行,可見~(~p→~g)與~八g的 真值表相同,故~ (~p→g)=~png,又由上表最後一行 可見
~~p→~ [定理一]:PCp.g
3)→~pg 是一恒真命題
--當且僅當 V = Q ( p q r
----)是一恒真命題
JP ( p q r - J2 ) → → Q (4.8.4.
(E) + P( 4. g. ) = ( p q r
`-), Bj P( 4.9. r- 與Q(p.8.2...)的体表相同,因而
P ( p q r ) Q18, 2 小恒取TA其直值 故第一恒真命題
命題,則由“双條件合题(Biconditimal
→ 真值 表之規定必需P(pg.2)典
( ((b, q, r ----) Is 相同之真值,即其真值表相同故
P ( p q r
命题,或均同蔫矛盾命题,到
証明顯而易見,不贅
利用恒真命題的代替原則上述定理可得如下结果”
P(P.3) ≡ Q(PPT)無論PB為任命題函 數
又利用代替原則基本等僧公式可得下列各命題代數之
定律:
(la) PVP = P T (16) PAP=P. († 2 4 )
(22)(PvQ) VR = PV(QVR), (al (PAQ)MR = PA(QAR)
(1) The Associative Law)
(3a) PV Q = QVP (36) PAQ=RAR
(交换律 The Commons
(4a) PV(QAR) = (PVG)ATTU DANES |(46) PA(QVR)=(P^Q)V (PAR)
"5a) PV F=P
(6a) PV7=7
&stributive
(5P) PAJ = R (2018) (66) PAF = F
(7a) PV~P=7 (7b)PA~P=7 (7)
[8a) ~ (~P) = P • (8b) ~ J= X +F=J
HONGY
(b) Then they arrived at the football field, the gamo]
and sterben,
(d)] Everywhere I looked there were dirty footmarks?
(c) When I was younger, I thought us
(1)} Though 1} an poor I am hones、
(6) 111 co with you if you return"in tine.
(h) lle stayed home, so that he might finish the work.] (4)|2he could stay where alım nay.
Aldo smile, while one isįspeaking.
Following Bantonce
price ridse; the Adjective” Clauseoˇin` and say which word In the Principal Clause cách qualifies
She aLed In the house which was paint (b))lle gave me everything that I asked for,
(c) The prize that the nought, was fairly-won'
(6) Fo)met{a} woman who kno (your(oistek
进
she wanta.
(C) The cup that was broken is now useless,
(g)|Jane never does anything that is sill-
(h}}The)#ich which was broken was mine,
(1) The girl who was your friend was puriched,
Lallihelhahy! whoốc) nother hat, left.was.crving
嚳公美 資司高 献榮梅 T
(92) ~(PvQ)= ~PA~Q, (98) ~ (PAQ= ~ PV~Q
(16### 1 De Morgan's Law)
(記者附註)上列等價公式中PQR是代表布尔函數奖味 前公式中之中,不乏代表命题,不同,又表恒真命題 7表不 宿命题,以前之尤表“真值為真之命题”f表“直值弟假之命
2. 上列等價公式,為以前公式之擴-
Sea (Logical Implication)
定義:當布尔函数 P (pg 3.) → Q(pg)是
一恒真命題,則我們稱布尔函數PC82
輯蕴涵 (ogical Implies)布尔函數Q(p.92 )
定理如果P(p18.4-
函數 P.PP. 然成立
⇒Q(f85...),則对於任何 PCP, P2 P1. - - - ) ⇒ Q ( P1. P2 P; .
(誑)根據上一節之“系”如果等價式[PCpg.r)→Q(p.9.4) ....,我們有 =7成立,則對於任何布尔函數PP,PB...
[P(P., P., P; ··· ·) → Q (p. q. 2. - - -) ] = 7/
£ 1 § 1] D X ? ( p, q, r, - - · ) →→ Q (p,q. £ - - - ) £ k i ? (p, q; &, - - · · ) →→→→ Q ( p. 8, 2. - - - ) £ - 40 1 41 #34 等取任何一组指定的命题po.goo..... f p q r
Pa
-) →→→ Q (po. for 20--) £ - £65 411 A 這種情形下,我們仍然記作 PCpogoro)→Q(o.80. Wis 由此可知,邏輯蕴涵(⇒)分两種情況. (a) P( p, q, &; - • •) ⇒ Q ( p, q, x, - • • * z b b s k
-)→Q(p.g.3----)是一恒真命題亦即
··) →→7 Q (p. g. r. - - -)=7,
-)表示條件命題
[ PC p q r
h Bp [PC Po. 8o, ro 7).
→ Q (for)是一個真的命題
-)]= alpo, go.. [1122] P(4. q. 2, - - · · ) →→→ Q ( p, q, h, - - - ) 66 — 10 M ¥ 11 形為⇒8.這是表示中→g是一恒真命題,故對於任 何指定命題 及80,條件命題→我是一真的命題,我
們記
字 三個重要的推理規律:
(1)如果P ⇒ Q, 及P笃真時則Q亦為真 (2)如果Po=
→Q及Q⇒ R 則不 (3)如果P⇒Q則~Q =
注意引這裡的 P. Q.是P(po.go),Q(porn Jo)的簡 寫,在簡單情形下複合命题PQ.可代以簡單命题和 注意在數學證明中,我們通常是用認定題目時給出的 “假設為真故在应用)式時“营真時語可略去單是
1.一式即可推断Q写真
(例)投书代表“△ABC 的两底角分角线 AD和BE 相等
品代表“△ABC两腰 CA 和CB 相等”
則⇒‰就是代表定理“如果海形”?
的两底角角线 AD 和BE相等則
AABC的两腰 CA和CB相等
我們要証明這定理,亦即要一!
証明p⇒ 8.我們知道用直接證法來話明這定理較 難,应用間接証法(或稱反証法)較為容易
間接証法就是記明“如果AABC的两腰 CA和CB不等, 則AABC的两底角分角线 AD和距亦不等”這就是要証明
3)前提下:如果天氣了,我就去旅行
前提:我在家討書或“在家看電視典不去旅行但
我沒有在家讀書 結論:天氣不好...
設了代表“天氣好”,”代表“我去旅行
。代表“我在家讀書代表“我在家看電視”
因三段論律推出(p→g) A(g→) → (p→)是恒体題就要証明(和→J A[o V (do Azi) Ant 真命題,故[$→g)nce→2)]→($→气),
(正)由真值表
pg png pvq | (png) → (pvg))
TF
ENT
F F
T
可見(pag)→(pvg)是一恒真命題故PAY⇒ PV 我們应注意邏輯蕴涵是一個闻悟,因為P(0.8)→ Q(Pkg2,----)只有是恒真命題或不是恒真命題兩種可! 能救对於任意两布尔函數P(pg)及Q(P8A
十獎主佳節電柏榮佳度本堊紐
1.有男最影标霍片最年高戰的正宮樂臺舞利
無大鐵老営小色黑 市天石朝押品相珍
行如
精心
主領德洛 演銜嘉史
部崧尼套
印
製秀最高獨聲打諜特部之心長使來一安室自: 作之優格標中殺戰務一次惊人是以賊車一
The Pawnbroker
(証) (p→ga) a[MoV (So Ang)] A~。
⇒ (to→ Jo) ^ [ (doh whe) V (So Aogo Amo do)] Chic & Mit
(p→go)n[ fu(son~go Morto)] (B^_^~^=f}
\B & tv b = + ( k ž 2 } } )
(to go) Ango
(~fo Vgo) A~g. Ti
V
天賽球卷七
大盃糕都 决足世酞十早特起楼
中天
英對筱的
麗
大骤
式場別後明
踪仙野綠
雄英膽鐵
Wizard of Oz
美高梅
一七彩片
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