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PKLO1012 FLQaO知੫点班
Tutor:Hannah
PHIL1012 Final考嫖概况漣
1,开卷考试,只能带课本进考场,监考官会在 reading time 检查你们的书,所以书上不能夹带笔记,我提
供的 cheat sheet 切勿带入考场,一经发现算作弊处理
2,书上可以有小便利贴(不超过 5cm x 5cm)具体几个我不清楚
答枙方法
1,考试会有 answer book,请清晰地在 answer book 标明你回答的是哪个问题,不能省略步骤,
2,翻译第一步写 glossary
3,写完 tree 后如果有 counter example 要写 model,如果 no such model 也要说明(cheat sheet 里有说
明)
4,尽可能回答所有问题,空白算 0 分
考嫖结构漣
四大题,一题多个部分(以下为 2019 年 summer 考试题型,具体题型仅供参考,大致相同)
Q1: Translations. 10 parts, 一个 Part 3 分,一共 30 分.
Q2: Models. 10 parts, 一个 Part 3 分,一共 30 分.
Q3: Trees. 5 parts, Part A 4 分 ,Part B 4 分,Part C 6 分,Part D 6 分,Part E 10 分,共 30 分
Q4: Thinking Logically. 5 parts, 一个 Part 2 分,一共 10 分.
Translations
GPLI 的翻译
考点 1 瀟濘瀉濸瀅瀌瀇濻濼瀁濺 瀂瀇濻濸瀅 瀇濻濴瀁瀠, 瀟濦瀂瀀濸瀇濻濼瀁濺 瀂瀇濻濸瀅 瀇濻濴瀁瀠, 瀟濘瀋濶濸瀃瀇瀠,etc.
考点 2 瀟A瀇 濿濸濴瀆瀇 瀁瀠, 瀟A瀇 瀀瀂瀆瀇 瀁瀠, 濸瀇濶.
Models
考点:Truth values on models for GPLI. 要简单说明理由 (no such model 时)
Trees
考点: GPLI trees and reading off models.
要 Saturate paths (Universal Quantifier and Substitution of Identical).
Thinking Logically.
考点:Thinking logically about the properties of models and propositions.
检验你的逻辑思维,拓展题
是从来没见过的
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1.1 基础੯入ୈ:
Negation:否定 ¬
It is not the case that I do not like pineapple.
Conjunction:在句中र到 “and”, “but”ޞ使ऀ ∧
I like pineapple but it is sour = I like pineapple and
pineapple is sour.所以ऀ∧କ接
Disjunction:在句中र到”or”ޞ使ऀ ∨
I have apple for breakfast or lunch.
Conditional:在句中र到“if A then B” ޞ使ऀ →
(A → B)ୌङ A ީ Antecedent(前因),B ީ Consequent(后ߧ)
Only if 后வ接ৈߧ
Biconditional:र到”if and only if” ޞ使ऀ՞ 当且仅当
如: The grass is green if and only if snow is white.
怎么ࣲઆ?
ሺܽ ר ߚሻ:both true ޞ才 true,如ߧ߄▲个 F 就ީ F
ሺܽ ש ߚሻ:both false ङޞ候才ީ F,如ߧ߄▲个ީ T ৈߧ就ީ T
ሺܽ ՞ ߚሻ:both F 或ৱ both T 才 T
ሺܽ ՜ ߚሻ:只߄先决ߚ件(ও头左ଆୂ分,即ܽ)ީ T,但ީৈખ(ও头右ଆङୂ分,即ߚ)为 F,才ީ F
1.2.1 Tree 入ୈ:
如ߧ ߄不 অ合 ङߚ件 出࣫(ࡁ如 ࠒ validity ङ counterexample), tree ਈ फ接告 ઝ我们
counterexample ީ什么
1.2.2 Tree Rules (P315 ङ੮)
1. 操作之前先找 main connective
2. ߿据 main connective ࣁव੮套公式
3. 完成▲个任务,就打▲个√
4. Tree ߂后以߂এङ形式,只允ક存在,其他ङঅ号ୃ不可以存在
如ߧ▲个 path contain both a formula and its negation, we close the path with a ×
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如:
叉掉以后就不操作了,代੮个 path close
1.2.3 The Order of the Rules
Step 1.操作 double negation
Step 2 操作 any non-branching rule(不会分叉ङ)
Step 3 操作 branching rule(߄分叉ङ)
如:
(߂୍) Test argument ީ否 valid:
如ߧ all path close 就ީ valid
如ߧ߄▲个 path ࡚߄ close,ऀ ↑ ੮ॐ,代੮ invalid,个 open ङ path ީ
counterexample
1.2 ଜੌ后习ங ଜੌ!必做!
Ex. 12.1.6, Ex. 12.1.9, Ex. 12.2.2, Ex. 12.3.1,
Ex. 13.2.2, Ex. 13.3.1, Ex. 13.4.3, Ex. 13.5.1, Ex. 13.6.1.1
2. Monadic Predicate Logic ੯
2.1 MPL 的特征: One name and one predicate ▲个大写字࠾加▲个小写字࠾
例: Bruce is a philosopher or Jane is a philosopher
Glossary:
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Px: x a philosopher.
b: Bruce
j:Jane
Pb ש Pj
Atomic proposition is a proposition made up from one name and one predicate. 如: Pb
Name:ऀ小写字母 a,b,c,d ঈ੮ॐ(不含 u,v,w,x,y,z,因为些叫 variable )
Predicate: ऀ大写字母加 variable 示,如 A,B,C,,,
如: Mary is happy.
Hx: x is happy
m:Mary
则 Hm: Mary is happy
Hm ީ▲个 Atomic proposition
ૠଛ的 glossary ଛ不ੌ写 x, y, z 之类的因为他们是 variable,不是▁个 name
2.2 Predicate 㓓么写?
1. Predicate ऀআ三人०单数形式
2. Predicate ऀ原句ࣁ抄
3. યொޞ找 main connective
4. ࡨ意倒,以句意为主
5. ৰਖ਼在哪加 negation: 如 Mary is not happy
Glossary: a: Mary
Hx:x is happy
例ொ:If Mary isn’t sailing, then unless she’s kite flying, John is sailing ੌ䎯:Unless=if not
Glossary:
j: John
m: Mary
Ax: x is sailing
Kx: x is kite flying
Am ՜ ሺKm ՜ Ajሻ
2.3 Variables and Quantifiers
2.3.1 Existential quantification x
当ொऩ中出࣫: something, anything(否定情况ङ something,如ߧ▲句ઢީ否定句৲且出࣫了 any,
就写成), at least one, there is a….ޞ使ऀ 澬ਙ少߄▲个ީ澭
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x ੮ॐ什么意思: there is at least on x such that….
例ொ: Something is good. Glossary: Gx:x is good ৬ઠ:xGx
ୌङ glossary 不ऀ写 x 因为我们已ऽଳx ੮ॐङީ “ਙ少߄▲个 x অ合个ࡌ”ङ意思
2.3.2 Universal quantification: x
当ொऩ中出࣫:all, everything, any (੮任何),,ঈޞ使ऀ 澬个ୃީ澭
x ੮ॐ什么意思: for all x, x is such that…
例ொ: Everything is good. Glossary: Gx:x is good ৬ઠ:x Gx
么ઓ:Existential 得૦很像,Universal 得૦很像
੯例ங Glossary: Gx : x is grey. Rx: x is round. Sx: x is special.
All round things are special.
ሺR ՜ Sሻ
Some round things are special
ሺR ר Sሻ
快ૻ方法:如果是ܠ就用→,如果是ܠ就用ר
为什么⪢?
All round things are special.
=For all things, if it is round, then it is special. (If then ऀ→੮ॐ)
为什么不ਈऀר?ሺR ר Sሻ就会变成:all things are round and special 所߄ङ东ୃީ又 round 又
special,不অ合句意
Some round things are special
=For something, it is round and special.ୌਙ少存在߮य़东,他ީ round ङ,也ީ special
ङ
常ি句型 (cheat sheet 上߄)
Something is round but not special. ሺR ר Sሻ
Something is neither round nor special ሺR ר Sሻ 或ৱ ሺR ש Sሻ
2.4 Restricted Quantification
Someone, anyone, everyone ঈঝࣔ指人ङ么写
Glossary: Hx: x is happy. Sx: x is sad. Px: x is a person
如: Everyone is happy. ሺP ՜ Hሻ
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Someone is sad. ሺP ר Sሻ
ୌ▲定先加▲个 Px ੮ॐ你指代ङ东ީ人
但ީொऩ中߄ girls, woman 之ঝङ指代ટޞ(你र到个ટ就ऽଳਂ定ީ人ঝ),只加 predicate 就੧,
不ऀ加 Px
2.5 Nothing =不存在“২少有▁个满ર”的㖊况= ¬x
No one ੮ଇސ式: 1. ¬x(Pxר▲切属性)
2. x(Px՜▲切ब反ङ属性)
属性之ऀרକ接੮ॐ并列
例ொ: No one who isn’t happy is laughing
(ঈ于 Everyone who isn’t happy is not laughing.)
߄两य़੮ଇސ式
1. ሺP ר L ר Hሻ
2. ൫ሺP ר Hሻ ՜ L൯
(ৰડ写其中▲个就可以了,我个人推਼আ二य़)
2.6 汉娜১创ଝ淪੯方法名੮ଚ:
主关ટ: 个句子主想੮ଇङ两个 name 之ङ关,
ࡁ如 a ࡁ b 劅,▲ਢީ后ߧ(ও头右ଆ)
修ટ:形容个 name ङ各य़形容ટ,ࡁ如 a ީङ
例ொ: Translate the following into MPL:
Glossary:
Cx: x is certain Rx: x is red Ex: x is expensive Hx: x is heavy Fx: x is probable
1. All red things that are not heavy are expensive
ሺሺR ר Hሻ ՜ Eሻ
2. If something is certain, then it’s probable
ሺC ՜ Fሻ
为什么ୌޢޢީ something ଐऀও头⪢?因为ୌީ if then, 我们ऀ→੮ॐ,ީ和→对ङ,因࠴前
வङ quantifier ऀ
3. Everyone runs only if someone does not run
ሺP ՜ Gሻ ՜ ሺP ר Gሻ
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3. General Predicate Logic 的੯
GPL:a two-place predicate: it requires two names to make a proposition ▲句ઢ中߄两个 name,
(ࡁૻ)▲个大写字࠾加两个小写字࠾,੮ॐ x 和 y 之ङ关
3.1 GPL ੯程
1, one place or two place
2, 写 Glossary
3, 代入 names, ऀ connective 或 quantifiers ৬ઠ
如: Ben lives in London
Lxy: x lives in y
b: Ben
l: London Lbl
两个 name 的எ序不ৗ互换,ࡁ如 Llb ङ意思ީ London lives in Ben.
1. Alfred can solve every puzzle.
a: Alfred
Px: x is a puzzle.
Sxy: x can solve y x(Px → Sax)
2. Alfred cannot solve any puzzle. ୌङ any 指ङީ some ङ否定形式
ሺP ՜ Saሻ 或ৱ ሺP ר Saሻ
3. Alfred cannot solve every puzzle.
ሺP ՜ Saሻ 或ৱ ሺP ר Saሻ
例ொ 1 :Although the Eiffel tower is taller, Clare prefers Dave.
Glossary:
t: the Eiffel tower
Txy : x is taller than y
Pxyz : x prefers y to z
Ttd ר Pcdt
ୌङ Although 和 But ▲߽,ୃީרङ意思,后半句ઢ: Clare prefers Dave ொ干中मऋ了
(prefers Dave to Eiffel tower)
例ொ 2 :Alice is taller than Bob and vice versa (反之亦ࣀ)
(ঈ于 Alice is taller than Bob and Bob is taller than Alice)
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Glossary a: Alice
b: Bob
Txy: x is taller than y.
Tab ר Tba
例ொ 3 : If Alice is taller than everything that Bob is taller
than, and Bob is taller than Alice, then Alice is taller than
herself.
1,拆分句子找关୰ટ
If 澬Alice is taller than everything(that Bob is taller than), and Bob is taller than Alice澭,then
澬Alice is taller than herself.澭
ሺሺTb ՜ Taሻ ר Tbaሻ ՜ Taa
3.2 Multiple Quantifier
关于 x 和 y ङ关,加上 quantifier ङ排列ু合߄以下几य़:
1. xySxy 2. yxSxy 3. xySxy 4. yxSxy
5. xySxy 6. yxSxy 7. yxSxy 8. xySxy
分三ঝ:
1., some-some (আ 3,4 य़情况)
如:Some chihuahuas are bigger than some beagles
2, some-any/ any-some (আ 5,6,7,8 य़情况)
如:Some timid dog growls at every grey cat
3, any-any (every-every) (আ 1,2 य़情况)
如:Every dog growls at every timid cat
★3.2.1 some-some ৬ઠސࡣ:
1, 找出,,▲放在句子ङ前வ
2, 把修ટऀ ר କ接
如:Some chihuahuas are bigger than some beagles.
Glossary: Bx: x is a beagle
Cx: x is a chihuahua
Bxy: x is bigger than y ሺC ר B ר Bሻ
★3.2.2 any-any (every-every)ސࡣ
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1 े定 quantifier ީ,把x放在句子前வ
2, 把两个࣒体之ஔ了主关ટङ修ટ放在▲ऀ ר କ接
3. 把主关ટ放句ߍ,ऀ → କ接
如:The bigger the dog, the happier it is.
ሺሺD ר D ר B ՜ Hሻሻ
આߤ:
1, 如ߧ࡚߄ޢे指ޢ(some/a)之ঝङஒ制ޞऀ any-any(对任何情况ୃ成ॹ)
对于任何ङࣝ子ߛપ,大ङ开心
因为ީ两个࣒体之ङࡁૻ,所以ऀ x 与 y ੮ॐ 比级▁ৱ的格式是 any-any
2,找出修ટ:
x 和 y ୃީࣝ 澬Dx Dy澭
x ࡁ y 大 (前因) 澬Bxy澭
3, 找出主关ટ
x ࡁ y 开心 (后ߧ) 澬Hxy澭,大ङࣝ子开心,开心ީ我们想੮ଇङ意思
例ொ:Elvis prefers any top-twenty song that the Rolling Stones recorded to any song that he
himself recorded.
ሺሺS ר T ר Rr ר S ר Reሻ ՜ Peሻ
1,判ލީ什么句型:any-any
2,划句子,找出 prefer 什么 to 什么
Elvis prefers (any top-twenty song that the Rolling Stones recorded) to (any song that he
himself recorded).
e: Elvis
r: the Rolling Stones
Sx: x is a song
Tx: x was in the top twenty
Rxy: x recorded y
Pxyz: x prefers y to z
3. े定 quantifier ީ,把x放在句子前வ
4,找出修ટ,ऀ ר କ接
x ީ▲个 song 澬Sx澭
y ީ▲个 song 澬Sy澭
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x ީ▲个 top-twenty 澬Tx澭
Rolling Stones record x 澬Rrx澭
Elvis record y 澬Rey澭
5,找出主关ટ,ऀ ՜ କ接
Elvis prefers x to y 澬Pexy澭
★3.2.3 some-any/ any-some ৬ઠސࡣ
1,े定 x 和 y 前வङ quantifier, (ީxଐީ?)
2,找出两个࣒体ङ修ટ
3,߿据 quantifier 加କ接ટ,主关▲ਢ在 y 后வ੮ॐ(如ߧީ x 和 y 之ङ关就放在 y 后வ,
如ߧީ x 和 z ङ关就在 z 后வ)
如:Every person owns a dog.
Px: x is a person
Dx: x is a dog
Oxy: x owns y
ሺP ՜ ሺD ר Oሻሻ
1,ީ any-some ࠀ式()
2,找出修ટ
x ީ▲个 Person 澬Px澭
y ީ▲个 dog 澬Dy澭
x owns y 澬Oxy澭
3,߿据 quantifier 加କ接ટ,
y 前வީ,所以后வऀר,主关▲ਢ在 y 后வ੮ॐ,所以ୌީD ר O
x 前வީ,କ接 x 与 y,所以后வऀ՜
ࡨ意:
▲ਢ any-any 或ৱ some-some ङޞ候 x 和 y 才▲放在句子ङ前வ,some-any/any-some
ߛપ߿据句子意思按ி序৬ઠ,ࡁ如ୌ就把放在句子ङ前வ
ୌङ quantifier 指定了 variable ङਸ围,ࡁ如ሺP ՜ ሺD ר Oሻሻ ୌङ Oxy 和 Px ୃੴ
前வङव
例ொ:There is someone who is everyone’s father
Px: x is a person.
Fxy: x is a father of y ሺP ר ሺP ՜ Fሻሻ
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4. GPLI 的੯
4.1.1 GPLI ৬ઠொ型 1:
和 GPL 基ߎब同,只ީ多了加ঈ号ङ功ਈ,在ொऩ中र到 other than, only, except ঈ߿据句意使ऀ
例:No one other than Mary owns anything that Mary wants.
ஔ了 Mary 以外࡚人ਈ够拥߄ Mary 想ङ任何东
Glossary: Px: x is a person Wxy: x wants y Oxy: x owns y
ሺሺP ר ് m ר Wmሻ ՜ Oሻ
No one other than Mary=ୣ了 Mary 以外没有人=Everyone who is not Mary can not
句ઢ৬ઠ为:Everyone who is not Mary doesn’t own anything that Mary wants.
ީ Any-any 句型
1,把放在句子开头
2,把所߄修ટ放在▲,ऀרକ接
a) x ީ个 Person
b) x 不ީ Mary
c) Mary wants y
3,把主关放在句子ৈ尾,ऀ՜କ接 x 不 owns y
4.1.2 GPLI ৬ઠொ型 2: 最ள级ङ৬ઠސࡣ
1,左ଆ写߂࣒品ङ所߄修ટ
2,右ଆ三ୂ分: a) 右ଆ࣒品ङ所߄修ટ
b) 右ଆ不ঈ于左ଆ
c) 主ࡁૻ,ऀ quantifier 对应ङঅ号କ接(ऀ՜,ऀר)
▲ਢ quantifier ୃީ,因为߂ީ对于所߄东ୃ成ॹङ
如:Diane is the tallest woman.
Glossary: Wx: x is a woman.
d: Diane y: x is taller than y
Wd ר ሺሺW ר ് dሻ ՜ Tdሻ
1,左ଆ写߂࣒品ङ所߄修ટ (Diane ީ▲个 woman)
2,右ଆ三ୂ分:
a) 右ଆ࣒品ङ所߄修ટ(ੴࡁૻङ东 x ީ▲个 woman)
b) 右ଆ不ঈ于左ଆ (x 不ީ Diane)
c) 主ࡁૻ,ऀ quantifier 对应ङঅ号କ接(ऀ՜,ऀר)(主关: Diane ࡁ x )
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例ொ:Alice is neither the youngest nor the oldest in her class
Glossary: a:Alice Yxy: x is younger than y. Cxy: x is in y’s class
ሺሺCa ר ് aሻ ՜ Yaሻ ר ሺሺCa ר ് aሻ ՜ Yaሻ
Neither nor ੮ଇސ式:
ܣ ר ܤ或ৱሺA ש Bሻ
1,र到߂ऀ
2,ও头左ଆ:x 在 a ङ class, x 不ީ a
3,a ࡁ x 小
前三࠵৬ઠ出ङީ A,
4,र到߂ऀ
5,ও头左ଆ:x 在 a ङ class, x 不ީ a
6,x ࡁ a 小
后三࠵৬ઠ出ङީ B
因为ީ Neither nor,所以变成ܣ ר ܤ
5. MPL models
5.1.1 Model 内包括:
1. a domain (a set of objects)—this specifies what
‘everything’ means according to the model.
2. a specification of a referent (an object) for each name.
3. a specification of an extension (a set of objects) for each predicate.
5.1.2 Model ࢵ:
Domain 不ਈީॱத∅,因为所߄ङ数据ୃߛ࢛于 domain,但 predicate
可以ީ empty (∅)
∅ 䥹旅代坧廘个旅合惋晡什么忼没有
5.1.3 怎么判ލ atomic proposition ङ True or False?
An atomic proposition is true if and only if the name’s referent is in the predicate’s
extension.
例ொ: State whether Pa is true or false
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1. Domain: {1, 2, 3, . . . } Referent of a: 1 Extension of P: {1, 3, 5, . . . }
2. Domain: {1, 2, 3, . . . } Referent of a: 1 Extension of P: {2, 4, 6, . . . }
提ॐ:如ߧ出࣫什么{1,2,3,4…}则代੮所߄࠳实数,{2,4,6,8…}代੮所߄偶数,{1,3,5,7…}代੮所߄奇数
如ߧ a 在 P 内则 true,写为:Referent a is in the extension of P
例ொ:Domain:{1,2,3,4…} Extension: F: {2,4,6,8…}
xFx is true in a model if and only if everything in the domain of the model is in the
extension of F on that model.
在个 extension 中 F 个 extension 如ߧ包含了 domain 内全ङ object ङઢ,xFx 就ީ True,
࠴ޞxFx 为 False
xFx is true in a model if and only if something in the domain of the model is in the
extension of F on that model.
在个 extension 中 F 个 extension 如ߧ包含了 domain 内২少▁个 object ङઢ,xFx 就ީ
True, ࠴ޞxFx 为 True
িੰங型 1: ো model 判ލ proposition ީ不ީ true
ࢵ:看 domain 中是不是所有/২少存在▁个 object 在 Extension, 组合㖊况参ি truth table
例ொ:Here is a model: Domain: {1, 2, 3, 4}
Extensions: E: {2, 4} O: {1, 3}
State whether each of the following propositions is true or false in this model.
(i) xEx
False 因为 Domain 中߄ 1 和 3,但ީ E 中࡚߄
(ii) ሺE ש Oሻ
True Domain 中所߄ङ object 么在 E 中么在 O 中
(iii) E
True E 中包含了 Domain 中ਙ少▲个ङ object (如 2 ޘ在 domain 中也在 E 内)
(iv) ሺE ר Oሻ
False 因为不存在߽▲个 object ޘ在 E 中也在 O 中
(v) ሺE ՜ Oሻ
True 如ߧE为 T,Ox 为 F ৈߧ就ީ F,但ީ不存在߽ङ x 不在 E 中也不在 O 中,因࠴为
True
(vi) E ש E
True E为 False 因为 E 中࡚߄包含 domain 中所߄ङ object,但ީE ީ True 因为
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存在▲个 x 使得它在 domain ୌ但ީ不在 E 中(如 1, 3)
例ொ 2: Here is a model:
Domain: {Alex, Ben, Carol, Diana}
Referents: a: Diana b: Alex
Extensions: O: {Alex, Ben} Y: {Carol, Diana}
Is the proposition Oa ש ሺO ר Y ሻ true or false in this model? Explain your answer
1. Diana 不在 O 个 extension 内, 所以 Oa ީ False
2. ߄࡚߄ਙ少存在▲个 x ࢠૡሺO ר Y ሻ▲ߚ件,即ޘ在 O 中也在 Y 中----࡚߄,则ሺO ר Y ሻ为
False
Disjunction 左ଆީ False 右ଆ也ީ False 则整个 proposition 为 False
িੰங型 2:describe a model એ proposition to be true 再 describe a model એ proposition to be
false (ঋࠄ不ީ唯▲ङ,এ单好)
ੌ䎯:1,Domain 尽এ单
2,referent 如ߧ߄就写,࡚߄就不写
3,Extension 写ࢎࠟ,懒得操作फ接写∅或ৱ就写▲两个 object
例ொ:For each of the following propositions, describe
(a) a model in which it is true, and
(b) a model in which it is false. If there is no model of one of these types, explain why
如果不存在ૠ样的 model 的答ங模板
No such model. For the formula to be true on a model, it would have to be the case that 澬把એ
个 model true/false ङߚ件写出ߛ澭
1. ሺF ר Gሻ
想个 proposition True,必ீ在 domain 中২少存在▁个 x 满ર “ሺF ר Gሻ”▲ࡌ
想个 proposition False,必ீ在 domain 中所有的 x 不ৗ满ર “ሺF ר Gሻ” ▲ࡌ
(a) Domain: {1, 2, 3, . . . } Extensions: F : {1, 2, 3, . . . } G : {2, 4, 6, . . . }
(b) Domain: {1, 2, 3, . . . } Extensions: F : {1, 3, 5, . . . } G : {2, 4, 6, . . . }
2. ሺF ש Gሻ
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想个 proposition True,必ீ在 domain 中所有的 x ੌ满ર “ሺF ש Gሻ”▲ࡌ
想个 proposition False,必ீ在 domain 中২少有▁个 x 不ৗ满ર “ሺF ש Gሻ” ▲ࡌ
(a) Domain: {1, 2, 3, . . . } Extensions: F : {1, 3, 5, . . . } G : {2, 4, 6, . . . }
(b) Domain: {1, 2, 3, . . . } Extensions: F : {1} G : {2}
3. ሺF ר Fሻ
想个 proposition True,必ீ在 domain 中所有的 x ੌ满ર “ሺF ר Fሻ”▲ࡌ
想个 proposition False,必ீ在 domain 中২少有▁个 x 不ৗ满ર “ሺF ר Fሻ” ▲ࡌ
(a) No such model. For the formula to be true on a model, it would have to be that each
member of the domain was both in the extension of F and also not in the extension of F. 不可ਈ存
在߽ङ x ޘ在 F 中也在F中
(b) Domain: {1, 2, 3, . . . } Extensions: F : {1}
6. GPL Models
GPL Model ொ型 1 1,ো model 和 proposition,叫你判ލ T or F
ੌ䎯:对于ܠ,我们找 True,即找出▁个满ર条件的 variable
对于ܠ,我们找 False,即找出▁个不满ર条件的 variable
GPL ߄ domain, referent, extension
࠴ޞ对于▲个 n-place predicate, extension 应ީ▲个 set, ऀ<>੮ॐ,如<1, 2>
如: Domain: {Amy, Bob, Carol}
Referent: a: Amy, b: Bob, c: Carol
Extension: P: {
, }
么Pxa 就ީ F,因为 P 中ङ set ࡚߄以 Amy 做আ二ாङ
判ލ TF ޞ不र之前 tree ङ东,只察 extension ީ否成ॹ即可
例:Here’s a model:
Domain: {Alice, Bob, Carol, Dave, Edwina}
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Referents: a: Alice, b: Bob, c: Carol, d: Dave, e: Edwina
Extensions: M: {Bob, Dave}, F: {Alice, Carol, Edwina}, T: {, , ,
, , , , , ,
}
1. Tba
False. 不在 T 内
2. Tba → Taa
True. ও头左ଆ为 False,ও头右ଆ为 False,ৈߧ为 True
3. xTcx
True. You can pick an object from the domain such that the referent of c, namely, Carol, and that
object, taken in that order, are in the extension of T. (You can pick Edwina, for instance).
当 x 为 Dave ޞৈߧ为 True
4. ሺF ר Tdሻ
True. 当 x 为 Edwina ޞF为 True 且 Tdx 为 True
5. xyTxy
False. 对于所߄ङআ▲ா为 x ߛપୃ߄▲个对应ङ y 作为আ二ா,
但ީ当 x= Edwina ޞ不存在对应ङ y,因࠴为 False
6. x(Mx → yTxy)
True. No matter what you pick first—call it x—if x is in the extension of M, then you can pick a
thing—call it y—such that x and y, taken in that order, are in the extension of T
不存在当 Mx 为 T ޞ,yTxy 为 False ङ情况
ொ型二 2,ো proposition 做出 model, 使 proposition Tor F
For each of the following propositions, describe (a) a model in which it is true and (b) a model in
which it is false. If there is no model of one of these types, explain why
ੌ䎯:对于ܠ,我们找 True,即找出▁个满ર条件的 variable
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对于ܠ,我们找 False,即找出▁个不满ર条件的 variable
ૠଛ的 model 如果 proposition ଛ没有小写字母 abcd 等 name 出现,就不写 referent
1. ሺF ՞ Fሻ
(a) Domain: {1, 2, 3} Extension of F : {<1, 1>,<2, 3>,<3, 2>}
(b) Domain: {1, 2, 3} Extension of F : {<1, 1>,<2, 3>}
2. F
(a) Domain: {1} Extension of F : {<1, 1>}
(b) Domain: {1} Extension of F : {∅}
3. F ר Faa
(a) No such model. For the proposition to be true, the ordered pair consisting of the referent of a,
followed by the referent of a, would have to both be in the extension of F and not in the extension
of F
(b) Domain: {1, 2, 3} Referent of a: 1 Extension of F : {<1, 1>}
7. GPLI Models
ொ型 1,ো model 和 proposition,叫你判ލ T or F
例:. Here is a model: Domain: {1, 2, 3, . . .}
Referents: a: 1 b: 1 c: 2 e: 4
Extensions: F: {1, 2, 3} G: {1, 3, 5, . . .}
R: {<1, 2>, <2, 3>, <3, 4>, <4, 5>, . . .}
State whether each of the following propositions is true or false in this model.
1, ሺRa ר Rbሻ
આொސࡣ 1,判断是ܠ是ܠ
对于,找出▲个ࢠૡߚ件ङ variable,થ proposition 即ީ True
2,找▁个 x 满ર所有的 ר
Rax 为 True ৲且Rb为 True [即 Rb为 False ]
3,ব小ਇ围,写出ੌ㖸每▁步条件满ર的取值
当 Rax 为 True ޞ,x=2, 当Rb为 False ޞ,x≠2
4,判断是否存在ૠ样的 x,如果存在▁个满ર所有条件的 x,ੴ proposition 则是 True,否则
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就是 False
因为ޗ x অ合થߚ件,所以થ proposition 为 False
2,ሺF ר ് e ר ሺF ר Rሻሻ
F 为 True Î x≠1, 2, 3
് e 为 True Î x≠4
F 为 True Î y=1, 2, 3
R 为 True Î 不存在,因为当 y=1 ޞ x 应થ为 2,y=2 ޞ x 应થ为 3,y=3 ޞ x 应થ为 4,但ީ߿据上
வङ原因,x≠1, 2, 3, 4
因࠴ޗ x ࢠૡ以上全ୂࡌ,થ proposition 为 False
3,ሺሺF ר Gሻ ՜ ൌ cሻ
આொސࡣ 1,判断是ܠ是ܠ
对于,找出▲个不অ合ߚ件ङ variable,થ proposition 即ީ False
2,看箭头方向的左右,如果㖸ੌ False,箭头左是 T 箭头右是 F
એF ר Gሻ为 True 且 ൌ c为 False
3,ব小ਇ围,写出ੌ㖸每▁步条件满ર的取值
当ሺF ר Gሻ为 T ޞ,F 为 True 且G为 True [即 Gx 为 False]
F 为 True Î x=1, 2, 3….
Gx 为 False Î x=2, 4, 6…
当 ൌ c ޞ,x≠c Î x≠2
4,判断是否存在ૠ样的 x,如果存在▁个满ર所有条件的 x,ੴ proposition 则是 True,否则
就是 False
因࠴ޗ x ࢠૡ以上全ୂࡌ,થ proposition 为 True (找不出▲个 x એ句ઢ不对)
ொ型二 2,ো proposition 做出 model, 使 proposition Tor F
如: ሺF ר ሺG ՜ ൌ ሻሻ
(a) Domain: {1, 2, 3} Extensions: F : {1, 2} G : {1}
(b) Domain: {1, 2, 3} Extensions: F : {1} G : {1, 2}
ሺሺF ר Raሻ ՜ ് aሻ
(a) Domain: {1, 2, 3} Referents: a : 1 Extensions: F : {1} R : ∅
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(b) Domain: {1, 2, 3} Referents: a : 1 Extensions: F : {1} R : {<1, 1>}
8. Trees for MPL
ऀ tree,书上 P220
Tree 提供了▲个ୃ为 True ङ example
If all paths close, there is in fact no
such a model
If at least one path open, there is such a model in which the
proposition at the top of the tree are all true, we can read off
a model from an open path
P220 为 tree ङ੮
先操作 non-branching,फ到变成 basic proposition 才ৈߘ
Order of application
1, non-branching rule
2, negated quantifier rules,
3, unnegated existential quantifier rule
4, unnegated universal quantifier rule
如:
主ৰ validity, satisfiable
在做߄ quantifier (和)ङ tree ޞ,代入 name (小写字࠾,如 a,b,c,d…)ߛઆ,
如ߧ前வީ,操作ޞ先打▲个√,只代入▲个字࠾即可,
如ߧ前வީ,操作ޞ先划▲个\, 把整个 tree 中出现的所有的小写字࠾ୃ代入▲ଲ,
如:
Test validity (把 conclusion 前加)
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4. All cows are scientists, no scientist can fly, so no cow can fly
9. Trees for GPL
1,Suppose we have the following open path . . .
1. Fa
2. ¬Fb
3. Gc
4. Ga
We read off a model like this . . .
Domain: {1, 2, 3}
Referents: a: 1, b: 2, c: 3
Extensions: F: {1}, G: {1, 3}
2, 1. Taa
2. ¬Tbc
3. Tab
Domain: {1, 2, 3}
Referents: a: 1, b: 2, c: 3
Extensions: T: {< 1,1 >, < 1,2>}
例ொ:
Determine using a tree whether the following proposition is a logical truth. If it is not, read off from
your tree a model on which it is false.
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1,R ՜ Ra
10. Trees for GPLI P315 ߄੮
GPLI ङ Tree ङ操作ސࡣ和 GPL 基ߎब同
ࣔ࠺操作:当出࣫如 a=b ङ情况ޞ,如ߧ tree 之前
出࣫了 Raa 么在 Tree 中写 Rbb,即把所߄ a ङ
情况ୃऀ b 代߁,把所߄出࣫ b ङ情况ୃऀ a 代߁,
ߛ寻找 close path
如:
ࢵ:
1,在ୌߎߛ出࣫了 Rab,把 a=b 代入所߄ङ式子得出 Raa, Rba, Rbb ङ情况,
2,已打ଋ勾ङ proposition,不对他 apply 个 rule
Saturate paths
即▲定把所߄ङ情况ୃ代入,ࡁ如已ऽ Rab 和 a=b,在 tree 中▲定把 Raa, Rba, Rbb 三य़情况写
ૡ去,ভ▁不可
什么时候 close:
1, र到个式子和它ङ negation (之前ઔଋ)
2,र到 a≠a ޞ
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Tree ৗ检ய的关系: [常ি Validity, satisfiability, logical truth (即 tautology) ]
例ொ 1:ࠒ validity
F
G
ൌ
ሺF ר Gሻ
例ொ 3:ࠒީ不ީ satisfiable
ሺ ് a ՜ Raሻ
Rb
a ് b
11. Numerical Quantifiers and Definite Descriptions in GPLI
P321 固定搭,划୍ࢵ,ৰડ▲ਢީ 2 个,߂多 3 个
At least n
There is at least one tiger.
T
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There are at least two tigers.
ሺT ר T ר ് ሻ
There are at least three tigers.
ሺT ר T ר T ר ് ר ് ר ് ሻ
理方法(不ଜੌ)
如:There are at least three tigers.
1,ୌ存在三个৯ x, y, z
2,他们ୃީ৯
3,他们互不बঈ
ࡨ意:߄几个৯就加几个 variable
At most n 最多只有 n 个
There is at most one tiger. 或ৱ ‘There is no more than one tiger
ሺሺT ר Tሻ ՜ ൌ ሻ
或ৱ ሺT ר T ר ് ሻ
There are at most two tigers.或ৱ ‘There are no more than two tigers
ሺሺT ר T ר Tሻ ՜ ሺ ൌ ש ൌ ש ൌ ሻሻ
或ৱ ሺT ר T ר T ר ് ר ് ר ് ሻ
There are at most three tigers.或ৱ ‘There are no more than three tigers
ሺሺT ר T ר T ר Tሻ ՜ ሺ ൌ ש ൌ ש ൌ ש ൌ ש ൌ ש ൌ ሻሻ
理方法(不ଜੌ):
There are at most three tigers
1,߄ n 个৯ variable 就写(n+1)个,࠴ޞୌ߄ 4 个 variable
2,他们ୃީ৯
3,他们互बबঈ
Exactly n 确定有 n 个
੯方法 1: Exactly n = at least n ר at most n
There is (exactly) one tiger.
T ר ሺሺT ר Tሻ ՜ ൌ ሻ
24
There are (exactly) two tigers.
ሺT ר T ר ൌ ሻ ר ሺሺT ר T ר Tሻ ՜ ሺ ൌ ש ൌ ש ൌ ሻሻ
There are (exactly) three tigers.
ሺT ר T ר T ר ൌ ר ൌ ר ൌ ሻ ר ሺሺT ר T ר T ר Tሻ ՜ ሺ ൌ ש
ൌ ש ൌ ש ൌ ש ൌ ש ൌ ሻሻ
੯方法 2:
There is (exactly) one tiger.
ሺT ՞ ൌ ሻ
There are (exactly) two tigers.
ሺ ് ר ሺT ՞ ሺ ൌ ש ൌ ሻሻሻ
There are (exactly) three tigers
ሺ ് ר ് ר ് ר ሺT ՞ ሺ ൌ ש ൌ ש ൌ ሻሻሻ
理方法:
There are (exactly) three tigers
1, ߄几个৯就加几个 variable, ୌ存在 3 个 tiger=ୌ߄ tiger x, y 和 z
2, x 和 y 和 z 互不बঈ
3, 对于▲个ޏङ tiger ߛપ,它么ީ tiger x 么ީ tiger y 么ީ tiger z
个人建ઑ大家在书上ऀ光ঃ划出ߛ,ৰડङޞ候৬र
Final 复习 tips
把 Cheat Sheet 抄在书上ॱग处
搞懂刷ொࣰङ所߄ࢵଐ߄ cheat sheet 上所߄内容
ਙ少做完所߄୍ࢵொऩ(Chapter 12 &13),߄可以做▲下之前ङર后ொ打基ॅ
做 Tree ઓ得打勾勾,不ঠ心
发୯汮大ࣦ
અઅ大家!!!䮾大家 HD!!!
25
Phil1012 Final Cheat sheet
此材料为参考使用漕考嫖前嫸务必将其誊写在嫿本内空白处漕切勿直接夹带廜考场漕一经发现算作弊处理漕后果
自尠漕与材料提供者无关。
Truth table 书 P50 做 Model 时参考
GPL 翻译套峰
Existential quantification x
当林目中出现漡 something, anything漏否定情况的 something 此时用 坧䠹漐澿澳濴瀇澳濿濸濴瀆瀇澳瀂瀁濸澿澳瀇濻濸瀅濸澳濼瀆澳濴瀖濁时使用
【僲少有一个是】
Universal quantification: x
当林目中出现漡all, everything, anything(坧瀡任何瀢),,䩈时使用 【每个忼是】
ܠ对应՜ ܠ对应ר
属性之擳用ר廝接 的括号内只俼出现一个՜
例林漡12.1.9 䨫 4 林(viii) Frank prefers any song recorded by the Rolling Stones to any song recorded by Elvis.
ሺሺS ר Rr ר S ר Reሻ ՜ P fሻ
Someone, anyone, everyone 䩈廘䭺特指人的墀加一个 Px: x is a person
例林漡8.3.5 䨫 4 林 If someone is sad, then not everyone is happy.
ሺP ר Sሻ ՜ ሺP ՜ Hሻ
翻译第一步先写 Glossary
★VRPH-VRPH 仺嫐方法漡
1, 找出漓漓一屶放在句子的前晡
2, 把修桯嫌用 ר 廝接
如漡SRPH cKLKXaKXaV aUH bLJJHU WKaQ VRPH bHaJOHV.
ሺC ר B ר Bሻ
★aQ\-aQ\ 漏HYHU\-HYHU\漐方法
1 䝭定 TXaQWLILHU 是漓把[放在句子前晡
2漓 把两个物体之擳散了主墀关䯺嫌的修桯嫌放在一屶用 ר 廝接
3. 把主墀关䯺嫌放句末漓用 ĺ 廝接
26
如漡TKH bLJJHU WKH GRJ, WKH KaSSLHU LW LV.
ሺሺD ר D ר B ՜ Hሻሻ
★VRPH-aQ\/ aQ\-VRPH 仺嫐方法
1漓䝭定 [ 和 \ 前晡的 TXaQWLILHU, (是[廗是?)
2漓找出两个物体的修桯嫌
3漓根据 TXaQWLILHU 加廝接嫌漓主墀关䯺一儫在 \ 后晡坧䠹漏如果是 [ 和 \ 之擳的关䯺就放在 \ 后晡漓如果是 [
和 ] 的关䯺就在 ] 后晡漐
如漡EYHU\ SHUVRQ RZQV a GRJ.
ሺP ՜ ሺD ר Oሻሻ
GPLI 翻译套峰
一儫的 GPLI 仺嫐同 GPL漓如果林目中出现
最槗䶦仺嫐方法漡
1漓左庸写最槗䶦物品的所有修桯嫌
2漓右庸三忧分漡 a) 右庸物品的所有修桯嫌
b) 右庸不䩈于左庸
c) 主墀比庂䶦漓用 TXaQWLILHU 对应的䨥号廝接漏用՜漓用ר漐
一儫 TXaQWLILHU 忼是漓因为最槗䶦是对于所有东塾忼成䧊的
如漡13.2.2 䨫 16 林 DLaQH LV WKH WaOOHVW ZRPaQ.
GORVVaU\: W[: [ LV a ZRPaQ.
G: DLaQH \: [ LV WaOOHU WKaQ \
Wd ר ሺሺW ר ് dሻ ՜ Tdሻ
GPLI Model 枙型一
1漓䷘ PRGHO 和 SURSRVLWLRQ漓叫你判断 T RU F
要点漣对于ܠ漕我们找 TUXe漕即找出一个满岴条件的 YaULabOe
对于ܠ漕我们找 FaOVe漕即找出一个不满岴条件的 YaULabOe
例林漡13.3.1 䨫 2 林
漏L漐 ሺRa ר Rbሻ
壢林方法 1漕判断是ܠ廙是ܠ
对于漓找出一个满岲条件的 YaULabOH漓嫤 SURSRVLWLRQ 即是 TUXH
2漕找一个 [ 满岴所有的 ר
Ra[ 为 TUXH 伋且Rb为 TUXH [即 Rb为 FaOVH ]
3漕缩小范围漕写出要想每一步条件满岴的取值
27
当 Ra[ 为 TUXH 时漓[=2, 当Rb为 FaOVH 时漓[2
4漕判断是否存在廚样的 [漕如果存在一个满岴所有条件的 [,嫦 SURSRVLWLRQ 则是 TUXe漕否则就是
FaOVe
因为无 [ 䨥合嫤条件漓所以嫤 SURSRVLWLRQ 为 FaOVH
(LL) ሺሺF ר Gሻ ՜ ൌ cሻ
壢林方法 1漕判断是ܠ廙是ܠ
对于漓找出一个不䨥合条件的 YaULabOH漓嫤 SURSRVLWLRQ 即是 FaOVH
2漕看箭头方向的左右漕如果想要 FaOVe,箭头左庺是 T 箭头右庺是 F
媨F ר Gሻ为 TUXH 且 ൌ c为 FaOVH
3漕缩小范围漕写出要想每一步条件满岴的取值
当ሺF ר Gሻ为 T 时漓F 为 TUXH 且G为 TUXH [即 G[ 为 FaOVH]
F 为 TUXH Î [=1, 2, 3«.
G[ 为 FaOVH Î [=2, 4, 6«
当 ൌ c 时漓[c Î [2
4漕判断是否存在廚样的 [漕如果存在一个满岴所有条件的 [,嫦 SURSRVLWLRQ 则是 TUXe漕否则就是
FaOVe
因此无 [ 满岲以上全忧墀求漓嫤 SURSRVLWLRQ 为 TUXH (找不出一个 [ 媨廘句嫜不对)
GPLI Model 枙型二
廚惍的 PRdeO 如果 SURSRVLWLRQ 惍晣没有小写字母 abcd 等 QaPe 出现漕就不写 UefeUeQW
No such model 䠹刂
如漡墀求 F ר Faa的 True model
(a) No such model. For the proposition to be true, ( the ordered pair consisting of the referent of a, followed by the
referent of a, would have to both be in the extension of F and not in the extension of F) 把想要让廚个 model be
true 的要求写一彎
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GPLI Tree rules P325
写完一个打√漈漈打漈漈漈勾漈漈漈勾漈漈漈
检查的时候检查有没有忾打勾漊漊漊
Read off Trees
1. 先写 Referent, tree 中出现几个小写字母就写几个 referent,
并䷘他们届值漡
RHIHUHQWV: a: 1, b: 2, c: 3
2. 写 Domain, 是所有 referent 的旅合
DRPaLQ: ^1, 2, 3`
3. 最后䘊 Tree 的任意一个 open path漓找到大写字母漓如 Rab,
Rba 为 True, 根据廘个写 Extension
Counter example 一定墀写 Referent!!!
Tree 的用法
主墀伂 Validity, satisfiability 和 logical truth
Logical truth 和 tautology 用法䗷同漓操作时忼是在前晡加
一个
Saturate paths
即一定墀把所有的情况忼代入漓比如已䛤 Rab 和 a=b漓在 tree 中一定墀把 Raa, Rba, Rbb 廘三䣌情况忾写廜去漕
缺一不可
什么时候 close:
Ixhxh 7 inthxha
ta 7⽐GN
ha thx
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1, 䘊到廘个式子和它的 negation (之前媱廆)
2漓䘊到 a≠a 时
翻译小 tips
If A then B -----澻濔澳→澳濕澼
A, if B ---------澻濕澳→澳濔澼
Only if A, B---- 澻濕澳→澳濔澼澳 澳 Only if 后晣接结果漊漊漊漊
例林漡8.2.1 13 林 Jenny is sailing only if both Mary and John are. Ae ՜ ሺAm ר Ajሻ
A, only if B -----澻濔澳→澳濕澼
A, if and only if B-----(A ՞ B)
Unless=if not
例林漡8.3.5 䨫 15 林 Everyone is in trouble unless Gary is happy. Hg ՜ ሺP ՜ Tሻ
Neither nor 坧庽方式漡 ܣ ר ܤ或伄ሺA ש Bሻ
BXW, TKRXJK 出现用 ר 廝接
other than Ben 坧䠹 x് Ben
例林 13.2.2 䨫 22 林 I know people other than Ben. ሺP ר Kn ר ് bሻ
Nothing =不存在瀂至少有一个满岴瀃的情况= 瀚x
No one 坧庽方式漡 1濁澳瀹x漏Pxר一切属性漐
2. x漏Px՜一切䗷反的属性漐
属性之擳用ר廝接坧䠹并列
例林 12.1.9 䨫 4 林(ii)漡No one admires Dave. ሺP ר Adሻ
句子中出现瀡if, then瀢一定是→漓就䪖是 someone 也墀改成
例林 8.3.2 䨫 20 林 II VRPHWKLQJ LV cHUWaLQ, WKHQ LW¶V SURbabOH. ሺC ՜ Pሻ
Only 两䣌坧庽方式
1漓ሺሺ廘个东塾的一切属性漓用 ר 廝接ሻ ՞ ൌ 廘个东塾ሻ
例林 13.2.2 䨫 24 林: The only happy person I know is Ben.
ሺሺP ר H ר Knሻ ՞ ൌ bሻ
2漓廘个东塾的属性 ר ሺሺP ר ് 廘个东塾ሻ ՜ 廘个东塾的属性ሻ
例林 13.2.2 䨫 15 林 Jonathon is the only person who watches Family Guy.
Vjf ר ሺሺP ר ് jሻ ՜ Vfሻ
大家加油漊祝大家 HD漊漊漊
Orderoftrees