廣義量詞系列：迭代量詞

2. 迭代<−,1>型量詞

John loves Mary.
John introduced Tom to everybody.
John offered something to Mary in exchange for Fido.

(John ... Mary)(−, −)(LOVE)
(John ... Tom ... everybody)(−, −, −)(INTRODUCE)
(John ... something ... Mary ... Fido)(−, −, −, −)(OFFER IN EXCHANGE FOR)

<<−,−>,2>
(John ... Mary)(−, −)(A)
A(j, m)
<<−,−,−>,3>
(John ... Tom ... everybody)(−, −, −)(A)
{x: A(j, t, x)} = U 或
PERSON ⊆ {x: A(j, t, x)}
<<−,−,−,−>,4>
(John ... something ... Mary ... Fido)(−, −, −, −)(A)
{x: A(j, x, m, f)} ≠ Φ 或
THING ∩ {x: A(j, x, m, f)} ≠ Φ

3. 迭代<1,1>型量詞

3.1 真值條件

「迭代<1,1>型量詞」除了「謂詞性論元」外，還有「名詞性論元」。此外，這類量詞的真值條件也 須反映所含限定詞的真值條件。當「迭代量詞」只含有一個限定詞時，情況較為簡單。舉例說，以下兩句

Every student likes Tom.
John introduced a boy to Mary.

(every ... Tom)(STUDENT, −)(LIKE)
(John ... a ... Mary)(−, BOY, −)(INTRODUCE)

<<1,−>,2>
(every ... Tom)(A, −)(B)
A ⊆ {x: B(x, t)}
<<−,1,−>,3>
(John ... a ... Mary)(−, A, −)(B)
A ∩ {x: B(j, x, m)} ≠ Φ

Every boy loves some girl.     (1)
The teacher recommended exactly one book to every student.

(在通常意義下)可以翻譯為

(every ... some)(BOY, GIRL)(LOVE)
(the ... exactly 1 ... every)(TEAHCER, BOOK, STUDENT)(RECOMMEND)

<<1,1>,2>
(every ... some)(A, B)(C)
A ⊆ {x: B ∩ {y: C(x, y)} ≠ Φ}
<<1,1,1>,3>
(the ... exactly 1 ... every)
(A, B, C)(D) (A為單數)
X ∩ A ⊆ {x: |B ∩ {y: C ⊆ {z: D(x, y, z)}}| = 1},
if |X ∩ A| = 1

3.2 轄域歧義

(every ... some)(BOY, GIRL)(LOVE) ⇔ BOY ⊆ {x: GIRL ∩ {y: LOVE(x, y)} ≠ Φ}     (2)

(some ... every)(GIRL, BOY)(LOVE−1) ⇔ GIRL ∩ {y: BOY ⊆ {x: LOVE(x, y)}} ≠ Φ     (3)

3.3 語義模型1

 語義模型1：設「論域」U = {a, c, e, j, m, r, l}，並有以下定義： a = Anne；c = Charles；e = Emily；j = John；m = Mary； BOY = {c, j}；GIRL = {a, e, m}；ROSE = {r}；LILY = {l}； LIKE = {(a, l), (c, l), (c, r), (j, r)}；HATE = {(e, l), {j, l), (m, r)}； LOVE = {(j, m), (c, a), (c, e), (e, c)}；INTRODUCE = {(a, m, c), (a, e, c)}

GIRL ∩ {y: LOVE(x, y)} ≠ Φ     (4)

BOY ⊆ {x: LOVE(x, y)}     (5)

4. 迭代結構化量詞

More boys than girls like the rose.
More boys like the rose than the lily.
More boys like than hate the rose.
More boys like the rose than girls the lily.
More boys like the rose than girls hate the lily.     (6)

John bought more roses than lilies.
At least three girls gave more roses than lilies to John.

(more ... than ... the)((BOY, GIRL), ROSE)(LIKE)
(more ... than ... the ... the)(BOY, (ROSE, LILY))(LIKE)
(more ... than ... the)(BOY, ROSE)(LIKE, HATE)
(more ... than ... the ... the)((BOY, GIRL), (ROSE, LILY))(LIKE)
(more ... than ... the ... the)((BOY, GIRL), (ROSE, LILY))(LIKE, HATE)
(John ... more ... than ...)(−, (ROSE, LILY))(BUY)
(at least 3 ... more ... than ... John)(GIRL, (ROSE, LILY), −)(GIVE)

<<12,1>,2>
(more ... than ... the)
((A, B), C)(D)
|A ∩ {x: X ∩ C ⊆ {z: D(x, z)}}| >
|B ∩ {y: X ∩ C ⊆ {z: D(y, z)}}|, if |X ∩ C| = 1
<<1,12>,2>
(more ... than ... the ... the)
(A, (B, C))(D)
|A ∩ {x: X ∩ B ⊆ {z: D(x, z)}}| >
|A ∩ {x: X ∩ C ⊆ {w: D(x, w)}}|, if |X ∩ B| = 1 ∧ |X ∩ C| = 1
<<1,1>,22>
(more ... than ... the)
(A, B)(C, D)
|A ∩ {x: X ∩ B ⊆ {z: C(x, z)}}| >
|A ∩ {x: X ∩ B ⊆ {z: D(x, z)}}|, if |X ∩ B| = 1
<<12,12>,2>
(more ... than ... the ... the)
((A, B), (C, D))(E)
|A ∩ {x: X ∩ C ⊆ {z: E(x, z)}}| >
|B ∩ {y: X ∩ D ⊆ {w: E(y, w)}}|, if |X ∩ C| = 1 ∧ |X ∩ D| = 1
<<12,12>,22>
(more ... than ... the ... the)
((A, B), (C, D))(E, F)
|A ∩ {x: X ∩ C ⊆ {z: E(x, z)}}| >
|B ∩ {y: X ∩ D ⊆ {w: F(y, w)}}|, if |X ∩ C| = 1 ∧ |X ∩ D| = 1
<−,12>,2>
(John ... more ... than ...)
(−, (A, B))(C)
|A ∩ {x: C(j, x)}| > |B ∩ {y: C(j, y)}|
<<1,12,−>,3>
(at least 3 ... more ... than ... John)
(A, (B, C), −)(D)
|A ∩ {x: |B ∩ {y: D(x, y, j)}| > |C ∩ {z: D(x, z, j)}|}| ≥ 3

|BOY ∩ {x: X ∩ ROSE ⊆ {z: LIKE(x, z)}}| > |GIRL ∩ {y: X ∩ LILY ⊆ {w: HATE(y, w)}}|,
if |X ∩ ROSE| = 1 ∧ |X ∩ LILY| = 1     (7)

X ∩ ROSE ⊆ {z: LIKE(x, z)}     (8)

|BOY ∩ {x: X ∩ ROSE ⊆ {z: LIKE(x, z)}}| = |{c, j}| = 2

|GIRL ∩ {y: X ∩ LILY ⊆ {w: HATE(y, w)}}| = |{e}| = 1

More boys than girls recommended the book to the librarian.
More boys recommended the book than the CD to the librarian.
More boys recommended the book to the librarian than to the teacher.
More boys than girls offered their coupons to the shop in exchange for a gift.
......

(more ... than ... the ... the)((A, B), C, D)(E) ⇔ |A ∩ {x: X ∩ C ⊆ {z: X ∩ D ⊆ {w: E(x, z, w)}}| >
|B ∩ {y: X ∩ C ⊆ {z: X ∩ D ⊆ {w: E(y, z, w)}}|, if |X ∩ C| = 1 ∧ |X ∩ D| = 1

5. 迭代模糊量詞

 exp(−((|A ∩ B| − 78)/2)2), if |A ∩ B| < 78 μ[TRUTH]((about 80)(A)(B)) = 1, if 78 ≤ |A ∩ B| ≤ 82     (9) exp(−((|A ∩ B| − 82)/2)2), if |A ∩ B| > 82

 μ[TRUTH](almost every)(A)(B)) = 0, if |A ∩ B| / |A| < 0.67 (10) exp(−((|A ∩ B| / |A| − 1)/0.33)2), if |A ∩ B| / |A| ≥ 0.67

5.1 半模糊量詞的模糊化

About 80 students passed all exams.     (11)

μ[TRUTH]((about 80)(STUDENT)({x: EXAM ⊆ {y: PASS(x, y)}}))

Almost every boy met about 80 girls.     (12)

μ[TRUTH]((almost every)(BOY)({x: (about 80)(GIRL)({y: MEET(x, y)})}))

Bα = {x ∈ U: μ[B](x) ≥ α}

B0.5 = {a, b}

A = {α1/a1, ... αn/an}

μ[TRUTH](Q*(A)(B)) = Σ0 ≤ i ≤ n μ[TRUTH](Q(A)(Bαi)) × (αi − αi+1) (註4)     (13)

5.2 語義模型2

 x |GIRL ∩ {y: MEET(x, y)}| a b c d 77 85 75 100

B = {x: (about 80)(GIRL)({y: MEET(x, y)})} = {0.78/a, 0.11/c, 0.11/b, 0/d}     (14)

iαiBαiμ[TRUTH]((almost every)(BOY)(Bαi))αi − αi+1
0
1
Φ
0
0.22
1
0.78
{a}
0
0.67
2
0.11
{a, b, c}
0.56
0
3
0.11
{a, b, c}
0.56
0.11
4
0
{a, b, c, d}
1
0
5
0

0 × 0.22 + 0 × 0.67 + 0.56 × 0 + 0.56 × 0.11 + 1 × 0 = 0.06

5.3 "some"的模糊化

「模糊量詞」也可出現於另一「非模糊量詞」的「轄域」內。試看以下例句：

Some boy met about 80 girls.     (15)

⇔ BOY ∩ {x: (about 80)(GIRL)({y: MEET(x, y)})} ≠ Φ

some(A)(B) ⇔ ∃x ∈ A B(x) ⇔ B(a) ∨ B(b) ∨ B(c) (註5)

some(A)(B) ⇔ ∃x ∈ A B(x) ⇔ ∨x ∈ A B(x)     (16)

μ[TRUTH](∨x ∈ A x) = maxx ∈ A μ[TRUTH](x)

μ[TRUTH](some*(A)(B)) = maxx ∈ A μ[TRUTH](B(x)) = maxx ∈ A μ[B](x) (註7)

maxx ∈ BOY μ[{x: (about 80)(GIRL)({y: MEET(x, y)})}](x)     (17)

iαiBαi μ[TRUTH](some(BOY)(Bαi))αi − α i+1
0
1
Φ
0
0.22
1
0.78
{a}
1
0.67
2
0.11
{a, b, c}
1
0
3
0.11
{a, b, c}
1
0.11
4
0
{a, b, c, d}
1
0
5
0

0 × 0.22 + 1 × 0.67 + 1 × 0 + 1 × 0.11 + 1 × 0 = 0.78

5.4 其他非模糊量詞的模糊化

every(A)(B) ⇔ ∀x ∈ A B(x) ⇔ ∧x ∈ A B(x)

μ[TRUTH](∧x ∈ A x) = minx ∈ A μ[TRUTH](x)

μ[TRUTH](every*(A)(B)) = minx ∈ A μ[B](x)

(at least 2)(A)(B) ⇔ ∃x, y ∈ A (x ≠ y ∧ B(x) ∧ B(y))

μ[TRUTH]((at least 2)*(A)(B)) = minimax2x ∈ A μ[B](x)

min({0.78, 0.11, 0.11, 0}) = 0
minimax2({0.78, 0.11, 0.11, 0}) = 0.11

no(A)(B)
~some(A)(B)
only(A)(B)
every(B)(A)
(more than n)(A)(B)
(at least n + 1)(A)(B)
(at most n)(A)(B)
~(more than n)(A)(B)
(fewer than n)(A)(B)
~(at least n)(A)(B)
(exactly n)(A)(B)
(at least n)(A)(B) ∧ (at most n)(A)(B)
(all except n)(A)(B)
(exactly n)(A)(~B)
(at least q)(A)(B) (q為分數)
(at least m)(A)(B)

the(A)(B) (A為單數)
every(X ∩ A)(B), if |X ∩ A| = 1
John(−)(B)
every({j})(B)

μ[TRUTH](no*(A)(B)) = 1 − maxx ∈ A μ[B](x)

~B = {(1 − 0.78)/a, (1 − 0.11)/c, (1 − 0)/b, (1 − 0)/d} = {0.22/a, 0.89/c, 1/b, 1/d}

μ[TRUTH]((exactly n)*(A)(B)) = min({μ[TRUTH]((at least n)*(A)(B)), μ[TRUTH]((at most n)*(A)(B))})

μ[TRUTH]((exactly n)*(A)(B)) = min({minimaxnx ∈ A μ[B](x), 1 − minimaxn+1x ∈ A μ[B](x)})

6. 迭代疑問量詞

6.1 多重疑問句

Who gave what to whom?
Which boys love which girls?

(who ... what ... whom)(−, −, −)(GIVE)
(which ... which)(BOY, GIRL)(LOVE)

「單式疑問量詞」的「解答論元」相當於一個由個體組成的集合。當我們把n個「單式疑問量詞」組合成「迭代 疑問量詞」時，該「迭代疑問量詞」的「解答論元」便是一個由「有序n元組」組成的集合。舉例說，上面的 "(who ... what ... whom)"的「解答論元」便是由「有序三元組」組成的集合，例如(j, b, m) (其中 j、b和m分別代表"John"、"Bible"和"Mary")便可以是這個集合的元素之一。這一處理方式是合理的，因為當我 們問"Who gave what to whom?"時，我們期待的答案正是三個個體x、y和z，其中x把y送給z。下表列出上述「 迭代疑問量詞」的真值條件(以下為簡明起見，略去疑問句的「預設」)：

<<−,−,−>,3,3>
(who ... what ... whom)(−, −, −)(A)(B)
B = (PERSON × THING × PERSON) ∩ A
<<1,1>,2,2>
(which ... which)(A, B)(C)(D)
D = X2 ∩ (A × B) ∩ C

6.2 混合陳述-疑問量化句

「疑問量詞」亦可以與「陳述量詞」組成「混合陳述-疑問量化句」(Mixed Declarative-Interrogative Quantified Sentence)。舉例說，我們可以把以下兩句

Who likes the teacher?
Which boys introduced which girls to every guest?

(who ... the)(−, TEACHER)(LIKE)
(which ... which ... every)(BOY, GIRL, GUEST)(INTRODUCE)

<<−,1>,2,1>
(who ... the)(−, A)(B)(C)
C = PERSON ∩ {x: X ∩ A ⊆ {y: B(x, y)}},
if |X ∩ A| = 1
<<1,1,1>,3,2>
(which ... which ... every)(A, B, C)(D)(E)
E = X2 ∩ (A × B) ∩ {(x, y): C ⊆ {z: D(x, y, z)}}

Which girl does John love?

(which ... John)(GIRL, −)(LOVE−1)

(which ... John)(A, −)(B−1)(C) ⇔ C = X ∩ A ∩ {x: B(j, x)}

6.3 疑問句的歧義問題

Which girl does every boy love?

Every boy loves some girl.

<<1,1>,2,1>
(which ... every)(B, A)(C−1)(D)
D = X ∩ B ∩ {x: A ⊆ {y: C(y, x)}}

<<1,1>,2,2>
(every ... which)(A, B)(C)(D)
D = (A × (X ∩ B)) ∩ C

Which boys did two dogs bite? (註10)

D = {(a, d), (b, e)} ∨ D = {(a, d), (c, e), (c, f)} ∨ D = {(b, e), (c, e), (c, f)}

<<1,1>,2,1>
(which ... exactly 2)(B, A)(C−1)(D)
D = X ∩ B ∩ {x: |A ∩ {y: C(y, x)}| = 2}

<<1,1>,2,2>
(exactly 2 ... which)(A, B)(C)(D)
E ⊆ A ∧ |E| = 2 [D = (E × (X ∩ B)) ∩ C]

Which boys did which dogs bite?
Which books did most boys read?

7. 定語分句

「定語分句」(Attributive Clause)是指在句中作「定語」(即名詞修飾語)的分句。根據所含謂語的 形式，「定語分句」可分為「關係分句」(Relative Clause)(又稱「限定分句」Finite Clause)、「非限定分 句」(Non-Finite Clause)和「無動詞分句」(Verbless Clause)三種(有關這些分句的介紹請參閱拙文 《論「語結」》)。「定語分句」的語義功能類似「相交形容詞」，可以處理成 集合，並且與代表名詞中心語的集合構成「交集」。不過，由於「定語分句」的內部結構類似一個句子，它本 身也可包含各種量詞。下表列出一些「定語分句」及其集合論表達式：

whom some girl loves
{x: GIRL ∩ {y: LOVE(y, x)} ≠ Φ}

loved by John
{x: LOVE(j, x)}

fond of more than one activity
{x: |ACTIVITY ∩ {y: FOND-OF(x, y)}| > 1}

Every boy is happy.
BOY ⊆ HAPPY(e(X'))
Anne introduced the girl to Charles.
X ∩ GIRL ⊆ {x: INTRODUCE(a, x, c)},
if |X ∩ GIRL| = 1
No member showed up.
MEMBER ∩ SHOW-UP = Φ

Every boy whom some girl loves is happy.
(BOY ∩ {x: GIRL ∩ {y: LOVE(y, x)} ≠ Φ})
⊆ HAPPY(e(X'))
Anne introduced the girl loved by John to Charles.
(GIRL ∩ {x: LOVE(j, x)}) ⊆ {x: INTRODUCE(a, x, c)},
if |GIRL ∩ {x: LOVE(j, x)}| = 1
No member fond of more than one activity showed up.
(MEMBER ∩ {x: |ACTIVITY ∩ {y: FOND-OF(x, y)}| > 1})
∩ SHOW-UP = Φ

|GIRL ∩ {x: LOVE(j, x)}| = |{m}| = 1

{x: INTRODUCE(a, x, c)} = {m, e}

∃x(A(x) ∧ B(x))

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