# 廣義量詞系列：量詞的迭代運算

## 1. 引言

《廣義量詞系列：迭代量詞》中，筆者討論了多種「迭代量詞」，這些「迭代量詞」的共通點是，它們都是由「單式量詞」組合而成的。可是，利用以往介紹的「單式量詞」，我們可以組合出無數的「迭代量詞」。如要一一列出這些「迭代量詞」的真值條件，這將是費時失事的。事實上，從上述網頁的表1至表4，讀者應能看到，「迭代量詞」的真值條件其實是從某些「單式量詞」的真值條件組合而成的。舉例說，上述網頁表3第二行的「迭代量詞」(the ... exactly 1 ... every)(A, B, C)(D)便是由"the"、"(exactly 1)"和"every"這三個「單式量詞」組合而成的，而這個「迭代量詞」的真值條件

X ∩ A ⊆ {x: |B ∩ {y: C ⊆ {z: D(x, y, z)}}| = 1}, if |X ∩ A| = 1     (1)

## 2. 減元算子

### 2.1 <1>型量詞與減一元算子的概念

John(−)(SING)

every(BOY)(SING)
(more ... than ...)(BOY, GIRL)(SING)

### 2.2 減一元算子的形式化定義

Ry1, ... yn = {xn+1: R(y1, ... yn, xn+1)}     (2)

{(x1, ... xn+1): R(x1, ... xn+1)}y1, ... yn = {xn+1: R(y1, ... yn, xn+1)}     (3)

Q(#)(R) = {(y1, ... yn): Q(#)(Ry1, ... yn)}，其中y為異於x的變項     (4)

## 3. <−,1>型量詞的迭代

It(Q', Q)(#', #)(R)

j ∈ {x: A(x)} ⇔ A(j)     (5)

 everybody(−)(A) = {(y1, y2): everybody(−)(Ay1, y2)} (根據(4)) = {(y1, y2): everybody(−)({x3: A(y1, y2, x3)})} (根據(2)) = {(y1, y2): PERSON ⊆ {x3: A(y1, y2, x3)}} ("everybody"的真值條件)

 Tom(−)(everybody(−)(A)) = {z1: Tom(−)({(y1, y2): PERSON ⊆ {x3: A(y1, y2, x3)}}z1)} (根據(4)) = {z1: Tom(−)({y2: PERSON ⊆ {x3: A(z1, y2, x3)}})} (根據(3)) = {z1: t ∈ {y2: PERSON ⊆ {x3: A(z1, y2, x3)}}} (專有名詞的真值條件) = {z1: PERSON ⊆ {x3: A(z1, t, x3)}} (根據(5))

 John(−)(Tom(−)(everybody(−)(A))) ⇔ j ∈ {z1: PERSON ⊆ {x3: A(z1, t, x3)}} (專有名詞的真值條件) ⇔ PERSON ⊆ {x3: A(j, t, x3)}} (根據(5))

John(−)(Tom(−)(everybody(−)(A))) ⇔ (John ... Tom ... everybody)(−, −, −)(A)

## 4. <1,1>型量詞的迭代

 every(C)(D) = {(y1, y2): every(C)(Dy1, y2)} (根據(4)) = {(y1, y2): every(C)({x3: D(y1, y2, x3)})} (根據(2)) = {(y1, y2): C ⊆ {x3: D(y1, y2, x3)}} ("every"的真值條件)

 (exactly 1)(B)(every(C)(D)) = {z1: (exactly 1)(B)({(y1, y2): C ⊆ {x3: D(y1, y2, x3)}}z1)} (根據(4)) = {z1: (exactly 1)(B)({y2: C ⊆ {x3: D(z1, y2, x3)}})} (根據(3)) = {z1: |B ∩ {y2: C ⊆ {x3: D(z1, y2, x3)}}| = 1} ("(exactly 1)"的真值條件)

 the(A)((exactly 1)(B)(every(C)(D))) ⇔ X ∩ A ⊆ {z1: |B ∩ {y2: C ⊆ {x3: D(z1, y2, x3)}}| = 1}, if |X ∩ A| = 1 ("the"的真值條件)

the(A)((exactly 1)(B)(every(C)(D))) ⇔ (the ... exactly 1 ... every)(A, B, C)(D)

## 5. 結構化量詞的迭代

(more ... than ...)(A)(the(B)(C), the(B)(D))

 the(B)(C) = {y1: the(B)(Cy1)} (根據(4)) = {y1: the(B)({x2: C(y1, x2)})} (根據(2)) = {y1: X ∩ B ⊆ {x2: C(y1, x2)}}, if |X ∩ B| = 1 ("the"的真值條件)

the(B)(D) = {y1: X ∩ B ⊆ {x2: D(y1, x2)}}, if |X ∩ B| = 1

(more ... than ...)(A)(B, C) ⇔ |A ∩ B| > |A ∩ C|     (6)

 (more ... than ...)(A)(the(B)(C), the(B)(D)) ⇔ |A ∩ {y1: X ∩ B ⊆ {x2: C(y1, x2)}}| > |A ∩ {y1: X ∩ B ⊆ {x2: D(y1, x2)}}|, if |X ∩ B| = 1 (根據(6))

(more ... than ...)(A)(the(B)(C), the(B)(D)) ⇔ (more ... than ... the)(A, B)(C, D)

## 7. 模糊量詞的迭代

### 7.1 兩個量詞的迭代

μ[TRUTH](Q*(A)(B)) = Σ0 ≤ i ≤ n μ[TRUTH](Q(A)(Bαi)) × (αi − αi+1)     (7)
μ[TRUTH](some*(A)(B)) = maxx ∈ A μ[B](x)     (8)

Some boy met about 80 girls.     (9)

(some* ... (about 80))(BOY, GIRL)(MEET) (註4)

maxx ∈ BOY μ[{y1: (about 80)(GIRL)({x2: MEET(y1, x2)})}](x)

### 7.2 三個量詞的迭代

Some boy introduced almost every girl to about 80 visitors.     (10)

(some* ... (almost every)* ... (about 80))(B, G, V)(I)

(almost every)*(G)(about 80(V)(I)) = {z1: (almost every)*(G)({y2: (about 80)(V)({x3: I(z1, y2, x3)})})}

some*(B)({z1: (almost every)*(G)({y2: (about 80)(V)({x3: I(z1, y2, x3)})})})     (11)

 (z, y) |V ∩ {x: I(z, y, x)}| (a, c) (a, d) (a, e) (b, c) (b, d) (b, e) 77 81 84 20 79 88

 exp(−((|A ∩ B| − 78)/2)2), if |A ∩ B| < 78 μ[TRUTH]((about 80)(A)(B)) = 1, if 78 ≤ |A ∩ B| ≤ 82     (12) exp(−((|A ∩ B| − 82)/2)2), if |A ∩ B| > 82

 μ[TRUTH](almost every)(A)(B)) = 0, if |A ∩ B| / |A| < 0.67 (13) exp(−((|A ∩ B| / |A| − 1)/0.33)2), if |A ∩ B| / |A| ≥ 0.67

B = {y2: (about 80)(V)({x3: I(a, y2, x3)})} = {0.78/c, 1/d, 0.37/e}

iαiBαiμ[TRUTH]((almost every)(G)(Bαi))αi − αi+1
0
1
{d}
0
0
1
1
{d}
0
0.22
2
0.78
{c, d}
0.37
0.41
3
0.37
{c, d, e}
1
0.37
4
0

μ[TRUTH]((almost every)*(G)(B)) = 0 × 0 + 0 × 0.22 + 0.37 × 0.41 + 1 × 0.37 = 0.52     (14)

C = {y2: (about 80)(V)({x3: I(b, y2, x3)})} = {0/c, 1/d, 0/e}

μ[TRUTH]((almost every)*(G)(C)) = 0     (15)

D = {z1: (almost every)*(G)({y2: (about 80)(V)({x3: I(z1, y2, x3)})})} = {0.52/a, 0/b}

maxx ∈ B μ[D](x) = max({0.52, 0}) = 0.52

## 8. 格擴充算子

### 8.1 格擴充算子的概念

Ry1, ... yi−1, yi+1, ... yn+1 = {xi: R(y1, ... yi−1, xi, yi+1, ... yn+1)}     (16)

{(x1, ... xn+1): R(x1, ... xn+1)}y1, ... yi−1, yi+1, ... yn+1 = {xi: R(y1, ... yi−1, xi, yi+1, ... yn+1)}     (17)

[Q(#)]i(R) = {(y1, ... yi−1, yi+1, ... yn+1): Q(#)(Ry1, ... yi−1, yi+1, ... yn+1)}，其中y為異於x的變項     (18)

### 8.2 轄域歧義的表達

「格擴充算子」可用來表達某些「轄域歧義」問題。筆者在《廣義量詞系列：迭代量詞》中曾指出，語句

Every boy loves some girl.

every(BOY)([some(GIRL)]2(LOVE))

some(GIRL)([every(BOY)]1(LOVE)) (註6)

 [every(BOY)]1(LOVE) = {y2: every(BOY)(LOVEy2)} (根據(18)) = {y2: every(BOY)({x1: LOVE(x1, y2)})} (根據(16)) = {y2: BOY ⊆ {x1: LOVE(x1, y2)}} ("every"的真值條件)

 some(GIRL)([every(BOY)]1(LOVE)) ⇔ GIRL ∩ {y2: BOY ⊆ {x1: LOVE(x1, y2)}} ≠ Φ ("some"的真值條件)

some(GIRL)([every(BOY)]1(LOVE)) ⇔ (some ... every)(GIRL, BOY)(LOVE−1)

### 8.3 特殊句式的表達

There is a girl whom every boy loves. (存在句)

a(GIRL)([every(BOY)]1(LOVE))

John把一本書推薦給每名學生。

John向每名學生推薦一本書。

John(−)([every(STUDENT)]3([a(BOOK)]2(RECOMMEND)))

## 9. 疑問量詞的迭代

### 9.1 多重疑問句

[Q(#)]i(R) = {(y1, ... yi−1, Bi, yi+1, ... yn+1): Q(#)(Ry1, ... yi−1, yi+1, ... yn+1)(Bi)}，其中y為異於x但與B相同的變項     (19)

 [who(−)]3(A) = {(y1, y2, B3): who(−)(Ay1, y2)(B3)} (根據(19)) = {(y1, y2, B3): who(−)({x3: A(y1, y2, x3)}(B3))} (根據(16)) = {(y1, y2, B3): B3 = PERSON ∩ {x3: A(y1, y2, x3)}} ("who"的真值條件)

 [what(−)]2([who(−)]3(A)) = {(z1, B2, B3): what(−)({(y1, y2, B3): B3 = PERSON ∩ {x3: A(y1, y2, x3)}}z1, B3)(B2)} (根據(19)) = {(z1, B2, B3): what(−)({y2: B3 = PERSON ∩ {x3: A(z1, y2, x3)}})(B2)} (根據(17)) = {(z1, B2, B3): B2 = THING ∩ {y2: B3 = PERSON ∩ {x3: A(z1, y2, x3)}}} ("what"的真值條件)

 [who(−)]1([what(−)]2([who(−)]3(A))) = {(B1, B2, B3): who(−)({(z1, B2, B3): B2 = THING ∩ {y2: B3 = PERSON ∩ {x3: A(z1, y2, x3)}}}B2,B3)(B1)} = {(B1, B2, B3): who(−)({z1: B2 = THING ∩ {y2: B3 = PERSON ∩ {x3: A(z1, y2, x3)}}}(B1)} = {(B1, B2, B3): B1 = PERSON ∩ {z1: B2 = THING ∩ {y2: B3 = PERSON ∩ {x3: A(z1, y2, x3)}}}}     (20)

B = {(w1, w2, w3): w1 ∈ PERSON ∩ {z1: ∃y2, x3 ∈ U A(z1, y2, x3)} ∧ w2 ∈ THING ∩ {y2: ∃x3 ∈ U A(w1, y2, x3)}
∧ w3 ∈ PERSON ∩ {x3: A(w1, w2, x3)}

B = (PERSON × THING × PERSON) ∩ A

### 9.2 混合陳述-疑問量化句

[every(C)]3(D) = {(y1, y2): C ⊆ {x3: D(y1, y2, x3)}}

 [which(B)]2([every(C)]3(D)) = {(z1, E2): which(B)({(y1, y2): C ⊆ {x3: D(y1, y2, x3)}}z1)(E2)} (根據(19)) = {(z1, E2): which(B)({y2: C ⊆ {x3: D(z1, y2, x3)}})(E2)} (根據(17)) = {(z1, E2): E2 = X ∩ B ∩ {y2: C ⊆ {x3: D(z1, y2, x3)}} ("which"的真值條件)

 [which(A)]1([which(B)]2([every(C)]3(D))) = {(E1, E2): which(A)({(z1, E2): E2 = X ∩ B ∩ {y2: C ⊆ {x3: D(z1, y2, x3)}}E2)(E1)} = {(E1, E2): which(A)({z1: E2 = X ∩ B ∩ {y2: C ⊆ {x3: D(z1, y2, x3)}}(E1)} = {(E1, E2): E1 = X ∩ A ∩ {z1: E2 = X ∩ B ∩ {y2: C ⊆ {x3: D(z1, y2, x3)}}}

E = {(w1, w2): w1 ∈ X ∩ A ∩ {z1: ∃y2 ∈ U C ⊆ {x3: D(z1, y2, x3)}}
∧ w2 ∈ X ∩ B ∩ {y2: C ⊆ {x3: D(w1, y2, x3)}}

E = X2 ∩ (A × B) ∩ {(z1, y2): C ⊆ {x3: D(z1, y2, x3)}}

### 9.3 偶串列解的形式化推導

Which girl does every boy love?     (21)

[which(B)]2([every(A)]1(C))

[every(A)]1([which(B)]2(C))

[Q(A)]1([Q'(B)]2(C)) = {(D1, D2): D1 = E ∩ ([Q'(B)]2(C))D2}     (22)

E ⊆ A ∧ Q(A)(E)     (23)

Q(A)(E) ⇔ A − E = {j} ⇔ E = A − {j}

Q(A)(E) ⇔ every({j, m})(E) ⇔ {j, m} ⊆ E ⇔ E = {j, m}

 [which(B)]2(C) = {(y1, D2): which(B)(Cy1)(D2)} (根據(19)) = {(y1, D2): which(B)({x2: C(y1, x2)})(D2)} (根據(16)) = {(y1, D2): D2 = X ∩ B ∩ {x2: C(y1, x2)}} ("which"的真值條件)

 [every(A)]1([which(B)]2(C)) = {(D1, D2): D1 = E ∩ {(y1, D2): D2 = X ∩ B ∩ {x2: C(y1, x2)}}D2} (根據(22)) = {(D1, D2): D1 = E ∩ {y1: D2 = X ∩ B ∩ {x2: C(y1, x2)}}} (根據(17)) = {(D1, D2): D1 = A ∩ {y1: D2 = X ∩ B ∩ {x2: C(y1, x2)}}} (根據(23))

Q(A)(E) ⇔ A ⊆ E ⇔ E = A

D = {(w1, w2): w1 ∈ A ∩ {y1: ∃x2 ∈ U C(y1, x2)}} ∧ w2 ∈ X ∩ B ∩ {x2: C(w1, x2)}}

D = (A × (X ∩ B)) ∩ C

### 9.4 選答解的形式化推導

Which boys did two dogs bite?     (24)

[which(B)]2([(exactly 2)(A)]1(C))
[(exactly 2)(A)]1([which(B)]2(C))

E ⊆ A ∧ Q(A)(E) [D = ...]     (25)

D = (E × (X ∩ B)) ∩ C

(exactly 2)(A)(E) ⇔ |A ∩ E| = 2 ⇔ |E| = 2

E ⊆ A ∧ |E| = 2 [D = (E × (X ∩ B)) ∩ C]

## 10. 定語分句的形式化推導

John(−)([(more than 1)(ACTIVITY)]2(FOND-OF))

 [(more than 1)(ACTIVITY)]2(FOND-OF) = {y1: (more than 1)(ACTIVITY))(FOND-OFy1)} (根據(18)) = {y1: (more than 1)(ACTIVITY)({x2: FOND-OF(y1, x2)})} (根據(16)) = {y1: |ACTIVITY ∩ {x2: FOND-OF(y1, x2)}| > 1} ("(more than 1)"的真值條件)

 [some(GIRL)]1(LOVE) = {y2: some(GIRL)(LOVEy2)} (根據(18)) = {y2: some(GIRL)({x1: LOVE(x1, y2)})} (根據(16)) = {y2: GIRL ∩ {x1: LOVE(x1, y2)} ≠ Φ} ("some"的真值條件)

N1 + V + N2 + Prep + N3 (N、V和Prep分別代表名詞、動詞和介詞)