疑问量词的形式表达与推理模式 - WebRing

PERSON SING] = 1。根据以上计算结果,有(PERSON SING)1 = {a, d} ,
由此根据(89),可知[(at least who)(–)(SING)] = 1。其次,由于
(93) {Y U: Y (PERSON SING)1} = {{a}, {d}, {a, d}}
根据(90),我们有
(94) CA {{a}, {d}, {a, d}}
利用(94)以及附录 D,可以求出“(at least who)(–)(SING)”的 SA 为
(95) SA {a(–)(SING), d(–)(SING), (a and d)(–)(SING)}
在附录 B 和 C 中,我们列出了一些疑问量词,这些疑问量词之间的关系可
以概括为以下定理:
定理 2: (a) who(–)(~A) ≡ (everybody except who)(–)(A)
(b) whatn(–)(~A) ≡ (everything except whatn)(–)(A)
(c) whatd(A)(~B) ≡ (all except whatd)(A)(B)
(d) which(A)(~B) ≡ (all except which)(A)(B)
(e) (how many)(A)(~B) ≡ (all except how many)(A)(B)
(f) (what proportion of)(A)(~B) ≡ (all except what proportion
of)(A)(B)
3.8.3 析取疑问句与选答解
(g) who(–)(A) ≡ whatd(PERSON)(A)
(h) whatn(–)(A) ≡ whatd(THING)(A)
(i) which(A)(B) ≡ whatd(X A)(B)
(j) (at least who)(–)(A) ≡ (at least whatd)(PERSON)(A)
(k) (at least whatn)(–)(A) ≡ (at least whatd)(THING)(A)
(l) (at least which)(A)(B) ≡ (at least whatd)(X A)(B)
G&S (1989)认为除了合取疑问句外,疑问句还有另一种并列形式-“析取疑
问句”(disjoined interrogative),尽管并非所有学者都认为这是自然语言中一种重
要(或甚至真正存在)的现象。根据 G&S (1989),析取疑问句让人选择解答哪一个
析取项。因此,析取疑问句的解答并不唯一,从这一点看这类疑问句也具有非穷
尽性。比如说,以下疑问句便有三种可能解答 30 :
30 G&S (1989), (6), p. 431。
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