[试题] 106-1 傅皓政 逻辑 期末考

楼主: bung8833 (bangbangmang)   2018-01-08 22:49:26
课程名称︰逻辑
课程性质︰通识A4
课程教师︰傅皓政
开课学院:
开课系所︰
考试日期(年月日)︰2018/1/8
考试时限(分钟):100
是否需发放奖励金:是
(如未明确表示,则不予发放)
本文符号:
量词:∀∃
连接词:﹁ΛV→↔
等同符号:=
句式:φψ
推导:├
试题:
一、建构初阶逻辑语言(提示:包括符号与形构规则两个部份)。(10%)
(Construct a suitable language for first-order (or predicate) logic. Hint: two p
arts involved, alphabets and formation rules)
二、请以亚里斯多德的方法证明下列三段论的论证是否为有效论证,如为无效论证请写出具
有说服力的反例。(15%)
(Please prove whether the following syllogism is valid or not bt Aristotelian me
thod and specify a persuasive counterexample if it is invalid.)
(a) M I S
M O P
────
S O P
(b) M A S
P E M
────
S O P
(c) S E M
M E P
────
S A P
三、请将下列日常语言的语句翻译为初阶逻辑语言的表达式。(15%)
(Please translate the following sentences into first-order language expressions.
)
Tx:x是老师; Rxy:x尊敬y
(a) 没有老师会尊敬每个老师
(b) 每个老师都会尊敬某些老师
(c) 不是每个老师都会尊敬每个老师
四、请以真值树法证明下列语法序列是否为有效论证,若为无效论证请显示其反例结构。
(20%)
(Please use tableaux system to prove whether each of the following argument is v
alid and to specify a counterexample if it is invalid.)
(a) (∀x)(Ax→(∃y)(﹁ByΛCyx)), (∀x)(﹁Bx→Dx), (∀x)(Dx→Ex)├ (∀x)(Ax→(∃
y)(EyΛCyx))
(b) (∀x)(Px→Mx), (∀x)(Mx→﹁Sx) ├ (∃x)(SxΛ﹁Px)
五、请以实例说明下列谬误,并说明推理过程。(10%)
(Please exemplify the following fallacies and explain the process of reasoning.)
(a) 以偏概全的谬误 (Fallacy of Improper Generalization)
(b) 稻草人谬误 (Straw Man Fallacy)
六、请完成下列演算,作答时需连同题目写在答案卷上。(30%)
(Please complete the following proofs. Notice: you should copy the whole questio
ns on your answer sheet.)
(a) A→C, ((AΛB)→C)→D ├ D
AΛB
———
A A→C
—————————
C
—————
(AΛB)→C ((AΛB)→C)→D
——————————————————
D
(b) K→﹁L, M→L ├ K→﹁M
K K→﹁L M M→L
—————— ——————
﹁L L
——————————————————

——————
﹁M
——————
K→﹁M
(c) (∀x)(∃y)(PyΛQxy), (∃x)﹁Px ├ (∃y)(PyΛ(∃x)(﹁PxΛQxy))
1. (∀x)(∃y)(PyΛQxy) Pr
2. (∃x)﹁Px Pr
3. ﹁Pa ______
4. (∃y)(PyΛQay) ______
5. PbΛQab ______
6. Qab ______
7. ﹁Pa ΛQab ______
8. (∃x)(﹁PxΛQxb) ______
9. Pb ______
10. PbΛ(∃x)(﹁PxΛQxb) ______
11. (∃y)(PyΛ(∃x)(﹁PxΛQxy)) ______
Appendix: Rules of inference
1. 树状自然演绎法推论规则
2. 线性自然演绎法推论规则
(i)等值规则
(1)笛摩根定律 (DeM) :
~(p^q)←→(~pv~q);
~(pvq)←→(~p^~q)
(2)交换律 (Comm) :
(pvq)←→(qvp);
(p^q)←→(q^p)
(3)结合律 (Assoc) :
(pv(qvr))←→((pvq)vr);
(p^(q^r)←→((p^q)^r)
(4)分配律 (Dist) :
(p^(qvr))←→((p^q)v(p^r));
(pv(q^r))←→((pvq)^(pvr))
(5)双重否定律 (DN) :
p←→~~p
(6)异质位换律 (Contra):
(p→q)←→(~q→~p)
(7)蕴涵律 (Impl) :
(p→q)←→(~pvq)
(8)等值律 (Equiv) :
(p←→q)←→((p→q)^(q→p));
(p←→q)←→((p^q)v(~p^~q)
(9)移出律 (Exp) :
((p^q)→r)←→(p→(q→r))
(10)重言律 (Taut) :
p←→pvp;
p←→p^p
(ii)蕴涵规则
(1)肯定前项律 (MP): p→q, p├ q
(2)否定后项律 (MT): p→q, ~q├ ~p
(3)假言三段论 (HS): p→q, q→r├ p→r
(4)选言三段论 (DS): pvq, ~p├ q; pvq, ~q├ p
(5)简化律 (Simp) : p^q├ p; p^q├ q
(6)添加律 (Add) : p├ pvq
(7)连言律 (Conj) : p,q├ p^q
(8)建构两难律 (CD): (p→q)^(r→s), pvr├ qvs; p→q,r→s,pvr├ qvs
作者: daniel880123 (圈內小高一)   2018-01-09 16:44:00
这次好像没有惊喜题

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