是非題
- [(A∧¬B)∨(B∧¬C)∨(C∧¬A)]→[(A∧B∧C∧D)→(E↔F)] is a tautology.
T。
要讓條件句為假,(A∧¬B)∨(B∧¬C)∨(C∧¬A)得為T,(A∧B∧C∧D)→(E↔F)得為F。要讓(A∧B∧C∧D)→(E↔F)是F,那麼A∧B∧C∧D要為T、E↔F為F。A∧B∧C∧D要為T的話,(A∧¬B)∨(B∧¬C)∨(C∧¬A)就會為F,所以找不到讓題幹的句子為假的真值設定。
. - ∃x(P(x)↔R(x)) is logically equivalent to ∃xP(x)↔∃xR(x).
F。反例:Domain = {0,1} P = {0} R = {1},此時∃xP(x)↔∃xR(x)為T,∃x(P(x)↔R(x))為F。
. - Assume that only one of the following two sentences is true: (1) Pigs can fly unless Kant is not right; (2) Kant is not right only if pigs can fly. Based on this assumption, it is true that (3) if pigs can fly, then I will cry.
T。
P:Pigs can fly。K:Kant is right。I:I will cry。
(1)P∨¬K (2)¬K→P (3)P→I
因為(1)和(2)裡只有一句話為真,所以P要為假(P為真的話兩句話都會為真)。因此P→I為真。
. - If P and S are consistent and S and Q are inconsistent, then P cannot imply Q.
T。
在P和Q是一致的,而且S和Q是不一致的,而且P蘊含Q的情況下,當P為真時Q也會為真(P蘊含Q)、P和S可以同時為真(P和Q是一致的),所以會有個情況是P、Q和S同時為真。但是這結果和S和Q是不一致的預設矛盾,所以P不蘊含Q。
. - Suppose that most philosophers are truth-pursuers and that most truth-pursuers are smart. Then we can conclude that most philosophers are smart.
F。反例:
給反例
- ∃x(Px→∀yRy) /\∃xPx→∀yRy
Model = (D, PM, RM), Domain = {0,1}, PM = {0}, RM = ∅
. - ∀x¬R(x, x)∧∀x∃yR(x, y)∧∀x∀y∀z(R(x, y)→(R(y, z)→R(x, z))) /\∃x∀y(¬x=y→R(x, y))
Model = (D, RM),Domain = 整數的集合,RM = 我們平常對小於符號「<」的解釋
Let “Lxy”stand for “x loves y”,
“Hxy”stand for “x hates y”and
“Px”stand for “x is a philosopher”.
Please symbolize the following sentence.
There is some philosopher who hates exactly two persons who are not philosophers and who love each other but no one else.
∃x(Px∧∃y∃z(Hxy∧Hxz∧∀w(Hxw→(w=y∨w=z))∧¬Py∧¬Pz∧Lyz∧Lzy∧∀w(Lyw→w=z)∧∀w(Lzw→w=y)))
證明
1. ∀x¬[(Px↔Rx)↔Qx]
2. ∃x∃y(¬Rx∨Sxy) /\∃x∃y[Qx→(¬Sxy→Px)]
3. ¬∃x∃y[Qx→(¬Sxy→Px)] AIP
4. ∀x∀y¬[Qx→(¬Sxy→Px)] 3,QN
5. ¬Rx∨Sxy 2,EI
6. ¬[(Px↔Rx)↔Qx] 1,UI
7. ¬[Qx→(¬Sxy→Px)] 4,UI
8. Qx∧¬Sxy∧¬Px 7,DN, Impl, DeM
9. ¬Sxy 8,Simp
10. ¬Rx 5,9,DS
11. (Px↔Rx)↔¬Qx 6,¬(A↔B)≡A↔¬B
12. ((Px→Rx)∧(Rx→Px))→¬Qx 11,Equiv, Simp
13. (Px→Rx)→((Rx→Px)→¬Qx) 12,Exp
14. ¬Px∨Rx 8,Simp, Add
15. Px→Rx 14,Impl
16. Rx→Px 10,Add, Impl
17. ¬Qx 13,15,16,MP
18. Qx 8,Simp
19. ¬Qx∧Qx 17,18,Conj
20. ∃x∃y[Qx→(¬Sxy→Px)] 3-19,IP
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九十九年中正哲學碩班甄試邏輯試答 - 啊啊哲學
給反例2
ReplyDeleteDomain= the set of real numbers
Rxy: x>y
(i) for all x, ¬(x>x) [no real number is less than itself]
(ii) for all x, there is an y such that x>y [every real number has a bigger one]
(iii) for all x,y,z, if x>y and if y>z, then x>z.
(iv) there is no x such that x>y for every y differs from x.
謝謝。這樣寫看起來好多了。
ReplyDelete不過(ii)應該是every real number has a smaller one?
yaya
ReplyDelete