Philosophy 2100: Introduction to Logic Final Exam I
[Pages:6]Name: __________________________
Philosophy 2100: Introduction to Logic Final Exam I
Sentence Logic (30 points)
Translations 6 points
Translate the following sentences using the dictionary provided. (2 point each)
P = Pablo buys a cat Q = The Queen declares a holiday R = Rafaela goes to the country
S = Sammie buys a cat. T = The trains run on time. V = Venus goes to the country
1. Either Pablo or Sammie bought a cat.
a) P v S
b) P & S
c) P -> S
d) S -> P
2. The trains won't run on time if the Queen declares a holiday.
a) ~T -> Q
b) ~Q -> ~T
c) Q -> ~T
d) Q ~T
3 . Venus and Rafaela will go to the country unless Sammie buys a cat.
a) S -> (V & R)
b) S v (V & R)
c) (V & R) -> S
d) ~S v (V & R)
1
Truth Tables (5 points):
I. Complete the truth table and determine whether the following formula is logically true, logically false, or contingent. (5 points)
(P v Q) -> (P & Q)
P Q
(P v
Q) -> (P & Q)
T T T F F T F F
T T T T F T F F
T T F F T F F T
TT T TF F FF T FF F
a) Logically True b) Logically False c) Logically Contingent
Natural Deductions in Sentence Logic (19 points):
I.. Supply the required justification for the derived steps in the following proof. (The justifications should only involve the 8 rules of implication and the 10 rules of equivalence.) (7 points)
1. (P & Q) v (P & R) Pr
2. P -> (~S & R) Pr
3. (Q & T) -> S Pr
4. P & (Q v R)
___________
5. P
___________
6. ~S & R
___________
7. ~S
___________
8. ~(Q & T)
___________
9. ~Q v ~T
___________
10. Q -> ~T
___________
1 Dist 4 Simp 2, 5 MP 6 Simp 3, 7 MT 8 DM 9 CE
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Name: __________________________
II. Use the first 8 rules of implication and 10 rules of equivalence to derive the conclusion of the following argument. (7 points)
(P v Q) -> R ~(S -> R) ~P
1. (P v Q) -> R 2. ~(S -> R) 3. ~(~S v R) 4. ~~S & ~R 5. ~R 6. ~(P v Q) 7. ~P & ~Q 8. ~P
Pr Pr 2 CE 3 DM 4 Simp 1, 5 MT 6 DM 7 Simp
3
III. Derive the conclusion of the following argument using any of the 20 rules. (5 points)
P -> Q ~P -> Q Q
1. P -> Q 2. ~P -> Q 3. ~Q -> ~P 4. ~Q -> Q 5. ~~Q v Q 6. Q v Q 7. Q
Pr Pr 1 Contr 2, 3 HS 4 CE 5 DN 6 Dupl
1. P -> Q 2. ~P -> Q
3. ~Q 4. ~P 5. ~~P 6. ~P & ~~P 7. ~~Q 8. Q
Pr Pr AIP 1, 3 MT 2, 3 MT 4, 5 Conj 3-6 IP 7 DN
4
Name: __________________________ Predicate Logic (58 points) Translations 10 points Translate the following sentences using the dictionary provided. (2 points each)
a = Ali b = Barry c = Cecilia
1. Ali and Cecilia like Dobermans.
a) Da v Dc
b) x(Ax & Cx)
Px = x likes Poodles. Sx = x feels secure. Dx = x likes Dobermans.
c) Da -> Dc
d) Da & Dc
2. If Barry likes Poodles, then he feels secure.
a) Pb -> Sb
b) Sb -> Pb
c) Pb & Sb
d) Pb -> xSx
3. No one likes both Poodles and Dobermans. a) x~(Rx & Px) b) ~x(Px & Dx)
c) ~x(Px & Dx)
d) x(~Px -> Dx)
4. Someone likes Dobermans, but it isn't Ali.
a) xDx v ~Da ~Da
b) ~Da -> xDx
c) xDx & ~Da
d) ~xDx &
5. Barry and Cecilia like poodles only if everyone does.
a) xPx -> (Pb & Pc) b) (Pb & Pc) -> xPx c) (Pb & Pc) -> ~xPx d) ~xPx -> (Pb & Pc)
5
Natural Deductions in Predicate Logic (48 points):
I. Supply the required justification for the derived steps in the following proof (10 points)
1. xPx -> xQx 2. x~(Qx v Rx) 3. ~(Qa v Ra)
4. ~Qa & ~Ra 5. ~Qa 6. x~Qx 7. ~xQx 8. ~xPx 9. x~Px 10. ~Pb 11. ~Pb v Sb
12. Pb -> Sb
Pr Pr
_ _ _
______
2 EE (flag a) 3 DM 4 Simp 5 EI 6 QE 1, 7 MT 8 QE 9 UE 10 Add 11 CE
II. Derive the conclusions of the following arguments. (13 points each)
x ((Qx v Rx) -> Px) x (Sx & ~Px) x (~Qx & ~Rx)
1. x ((Qx v Rx) -> Px) 2. x (Sx & ~Px) 3. (Qa v Ra) -> Pa 4. Sa & ~Pa 5. ~Pa 6. ~(Qa v Ra) 7. ~Qa & ~Ra 8. x (~Qx & ~Rx)
Pr Pr 1 UE 2 UE 4 Simp 3 MT 6 DM 7 EI
6
x(Px -> Qx) ~x~Qx -> ~x ~Rx ~x Px v xRx
1. x(Px -> Qx) 2. ~x~Qx -> ~x ~Rx 3. xQx -> ~x ~Rx 4. xQx -> x Rx
5. x Px 6. Flag a
7. Pa
8. Pa -> Qa
9. Qa 10. x Qx 11. x Rx 12. x Px -> x Rx 13. ~x Px v xRx
Pr Pr 2 QE 3 QE ACP FS UI 5 UE 1 UE 7, 8 MP 6-9 UI 4, 10 MP 5-11 CP 12 CE
Name: __________________________
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III. Derive the following theorem. (12 points)
. x ~Px v xPx
1. x Px 2. x Px -> x Px 3. ~x Px v x Px 4. x ~Px v xPx
1. ~(x ~Px v xPx ) 2. ~x ~Px & ~xPx 3. x Px & ~xPx 4. ~~(x ~Px v xPx ) 5. x ~Px v xPx
ACP 1-1 CP 2 C.E. 3 QE
or
AIP 1 DM 2 QE 1-3 IP 4 DN
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