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Unformatted text preview: UNIVERSITY OF TORONTO
Faculty of Arts and Science DECEMBER 2004 EXAMINATIONS
PHL 245H1F — Iain Martel
Duration — 3 hours
Examination Aid: Rule Sheets (attached)
Answer all questions in the exam booklets provided Part A: Translations (15 points) Translate each of the following sentences into a sentence in symbolic notation, using the
translation scheme provided. Make sure that your translations are as close to the logical structure
of the original sentences as possible (3 points each): 1. Only the best team will win the championship.
(Bxy: x is better than y; Tx: x is a team; Wx x will win the championship) 2. If the Labour party wins reelection then Tony Blair will remain Prime Minister, unless he
resigns ﬁrst. (Wx: x wins reelection; Px: x remains Prime Minister; Rx x resigns before the election; I: the
Labour Party; t: Tony Blair ) 3. Neither penguins nor ostriches can ﬂy, but both are birds.
(Px: x is a penguin; 0x: x is an ostrich; Fx: x can ﬂy; Bx: x is a bird) 4. No number is larger than itself, but for any pair of distinct numbers, one must be larger
than the other.
(Nx: x is a number; ny: x is larger than y) 5. All philosophers except the author of this exam are crazy fools!
(Px: x is a philosopher; Axy: x is the author of y; Cx: x is crazy; Fx: x is a fool; e: this exam)
Part B: Proving Invalidity (7 points)
6. (7 points) Prove that the following argument is invalid, either by constructing a
counterexample, or by using the ﬁnite universe method. Explain your answer. 1 . (3x)Px
2. (x)(QxD ~ Px) /(x) ~ Qx Page 1 of 4 Part C: Proofs (78 points)
In the following questions, only the proof rules on the attached pages may be used. 7. (12 points) Prove that the following argument is valid using natural deduction: 1. (X){(PX'QX)3(Y)[(”PY'QYPRXYH
2. (PaOQa)O ~Rab /Qb'—'>Pb 8. (12 points) Prove that the following argument is valid using natural deduction:
1 . ~ S D (TOU)
2. ~ RD ~ (TVU)
3. (T.=.U)3(~~SOR) /R08 9. (12 points) Prove that the following argument is valid using natural deduction: 1 (X)(Y)(Z)[(FXY'FYZ)3FXZ)]
2. (x)~Fxx /(x)(y)(nyD ~Fyx) 10. (12 points) Prove that the following argument is valid using natural deduction: 1 (x)[(LxVMx) 3 {[(NX'OXWPXl DQXH 2 (X){[(0X3QX)'~RX]3MX} 3. (3x)[Lx0~Mx] 4. (axllMxO~Lx] /(3x)[NxDRx] 11. (12 points) Prove that the following statement is a tautology using natural deduction:
{(X)[PX3(QX'RX)l'(X)[SX3(QX'RX)]}E(X)[(PXVSX)3 (QX°RX)l 12. (18 points) Translate the sentences of the following argument into symbolic notation, using
the translation scheme provided. Then prove that the argument is valid using natural deduction. Some criminal robbed the Casa Loma. Whoever robbed the Casa Loma either had an accomplice
among the staff or had to break in. To break in, one would either have to smash a window or pick
the door lock. Only an expert locksmith could have picked the door lock. Had anyone smashed
the door, he would have been heard, but nobody was heard. If the criminal who robbed the Casa
Lorna managed to fool the guard, he had to be a convincing actor. No one could rob the Casa
Loma unless he fooled the guard. No criminal could be both an expert locksmith and a
convincing actor. Therefore some criminal had an accomplice among the staff. (Cx: x is a criminal; Rx x robbed the Casa Loma; Sx: x had an accomplice among the staff" Bx: x
broke in; Sx x smashed a window; Px: x picked the door lock; Lx: x is an expert locksmith; Hx.‘ x
was heard; Fx.’ x managed to fool the guard; Ax: x is an accomplished actor) Total: 100 points Page 2 of 4 Rules of Inference l. Modus ponens (MP)
P 3 ‘l a.
q 5. Hypothetical syllogism (HS)
P 3 ‘1
5123."
p :> r
5. Constructive dilemma (CD)
(/23 q) '03:)
e v r
q v s
7. Conjunction (Conj)
P .‘L...
I) ‘ ll Axiom of replacement: Within the context of a proof, logically equivalent expressions may replace each other. 9. DeMorgan‘s rule (DM)
10. (Iommutativity (Com)
ll. Associativitymssoc)
I2. Distribution (Dist) 15. Double negation (DN)
14. Transposition (Trans)
1). Material implication (lmpl) 16. Material equivalence (Equiv) l7. Exportation (Exp)
18. Tautologyﬂnut) 2. Modus tollcns (MT)
[1 3 ‘1
LL
‘1) 4. Disiunctive syllogism (DS)
P V ‘1
:P.
‘I 6‘ Simpliﬁcation (Simp)
L'ﬂ
P 8. Addition (Add) L.
P“! ~<p  q) <~p v ~q)
~ovq>==<~p~q) (pvq)::(qvp)
(IIq) == (4'11) [pv(qu)] ::[(pvq)vr]
[P'(q°r)l::l(P‘lI)'rl lp°(qu)l ::[(p'q)v(pr)]
[pv(q'r)l::l(pvq)°(pvr)l Paw/J
(qu)=:(~qD ~12)
(PDq)==(~qu) (p 5q) =: [(113 q)(.q:>p)l
(/1 5 r1) := [(1)  4) v <~p ”(1)1 [(pnp: r] :: [p3(q:>r)] p: (/2 v1»
1):: UPI?) Page 3 of 4 Conditional Proof Indirect Proof __ qur _. /q
q ACP ~q AI?
"' r~r Conj
r """I [P qu CP q DN Rules for Removing and Introducing Quantifiers (a. 5. r. . . . a, u. w are individual constants: v. y, x are individual variables) 1. Universal instantiation (UK) (.l)g.l (1)3";
5’ 3a
2. Universal generalization (UG) 92. not Sin.
(:t‘)§.1r allowed: (if)? .
Restﬂclions' (1) U6 must not be used within the scope of an indented se
(conditional and quence if the instantial variable occurs free in the ﬁrst line
Indirect proof) of that sequence.
(mterlapplng (2) 06 must not be used if 91; contains an existential name
quantiﬁes) and y is free in the line where that name is introduced.
3. Existential instantiation (El) (3.1991: not (Six)3.»
9‘4 allowed: 5‘;
Restriction: The existential name a must be a new name that has not occurred in
any previous line.
4. Existential generalization (E6) $1: 9? Z
(31‘) I (3.05.: Change of Quantifier Rules 00$! :: ~(3.r)~g;r (3.r)g:r :: ~(_t')~ gr
~(.r)§;.r :: (3.r)~$x ~30ng :: (;r)~ga
Identity Rules
LPrcm. 2. a=J::£=n 5. 2%
n =0 {1 = J Total pages = 4 Page 4 of 4 ...
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