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Unformatted text preview: UNIVERSITY OF TORONTO
Faculty of Arts and Science APRIL/MAY EXAMINATIONS 2004
PHL245HIS — Glen Hoffmann
Duration — 2 hours
April 26, 2004
Examination Aid: Sheet with rules provided  No other aids allowed
Name: Student #: 1. Define the following terms, providing an example of each (3 marks each): a) Validity in PD b) Mixed Quantiﬁer sentence c) TruthFunctional Compound (1) TruthFunctional Falsity e) Sentence of PL Name: Student #:
2. State whether the following sentences are true or false. Brieﬂy explain your answers (3 marks each): a) A theorem of SD is provable from no assumptions. b) An argument whose premises are all truthfunctionally true is deductively valid. c) One can always demonstrate that a sentence is quantiﬁcationally indeterminate by providing appropriate
interpretations. d) All formulas of predicate logic are sentences of predicate logic. e) An argument whose premises are inconsistent in PD is valid in PD. Name: Student #2 3. Use the abbreviation scheme provided to symbolize the following sentences (3 marks each): UD: Positive Integers
Ex: x is even Ox: x is odd Px: x is prime ny: x is larger than y
a: 1 :2
:3
:4 DOc' a) 1 and 2 are prime numbers, but 3 is odd if and only if 4 is even. b) No even number is prime. 0) Some even numbers are larger than any odd number. (1) All prime numbers are larger than 1. e) No number is larger than all numbers, but some number is smaller than all other numbers. Name: Student #: 4. Provide an interpretation which shows that the following argument is quantiﬁcationally invalid. Please
explain your answer (10 marks): Sb {VXHPXDSXL
Pb Name: Student #: 5. Construct a truthfunctional expansion of the following sentence for the set of constants {‘a’, ‘b’}. Use
your truthfunctional expansion to show that the sentence is quantificationally indeterminate. Please explain your answer (15 marks): [~ (3y) F y v ~(3y) Dy} V (W) (Fy & Dy) Name: Student #: 6. Construct a derivation to Show that the following sentence is a theorem in PD (15 marks): (Vy) (Dy v (32) Byz) D (Vy)(32) (Py v Byz) Name: Student #: 7. Construct a derivation to show that the following argument is valid in PD+ (15 marks): (VX) [(Elx) (Gyb & nyb) 3 FX] 13x2 ngb & Hxabl
(Vx) (be :3 ~Fx) 3 ~Gab Scrap Paper DERIVATION RULES OF SD
Reiteration (R) I P
D P
‘8c’ Rules
Conjunction Introduction (&I) Conjunction Elimination (&E)
P P & Q P & Q
or
Q > P D Q
D P & Q
‘3‘ Rules
Conditional Introduction (:71) Conditional Elimination (DE)
P P 3 Q
P
Q D
D P 3 Q Q
‘~’ Rules
Negation Introduction (~ I) Negation Elimination (~ E)
P
Q
~ Q
D ~ P D P
‘v’ Rules
Disjunction Introduction (VI) Disjunction Elimination (vE)
P P P v Q
or P
D P v Q D Q v P '_.
R
’3
R
D R
‘5’ Rules
Biconditional Introduction (51) Biconditional Elimination (5E)
' P P E Q P E Q
P or
Q Q
D D P
’i Q
P
D P E Q DERIVATION RULES OF SD+ All the Derivation Rules of SD and Rules of Inference Modus Tollens (MT) Hypothetical Syllogism (HS)
P D Q QDR
PDR D Disjunctive Syllogism (DS) PDQ
~Q
D ~P
PvQ
~P
D Q or D PVQ
~Q
P Rules of Replacement Commutation (Com) P&Q<1DQ&P
PvQ<1DQvP Implication (Impl)
P I) Q <1 D ~ P v Q De Morgan (DeM)
~(P&Q)<1D~PV~Q
~(PvQ)<lD~P8c~Q Transgositz'on (Trans)
P 3 Q <1 D ~ Q 3 ~ P Distribution (Dist) P&(QVR)<1D(P&Q)V(P&R) PV(Q&R)<1D(PVQ)&(PVR) Association (Assoc) P&(Q&R) <1D(P&Q)&R
Pv(QvR) <1l> (PVQ) VR
Double Negation (DN) P <1 D ~ ~ P Idempotence (Idem) P<1DP&P
P<lDPvP Exportatz'on (ExB)
PD (QDR) <1D (P8cQ) DR Eo'uivalence (Eguiv)
PEQ<ID(P:>Q)&(Q:>P)
PEQ<}D(P&Q)V(~P&~Q) DERIVATION RULES OF PD All the Derivation Rules of SD and Universal Introduction (VI)
P(a/x)
D (Vx)P Provided:
(i) a does not occur in an
undischarged assumption. (ii) a does not occur in (Vx)P. Existential Introduction (31)
P(a/x)
D (3x)? Universal Elimination (VE)
(Vx)P
D P(a/x) Existential Elimination (3E)
(3X) P ‘ P(a/x)
Q D Q
Provided:
(i) a does not occur in an
undischarged assumption.
(ii) a does not occur in (3x)?
(iii) a does not occur in DERIVATION RULES OF PD+ All the Derivation Rules of SD+ and of PD and Quantiﬁer Negation ( QN) ~ (Vx)P <1 D (3x) ~ P
~ (3x)P <1 D (Vx) ~ P DERIVATION RULES OF PDE All of the Derivation Rules of PD and Universal Elimination (VE)
(Vx)P
i> P(t/x) where t is a closed term Identity Introduction ( =1)
D l (Vx)x = x Existential Introduction (VE)
P(t/x)
I> (3x)P where t is a Closed term Identity Elimination (7 E) t1 = t2 t1 = t9
P or P
D Paw/ta) I> P(t2//t1) where t1 and t2 are closed terms ...
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