PHIL12A Section answers, 20 April 2011 - Philosophy

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Here are the FOL translations:(i) ∃x[Cube(x) ∧ Medium(x) ∧ ∀y((Dodec(y) ∧ Medium(y)) → Adjoins(x,y))](ii) ∀x[(Dodec(x) ∧ Medium(x)) → ∃y(Cube(y) ∧ Medium(y) ∧ Adjoins(x,y))]2. One of the above translations is strong, and the other weak, in the sense that the strong sentence impliesthe weaker one, but not vice versa. Indicate which translation is which.(i) is the stronger claim. Whenever (i) is true, (ii) is true too. In other words, every model that makes (ii) truemakes (i) true. But there are models that make (ii) true but not (i). So it is not always the case that (i) is truewhen (ii) is true.3. (Ex 11.27 and 11.28) Here are two arguments, each of whose first premise is ambiguous. Translate eachargument into FOL twice (using sensible predicates), corresponding to the ambiguity in the first premise.Under one translation the conclusion follows: prove it. Under the other it does not: describe a situationin which the premises are true but the conclusion false.(a) (Ex 11.27)1 Everyone admires someone who has red hair.2 Anyone who admires himself is conceited.3 Someone with red hair is conceited.Here’s a translation that makes the argument valid:1 ∃x(Red(x) → ∀y(Admires(x,y)))2 ∀x(Admires(x,x) → Conceited(x))3 ∃x(Red(x) ∧ Conceited(x)The argument would be invalid if we had translated the first premise as the weaker:∀x∃y(Red(y) ∧ Admires(y,x)).(b) (Ex 11.28)1 All that glitters is not gold.2 This ring glitters.3 This ring is not gold.The following translation makes the argument valid:2
1 ∀x(Glitters(x) → ¬Gold(x))2 Glitters(ring)3 ¬Gold(ringThe following weaker translation of the first premise would make the argument invalid:¬∀x(Glitters(x) → Gold(x))3 Proofs with quantifiers1. Some of the following arguments are valid, some are not. Give an informal proof for those which are valid; forthe others, give counterexamples. I’ve done the first one.(a) (Ex 12.4)1 ∀y[Cube(y) ∨ Dodec(y)]2 ∀x[Cube(x) → Large(x)]3 ∃x¬Large(x)4 ∃xDodec(x)Proof. The third premise tells us that there is something which is not large. Let’s call this object d. Nowby universal instantiation we have the following:(1) Cube(d) ∨ Dodec(d)(2) Cube(d) → Large(d)By modus tollens on (2) above, we have that d is not a cube. But then by (1) above it must be the case thatd is a dodecahedron.Since d was an arbitrary object in the domain of discourse, we can by existential instantiation obtain∃xDodec(x), which is our desired conclusion.3
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Page 5: c. So this shows that if j is small

PHIL12A Section answers, 20 April 2011 - Philosophy

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