PHIL12A Section answers, 20 April 2011

Julian Jonker

1 From last time...

Compare especially (a), (d) and (e) below.

1. (Ex 11.19)

(c) No cube with nothing to its left is between two cubes.

∀x((Cube(x) ∧ ¬∃yLeftOf(y,x)) → ¬∃z∃wBetween(x,z,w))

(d) The only large cubes are b and c.

Large(b) ∧ Cube(b) ∧ Large(c) ∧ Cube(c) ∧ ∀x((Large(x) ∧ Cube(x)) → (x = b ∨ x = c))

(e) At most b and c are large cubes.

NB: last time I omitted the following important tip printed in the textbook. ‘There is a significant difference between this sentence and the previous one. This one does not imply that b and c are large cubes, while the previous one does.

∀x((Large(x) ∧ Cube(x)) → (x = b ∨ x = c))

2 FOL translation exercises

1. The following sentence is ambiguous. I show you how to translate it twice, showing how the English supports two different FOL translations.

Every medium dodecahedron is adjacent to a medium cube.

There are two ways to translate this sentence. (Note how I use constructions like ‘for all’ and ’there is some’ in order to make explicit the role of quantifiers in the sentence.)

(i) There is some medium cube such that all the medium dodecahedra are next to it. (ii) For every medium dodecahedron there is some medium cube that it is next to.

1 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 implies the 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) true makes (i) true. But there are models that make (ii) true but not (i). So it is not always the case that (i) is true when (ii) is true.

3. (Ex 11.27 and 11.28) Here are two arguments, each of whose first premise is ambiguous. Translate each argument 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 situation in 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(ring

The following weaker translation of the first premise would make the argument invalid: ¬∀x(Glitters(x) → Gold(x))

3 Proofs with quantifiers

1. Some of the following arguments are valid, some are not. Give an informal proof for those which are valid; for the 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. Now by 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 that d 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 (b) (Ex 12.5)

1 ∀y[Cube(y) ∨ Dodec(y)]

2 ∀x[Cube(x) → Large(x)]

3 ∃x¬Large(x)

4 ∃x[Dodec(x) ∧ Small(x)]

This argument is not valid. Suppose that there is a medium dodecahedron only. Then the premises are true (check this!) but the conclusion is false.

(c) (Ex 12.8)

1 ∀x[Cube(x) ∨ (Tet(x) ∧ Small(x))]

2 ∃x[Large(x) ∧ BackOf(x,c)]

3 ∃x[FrontOf(c,x) ∧ Cube(x)]

Proof: By the second premise there is something which is large and behind c. Let’s call this object e. By Existential Instantiation we get that e is large and that it is behind c. But since BackOf and FrontOf are inverses, we know that c is in front of e. Now by the first premise everything is either a cube or a small tetrahedron, so by Universal Instantiation, e is either a cube or a small tetrahedron. But we know that e is large – so it must be cube. So c is in front of e and e is a cube. But now we use Existential Generalization in order to obtain our conclusion: that there is something that c is in front of and that is a cube.

(d) (Ex 12.9)

1 ∀x[(Cube(x) ∧ Large(x)) ∨ (Tet(x) ∧ Small(x))]

2 ∀x[Tet(x) → BackOf(x,c)]

3 ∀x[Small(x) → BackOf(x,c)]

Proof: We’re going to reason about an arbitrary object – it could be any one. For convenience, we’ll call it j. We’ll show by Conditional Proof that if j is small then it is behind c. But since j is arbitrary, this will be true of every object. Suppose j is small. By Universal Instantiation on the first premise we get that j is either a large cube or a small tetrahedron. Since it is small, we know it is a tetrahedron. By Universal Instantiation on the second premise we get that if j is a tetrahedron then it is behind c. So, by modus ponens, j is behind

4 c. So this shows that if j is small, then it is behind c. (In other words, we have proved Small(j) → BackOf(j,c).

But j was arbitrary, so by Universal Generalization, we get that any small object is behind c, which is the desired conclusion.

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