philosophy 2340 by lindash

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									                                                                      Name____________________________
                                            PHILOSOPHY 2340
                                              Symbolic Logic
                                                Fall, 2001

                                          SECOND EXAMINATION


Part I: Translations

Please translate the following sentences into FOL, using the Tarski's World abbreviations. (Note: for
these questions, interpret "some" to mean "at least one.") (4 points each.)


1. if a is not a dodecahedron, then it is a cube.

        ¬Dodec(a) → Cube(a)

2. b is a cube only if it is not large.

        Cube(b) → ¬Large(b)

3. All the medium cubes are to the right of b.

        ∀x ((Medium(x) ∧ Cube(x)) → RightOf(x,b))

4. f is either a dodecahedron or a tetrahedron, if it is not large.

        ¬Large(f) → (Dodec(f) ∨ Tet(f))

5. d is a large dodecahedron unless a is a small cube.

        ¬(Small(a) ∧ Cube(a)) → (Large(d) ∧ Dodec(d))

6. a is a tetrahedron if and only if all the large cubes are in front of b.

        Tet(a) ↔ ∀x ((Large(x) ∧ Cube(x)) → FrontOf(x,b)

7. No small dodecahedron is in front of a.

        ∀x ((Small(x) ∧ Dodec(x)) → ¬FrontOf(x,a))

8. Everything is either small, medium, or large, but nothing is all three.

        ∀x (Small(x) ∨ Medium(x) ∨ Large(x)) ∧ ∀x ¬(Small(x) ∧ Medium(x) ∧ Large(x))

9. Some things that are neither tetrahedra nor dodecahedra are large.
        ∃x (¬(Tet(x) ∨ Dodec(x)) ∧ Large(x))

10. Every cube is in a different row from every other cube.

        ∀x∀y ((Cube(x) ∧ Cube(y) ∧ x ≠ y) → ¬SameRow(x,y))

11. There are different tetrahedra that are the same size.

        ∃x ∃y (Tet(x) ∧ Tet(y) ∧ x ≠ y ∧ SameSize(x,y))

Part II: Basic Concepts

List two of the DeMorgan’s equivalences for quantifiers, and explain why they are called by this
name, that is, how they are related to the DeMorgan’s equivalences for propositional logic. (10 points)

Here are all four of the equivalences:

        ¬∃x P(x) ⇔ ∀x ¬P(x)
        ¬∀x P(x) ⇔ ∃x ¬P(x)
        ¬∃x ¬P(x) ⇔ ∀x P(x)
        ¬∀x ¬P(x) ⇔ ∃x P(x)

They are called “DeMorgan’s Equivalences” because they have a close relation to the DeMorgan’s
equivalences for propositional logic. ∃x P(x) is similar to a potentially infinite disjunction: P(a)∨ P(b)
∨ ...   Similarly, ∀x P(x) is similar to a potentially infinite conjunction: P(a) ∧ P(b) ∧ . . .   So, for
example, ¬∃x P(x) is closely related to ¬ (P(a)∨ P(b) ∨ . . . ) which, by the DeMorgan’s laws for
propositional logic, is equivalent to ¬P(a) ∧ ¬P(b) ∧ . . . This, in turn, by virtue of the connection
between potentially infinite conjunction and universal quantification, amounts to saying ∀x ¬P(x) --
and there is our first “DeMorgan’s equivalence” for quantifiers!
Part III: Truth-Functional Form

Write down the truth-functional form of the following sentence (as determined by the algorithm in the
text). Then construct a truth table and use it to determine whether the sentence a tautology. Is it a first-
order validity (FO validity)? Explain your answer. (10 points.)

(x (Tet(x)  Large(x))  x Tet(x))  x Large(x)

(x (Tet(x)  Large(x))  x Tet(x))  x Large(x)
                     A             B               C
(A  B)  C

A   B   C (A  B)  C
T   T   T    T    F F
T   T   F    T    T T
T   F   T    F    T F
T   F   F    F    T T
F   T   T    F    T F
F   T   F    F    T T
F   F   T    F    T F
F   F   F    F    T T

Not a tautology (because false in the first row).
The sentence is a FO validity. We can see this by seeing that it is impossible to make the sentence false.
The sentence would be false only if the antecedent is true and the consequent is false. The antecedent
says that everything is either a tetrahedron or large, and something is not a tetrahedron. If the antecedent
is true then, since everything is either a tetrahedron or large, the thing that isn’t a tetrahedron must be
large. So if the antecedent is true, there must be something that is large. Now, can we make the
consequent false? The consequent says that it is not true that everything is not large. This is equivalent
(by the DeMorgan’s equivalences for quantifiers) to saying that something is large. But we have just
seen that if the antecedent is true, then this must be true also. Thus it is impossible for the antecedent to
be true and the consequent false.


Part IV: Proofs

Please construct proofs in the formal system F for the following arguments. You may use any of the
following principles in addition to the introduction and elimination rules: the DeMorgan's equivalences;
Disjunctive Syllogism (from P  Q and P, infer Q); Excluded Middle (write any sentence of the form
P v ~P at any point in a proof). As justification just write “taut con:” followed by “DM,” “DS,” or
“EM” depending on which principle you use, followed by the appropriate line numbers (none needed for
EM, of course). (However, it should be possible to do the proofs that follow pretty straightforwardly
without using any of these principles, so I wouldn't waste a lot of time trying to figure how if at all they
can be used!) (12 points each)

1. Prove (P  Q)  ((R  P)  (R  Q)) using no premises. (Took me 8 steps using only the
introduction and elimination rules.) [Note: If we count the first line (empty) as line 1, it’s 9 steps.]
2. Prove A ↔ B from the following premises. (Took me 15 steps, using no previous theorems, or 10
steps using a previous theorem.)

       1. A  B
       2. (A  B)
3. Prove Q  (P  R) from the following premise. (Took me 8 steps, using only the introduction
and elimination rules.)

       1. P  Q

								
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