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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