Exercise Week 2 Flashcards
- Suppose that the domain of the propositional function P (x) consists of the integers 0, 1, 2, 3, and 4.
Write out each of these propositions using disjunctions, conjunctions, and negations.
Answer Questions Below
1a(Thursday). ∃x P(x)
P(0) ∨…. ∨P(4)
1b(Thursday) .∀x P(x)
P(0)∧…∧P(4)
- Let P (x) be the statement “x can speak Russian” and let Q(x) be the statement “x knows the computer language C++.” Express each of these sentences in terms of P (x), Q(x), quantifiers, and logical connectives. The domain for quantifiers consists of all students at your school.
Answer Questions Below
2a (Thursday). There is a student at your school who can speak Russian and who knows C++.
∃x (P(x) ∧ Q(x))
1c. ∃x ¬ P(x)
¬P(0) ∨… ∨¬P(4)
2b (Thursday). There is a student at your school who can speak Russian but who doesn’t know C++.
∃x (P(x) ∧ ¬Q(x))
2c (Thursday). Every student at your school can speak Russian or knows C++.
∀x (P(x) ∨ Q(x))
2d (Thursday). No student at your school can speak Russian or knows C++.
¬∃x (P(x) ∨ Q(x)) == ∀x (¬P(x) ∧ ¬Q(x))
3.Translate in two ways each of these statements into logical expressions using predicates, quantifiers,
and logical connectives. First, let the domain consist of the students in your class and second, let it
consist of all people.
Answer questions next
3a (Thursday). Someone in your class can speak Hindi
- ∃x(H(x))
- ∃x(C(x) ∧ H(x))
3b (Thursday). Everyone in your class is friendly
- ∀x (F(x))
- ∀x (C(x) → F(x))
3c (Thursday). There is a person in your class who was not born in California
1.∃x ¬B(x)
2.∃x(C(x) ∧ ¬B(x))
3d (Thursday). A student in your class has been in a movie
- ∃x M(x)
- ∃x (C(x) ^ M(x))
3e (Thursday). No student in your class has taken a course in logic programming
- ¬∃x P(x) or ∀x ¬P(x)
- ¬∃x (C(x) ^ P(x)) or ∀x (C(x) → ¬P(x))
- Translate each of these statements into logical expressions using predicates, quantifiers, and logical
connectives.
Answer Questions next
4a (Thursday). No one is perfect
¬ ∃x P(x) = ∀x ¬ P(x)
4b (Thursday). Not everyone is perfect
c ∀x P(x) = ∃x ¬ P(x)
4c (Thursday). All your friends are perfect.
∀x (F(x) → P(x))
4d (Thursday). At least one of your friends is perfect.
∃x (F(x) ^ P(x))
4e (Thursday). Everyone is your friend and is perfect
∀x (F(x) ^ P(x))
4f (Thursday). Not everybody is your friend or someone is
not perfect.
¬ ∀x F(x) ∨ ∃x ¬ P(x)
- Express each of these statements using quantifiers. Then form the negation of the statement, so that no negation is to the left of a quantifier. Next, express the negation in simple English. (Do not simply use the phrase “It is not the case that.”) Domain: All Creatures
Answer next questions
5a. Some old dogs can learn new tricks.
- ∃x (D(x) ^ T(x))
- ∀x ( D(x) → ¬ T(x))
- For all creatures. if you are an old dog you cannot learn new tricks.
5b. No rabbit knows calculus.
- ¬ ∃x (R(x) ^ C(x))
2.∃x (R(x) ^ C(x))
5c. Every bird can fly.
- ∀x (B(x) → F(x))
- ∃x( B(x) ^ ¬ F(x))
5d. There is no dog that can talk.
- ¬ ∃x ( D(x) ^ T(x))
- ∃x ( D(x) ^ T(x))
5e. There is no one in this class who knows French and Russian
- ¬∃x(C(x) ^ (F(x) ^ R(x)))
- ∃x(C(x) ^ (F(x) ^ R(x)))
- Express the negation of each of these statements in terms of quantifiers without using the negation
symbol.
Answer Questions Below
6a.¬∀x(x > 1)
∃x( x <= 1)
6b.
∃x( x > 1)
6c.
∀x( x < 4)
6d.
∀x( x >= 0)
6e.
∃x((x >= 1) ^ (x <= 2))
6f
∀x(( x >= 4) ^ (x <= 7))
Bonus: ¬∀x(1 <= x < 2)
∃x ¬( (x >= 1) ^ (x < 2))
