Connect 1

Taking into account rules of precedence, which of the following parenthesized expressions is equivalent to ¬p ∧ r → q ∧ s?

(-p^r) -> (q ^ s) Reason: The operator ¬ has the highest precedence, next ∧ and after that →. You might find the following mnemonic useful for remembering the precedence order: "Never Copy Ideas" = Negation, Conjunction, Implication.

Which of the following compound propositions are satisfiable?

(¬p ∨ ¬q ∨ ¬r) ∧ (p ∨ q) ∧ (¬p ∨ ¬q) (¬p ∧ q ∧ ¬r) ∧ (p ∨ q) ∧ (¬p ∨ ¬q) (p ∧ q) → (p ∨ q)

Which of the following is a correct representation of the specification "If every user in the department is logged-in, then there must be at least two servers running" in predicate logic. Let the domain of variable x be all users and the domain of variables y and z be all pieces of computing machinery at the university. Use D(x) for "x is a user in the department," L(x) for "x is logged-in," S(y) for "y is a server," and R(y) for "y is running."

(∀x (D(x) → L(x))) → (∃y∃z (S(y) ∧ S(z) ∧ (y≠z) ∧ R(y) ∧ R(z))) Reason: The antecedent of the conditional states that all users from the department are logged on and the consequence states that there are two different servers running.

Which of the following sentences in propositional logic are tautologies? -p -> p -p V p p p -> p

-p V p p -> p

Which of the following English sentences represents ∀x P(x), where P(x) is the statement that "x has more than 1 GB RAM" and the domain is all of the computers at a university?

Each computer has more than 1 GB RAM. All the computers have more than 1 GB RAM.

Match the logical equivalence on the left with its name on the right. p ∧ T ≡ p p ∨ T ≡ T p ∨ p ≡ p ¬(¬p) ≡ p p ∨ (p ∧ q) ≡ p

Identity Law Domination Law Idempotent Law Double negation law Absorption law

Match the compound proposition on the left with the equivalent compound proposition on the right. -p V q -(pvq) (p->q)->r pv(q^r)

q = T p = F and q = F r = T p = T

Match each proposition on the left with the word describing it on the right. p <-> -p -p -> -p p -> -p

tautology contradiction contingency

Match the sentence on the left with the domain on the right that will make the sentence true. ∃x (x > 5) ∃x (x ≤ 0) ∃x (x^2 = 1)

{2, 4, 7} {-5, 5} {1, 2, 3, 4, 5}

Which of the following are valid logical equivalences? p ∨ (q ∧ r) ≡ (p ∨ q) ∧ r p ∧ F ≡ p ¬(p ∧ q) ≡ ¬q ∧ ¬p ¬(p ∧ q) ≡ ¬q ∨ ¬p p ∨ (p ∧ q)≡ p

¬(p ∧ q) ≡ ¬q ∨ ¬p p ∨ (p ∧ q)≡ p

Which of the following is the correct translation of the sentence "All the animals in the zoo are from South America"? You can assume that the domain of all variables is all physical objects in the world. Use A(x) for "x is an animal," Z(x) for "x is in the zoo," and S(x) for "x is from South America."

∀x ((A(x) ∧ Z(x)) → S(x)) ∀x (A(x) ∧ Z(x) ∧ S(x)) ∀x (¬A(x) ∨ ¬Z(x) ∨ S(x))

Which of the following is the correct translation of the sentence "All the animals in the zoo are from South America"? You can assume that the domain of all variables is all physical objects in the world. Use A(x) for "x is an animal," Z(x) for "x is in the zoo," and S(x) for "x is from South America."

∀x (¬A(x) ∨ ¬Z(x) ∨ S(x)) ∀x ((A(x) ∧ Z(x)) → S(x)) ∀x (A(x) → S(x))

Which of these existential quantifications are true, where the domain of x is the positive integers?

∃x (x > 0) ∃x (x2 > x)

In which of the following expressions is the variable x free?

∃y P(y) ∧ Q(x) ∀y (P(x) ∧ Q(y))

Match the fully parenthesized expression on the left to that on the right. 1. ((p → q) ∨ r) 2. (p → (q ∨ r)) 3. ((p ∧ q) ∨ r) 4. (p ∧ (q ∨ r))

1. This cannot be expressed using the operators → and ∨ and no parentheses. 2. p → q ∨ r 3. p ∧ q ∨ r 4. (p∧ (q∨ r)) Drop zone This cannot be expressed using the operators ∧ and ∨ and no parentheses. correct Toggle Button Unavailable. This cannot be expressed using the operators ∧ and ∨ and no parentheses.

Which of the following sentences in propositional logic are contradictions?

-p ^ p -(p -> p)

Match each bit string on the left with the operation on the right used to obtain it from the bit strings 1100 and 0110. 1. 0100 2. 1110 3. 1010 4. 1011

1. AND 2. OR 3. XOR 4. This string is not the AND, OR, or XOR of the two strings.

For which of these universes is ∃x (x^3 ≤ x^2) true? All negative integers All positive integers greater than 1 {0, 2, 3, 4, 5} {2, 3, 4, 5} All positive integers

All negative integers {0, 2, 3, 4, 5} All positive integers

Which of the following are correct ways of determining if two compound propositions p and q are equivalent. (Select all that are correct.) Construct a truth table for (p → q) ∧ (q → p). If the proposition is tautology, then p and q are equivalent. Construct a truth table for both of the compound propositions p and q. If the two are true in the same rows, then p ≡ q. Construct a truth table for p ↔ q. If it is a tautology, then p and q are equivalent. Construct a truth table for p → q. If it is a tautology, then p and q are equivalent.

Construct a truth table for (p → q) ∧ (q → p). If the proposition is tautology, then p and q are equivalent. Construct a truth table for both of the compound propositions p and q. If the two are true in the same rows, then p ≡ q. Construct a truth table for p ↔ q. If it is a tautology, then p and q are equivalent.

For which predicates P is the statement ∀x P(x) true, where the domain is the positive integers?

P(x) is the statement "x^2 ≥ x." P(x) is the statement "x > 0."

In which of these rows of the truth table is the compound proposition (p ∧ r) → ¬(q ∨ p) true?

The row where both p and q are true, but r is false. The row where p, q, and r are all false.

Match the English on the left with the predicate calculus translation on the right. Use C(x) for "x is in this class," and P(x) for "x has a perfect score." The domain for the variables is the collection of all students.

There is exactly one student in this class who has a perfect score. ⇔ ∃x ((C(x) ∧ P(x)) ∧ ∀y ((y ≠ x) → ¬(C(x) ∧ P(x)))) All students in the class have a perfect score. ⇔ ∀x (C(x) → P(x)) At least one student in the class has a perfect score. ⇔ ∃x (C(x) ∧ P(x)) There is no one in the class with a perfect score. ⇔ ∀x (¬C(x) ∨ ¬P(x)) There is exactly one student in this class, and that student has a perfect score. ⇔ ∃x ((C(x) ∧ P(x)) ∧ ∀y (C(y) → (y = x)))

Consider the following Prolog program using the predicates hasPet(a, b) to represent that person a owns a pet b and doesNotEat(a, b) to represent that animal a does not eat animal b. hasPet(joe, fish). hasPet(pierre, pig). hasPet(joe, chameleon). hasPet(selvi, cat). hasPet(rich, dog). hasPet(kumar, turkey). hasPet(rich, turkey). doesNotEat(X, Y) :- hasPet(Z, X), hasPet(Z, Y).

hasPet(X, fish) ⇔ joe doesNotEat(fish, chameleon) ⇔ yes doesNotEat(turkey, pig) ⇔ no hasPet(rich, Y) ⇔ dog

Match the sentence on the left with the equivalent statement on the right.

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