1. Which of the following is logically equivalent to the expression
p∨(p∧q)p \lor (p \land q)p∨(p∧q)?
A. p∧qp \land qp∧q
B. ppp
C. qqq
D. ¬p∨q\neg p \lor q¬p∨q
Answer: B) ppp
Rationale: This is an example of redundancy; p∨(p∧q)p \lor (p
\land q)p∨(p∧q) simplifies to ppp, as ppp alone makes the whole
expression true.
2. Which of the following represents the converse of the
statement "If ppp, then qqq"?
A. If ¬q\neg q¬q, then ¬p\neg p¬p
B. If qqq, then ppp
C. If ppp, then qqq
D. If ¬p\neg p¬p, then ¬q\neg q¬q
Answer: B) If qqq, then ppp
Rationale: The converse of p→qp \rightarrow qp→q is q→pq
\rightarrow pq→p, which reverses the direction of the
implication.
,3. Which of the following represents the negation of the
statement "∀x∈S,P(x)\forall x \in S, P(x)∀x∈S,P(x)"?
A. ∃x∈S,¬P(x)\exists x \in S, \neg P(x)∃x∈S,¬P(x)
B. ¬∃x∈S,P(x)\neg \exists x \in S, P(x)¬∃x∈S,P(x)
C. ∀x∈S,¬P(x)\forall x \in S, \neg P(x)∀x∈S,¬P(x)
D. ∃x∈S,P(x)\exists x \in S, P(x)∃x∈S,P(x)
Answer: A) ∃x∈S,¬P(x)\exists x \in S, \neg P(x)∃x∈S,¬P(x)
Rationale: The negation of a universal quantifier ∀\forall∀
becomes an existential quantifier ∃\exists∃ with the negated
predicate.
4. What is the logical negation of the statement "There exists an
xxx such that P(x)P(x)P(x) is true"?
A. For all xxx, P(x)P(x)P(x) is false.
B. For some xxx, P(x)P(x)P(x) is false.
C. There exists an xxx such that P(x)P(x)P(x) is false.
D. There exists an xxx such that P(x)P(x)P(x) is true.
Answer: A) For all xxx, P(x)P(x)P(x) is false.
Rationale: The negation of the existential quantifier ∃xP(x)\exists
x P(x)∃xP(x) becomes a universal quantifier ∀x¬P(x)\forall x
\neg P(x)∀x¬P(x).
, 5. Which of the following is a tautology?
A. p∧¬pp \land \neg pp∧¬p
B. p∨¬pp \lor \neg pp∨¬p
C. p→¬pp \rightarrow \neg pp→¬p
D. ¬p→p\neg p \rightarrow p¬p→p
Answer: B) p∨¬pp \lor \neg pp∨¬p
Rationale: The expression p∨¬pp \lor \neg pp∨¬p is always true,
regardless of the truth value of ppp, making it a tautology.
6. Which of the following statements is a contradiction?
A. p∧¬pp \land \neg pp∧¬p
B. p∨¬pp \lor \neg pp∨¬p
C. p→pp \rightarrow pp→p
D. p↔pp \leftrightarrow pp↔p
Answer: A) p∧¬pp \land \neg pp∧¬p
Rationale: A contradiction is a statement that is always false.
p∧¬pp \land \neg pp∧¬p is a contradiction because ppp and
¬p\neg p¬p cannot both be true.
p∨(p∧q)p \lor (p \land q)p∨(p∧q)?
A. p∧qp \land qp∧q
B. ppp
C. qqq
D. ¬p∨q\neg p \lor q¬p∨q
Answer: B) ppp
Rationale: This is an example of redundancy; p∨(p∧q)p \lor (p
\land q)p∨(p∧q) simplifies to ppp, as ppp alone makes the whole
expression true.
2. Which of the following represents the converse of the
statement "If ppp, then qqq"?
A. If ¬q\neg q¬q, then ¬p\neg p¬p
B. If qqq, then ppp
C. If ppp, then qqq
D. If ¬p\neg p¬p, then ¬q\neg q¬q
Answer: B) If qqq, then ppp
Rationale: The converse of p→qp \rightarrow qp→q is q→pq
\rightarrow pq→p, which reverses the direction of the
implication.
,3. Which of the following represents the negation of the
statement "∀x∈S,P(x)\forall x \in S, P(x)∀x∈S,P(x)"?
A. ∃x∈S,¬P(x)\exists x \in S, \neg P(x)∃x∈S,¬P(x)
B. ¬∃x∈S,P(x)\neg \exists x \in S, P(x)¬∃x∈S,P(x)
C. ∀x∈S,¬P(x)\forall x \in S, \neg P(x)∀x∈S,¬P(x)
D. ∃x∈S,P(x)\exists x \in S, P(x)∃x∈S,P(x)
Answer: A) ∃x∈S,¬P(x)\exists x \in S, \neg P(x)∃x∈S,¬P(x)
Rationale: The negation of a universal quantifier ∀\forall∀
becomes an existential quantifier ∃\exists∃ with the negated
predicate.
4. What is the logical negation of the statement "There exists an
xxx such that P(x)P(x)P(x) is true"?
A. For all xxx, P(x)P(x)P(x) is false.
B. For some xxx, P(x)P(x)P(x) is false.
C. There exists an xxx such that P(x)P(x)P(x) is false.
D. There exists an xxx such that P(x)P(x)P(x) is true.
Answer: A) For all xxx, P(x)P(x)P(x) is false.
Rationale: The negation of the existential quantifier ∃xP(x)\exists
x P(x)∃xP(x) becomes a universal quantifier ∀x¬P(x)\forall x
\neg P(x)∀x¬P(x).
, 5. Which of the following is a tautology?
A. p∧¬pp \land \neg pp∧¬p
B. p∨¬pp \lor \neg pp∨¬p
C. p→¬pp \rightarrow \neg pp→¬p
D. ¬p→p\neg p \rightarrow p¬p→p
Answer: B) p∨¬pp \lor \neg pp∨¬p
Rationale: The expression p∨¬pp \lor \neg pp∨¬p is always true,
regardless of the truth value of ppp, making it a tautology.
6. Which of the following statements is a contradiction?
A. p∧¬pp \land \neg pp∧¬p
B. p∨¬pp \lor \neg pp∨¬p
C. p→pp \rightarrow pp→p
D. p↔pp \leftrightarrow pp↔p
Answer: A) p∧¬pp \land \neg pp∧¬p
Rationale: A contradiction is a statement that is always false.
p∧¬pp \land \neg pp∧¬p is a contradiction because ppp and
¬p\neg p¬p cannot both be true.
