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. What is the truth value of the expression (p∧q)→(r∨s)(p \land
q) \rightarrow (r \lor s)(p∧q)→(r∨s) when
p=True,q=False,r=True,s=Falsep = \text{True}, q =
\text{False}, r = \text{True}, s =
\text{False}p=True,q=False,r=True,s=False?
A. True
B. False
C. Undefined
D. Cannot be determined
Answer: A) True
,Rationale: Since p∧qp \land qp∧q is False (because q=Falseq =
\text{False}q=False), the implication (p∧q)→(r∨s)(p \land q)
\rightarrow (r \lor s)(p∧q)→(r∨s) is always True, regardless of the
truth values of rrr and sss.
3. 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.
4. 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.
5. What is the truth value of the expression ¬(p∧q)∨(p→q)\neg(p
\land q) \lor (p \rightarrow q)¬(p∧q)∨(p→q) when p=Truep =
\text{True}p=True and q=Falseq = \text{False}q=False?
A. True
B. False
C. Undefined
D. Cannot be determined
Answer: A) True
Rationale: ¬(p∧q)\neg(p \land q)¬(p∧q) is true because p∧qp
\land qp∧q is false. The expression p→qp \rightarrow qp→q is
false. Since we have a disjunction (∨\lor∨), the entire expression is
true because ¬(p∧q)\neg(p \land q)¬(p∧q) is true.
6. Which of the following represents the negation of the
statement "∃x∈A,∀y∈B,P(x,y)\exists x \in A, \forall y \in B, P(x,
y)∃x∈A,∀y∈B,P(x,y)"?
A. ∀x∈A,∃y∈B,¬P(x,y)\forall x \in A, \exists y \in B, \neg P(x,
y)∀x∈A,∃y∈B,¬P(x,y)
B. ∃x∈A,∀y∈B,¬P(x,y)\exists x \in A, \forall y \in B, \neg P(x,
y)∃x∈A,∀y∈B,¬P(x,y)
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. What is the truth value of the expression (p∧q)→(r∨s)(p \land
q) \rightarrow (r \lor s)(p∧q)→(r∨s) when
p=True,q=False,r=True,s=Falsep = \text{True}, q =
\text{False}, r = \text{True}, s =
\text{False}p=True,q=False,r=True,s=False?
A. True
B. False
C. Undefined
D. Cannot be determined
Answer: A) True
,Rationale: Since p∧qp \land qp∧q is False (because q=Falseq =
\text{False}q=False), the implication (p∧q)→(r∨s)(p \land q)
\rightarrow (r \lor s)(p∧q)→(r∨s) is always True, regardless of the
truth values of rrr and sss.
3. 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.
4. 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.
5. What is the truth value of the expression ¬(p∧q)∨(p→q)\neg(p
\land q) \lor (p \rightarrow q)¬(p∧q)∨(p→q) when p=Truep =
\text{True}p=True and q=Falseq = \text{False}q=False?
A. True
B. False
C. Undefined
D. Cannot be determined
Answer: A) True
Rationale: ¬(p∧q)\neg(p \land q)¬(p∧q) is true because p∧qp
\land qp∧q is false. The expression p→qp \rightarrow qp→q is
false. Since we have a disjunction (∨\lor∨), the entire expression is
true because ¬(p∧q)\neg(p \land q)¬(p∧q) is true.
6. Which of the following represents the negation of the
statement "∃x∈A,∀y∈B,P(x,y)\exists x \in A, \forall y \in B, P(x,
y)∃x∈A,∀y∈B,P(x,y)"?
A. ∀x∈A,∃y∈B,¬P(x,y)\forall x \in A, \exists y \in B, \neg P(x,
y)∀x∈A,∃y∈B,¬P(x,y)
B. ∃x∈A,∀y∈B,¬P(x,y)\exists x \in A, \forall y \in B, \neg P(x,
y)∃x∈A,∀y∈B,¬P(x,y)
