"WGU D420 Discrete Math Logic Question Bank with Rationale"

1. 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: The first part of the expression, ¬(p∧q)\neg(p \land
q)¬(p∧q), is true because p∧qp \land qp∧q is false. The second
part, p→qp \rightarrow qp→q, is false, but the overall expression
is true due to the disjunction.


2. 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.


3. Which of the following represents the logical 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)\neg \forall 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. 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.


5. 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.


6. Which of the following is logically equivalent to the expression
p∨(q∧r)p \lor (q \land r)p∨(q∧r)?

A. (p∨q)∧(p∨r)(p \lor q) \land (p \lor r)(p∨q)∧(p∨r)

B. p∨q∨rp \lor q \lor rp∨q∨r

C. (p∧q)∨(p∧r)(p \land q) \lor (p \land r)(p∧q)∨(p∧r)

D. (p∨q)∧r(p \lor q) \land r(p∨q)∧r