Questions: Select the correct choices that complete the sentence below. De Morgan's laws state that ¬(p ∧ q) is equivalent to and ¬(p ∨ q) is equivalent to

Select the correct choices that complete the sentence below. De Morgan's laws state that ¬(p ∧ q) is equivalent to and ¬(p ∨ q) is equivalent to
Transcript text: Select the correct choices that complete the sentence below. De Morgan's laws state that $\sim(p \wedge q)$ is equivalent to $\square$ and $\sim(p \vee q)$ is equivalent to $\square$
failed

Solution

failed
failed

Solution Steps

To solve this problem, we need to apply De Morgan's laws from logic. De Morgan's laws provide a way to express the negation of a conjunction or disjunction in terms of the negation of its components. Specifically, the laws state that the negation of a conjunction is equivalent to the disjunction of the negations, and the negation of a disjunction is equivalent to the conjunction of the negations.

Solution Approach
  1. Use De Morgan's first law: The negation of a conjunction, \(\sim(p \wedge q)\), is equivalent to \(\sim p \vee \sim q\).
  2. Use De Morgan's second law: The negation of a disjunction, \(\sim(p \vee q)\), is equivalent to \(\sim p \wedge \sim q\).
Step 1: Apply De Morgan's First Law

According to De Morgan's first law, the negation of a conjunction is expressed as: \[ \sim(p \wedge q) \equiv \sim p \vee \sim q \] For the given values \( p = \text{True} \) and \( q = \text{False} \):

  • \( \sim p = \text{False} \)
  • \( \sim q = \text{True} \)

Thus, we have: \[ \sim(p \wedge q) = \sim(\text{True} \wedge \text{False}) = \sim(\text{False}) = \text{True} \] And: \[ \sim p \vee \sim q = \text{False} \vee \text{True} = \text{True} \] This confirms that: \[ \sim(p \wedge q) \equiv \sim p \vee \sim q \]

Step 2: Apply De Morgan's Second Law

According to De Morgan's second law, the negation of a disjunction is expressed as: \[ \sim(p \vee q) \equiv \sim p \wedge \sim q \] For the same values \( p = \text{True} \) and \( q = \text{False} \):

  • \( \sim p = \text{False} \)
  • \( \sim q = \text{True} \)

Thus, we have: \[ \sim(p \vee q) = \sim(\text{True} \vee \text{False}) = \sim(\text{True}) = \text{False} \] And: \[ \sim p \wedge \sim q = \text{False} \wedge \text{True} = \text{False} \] This confirms that: \[ \sim(p \vee q) \equiv \sim p \wedge \sim q \]

Final Answer

The correct choices that complete the sentence are: \[ \sim(p \wedge q) \equiv \sim p \vee \sim q \quad \text{and} \quad \sim(p \vee q) \equiv \sim p \wedge \sim q \] Thus, the final answer is: \[ \boxed{\sim(p \wedge q) \equiv \sim p \vee \sim q \text{ and } \sim(p \vee q) \equiv \sim p \wedge \sim q} \]

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of f(x) = 3x^2 / (x-7). Find the domain of f(x). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The domain is all real x, except x = . (Type an integer or a decimal. Use a comma to separate answers as needed.) B. The domain is all real x.
failedAnswered
∫ from 4 to 6 h(x) dx = □ ∫ from -5 to 6 h(x) dx = □ ∫ from -2 to 4 h(x) dx = □ ∫ from -5 to 4 h(x) dx = □ ∫ from 4 to 5 h(x) dx = □ ∫ from -2 to 2 h(x) dx = □ ∫ from -4 to 5 h(x) dx = □ ∫ from -4 to 4 h(x) dx = □
failedAnswered
(a) Using exponential notation, we can write the cube root of 10 as . (b) Using radicals, we can write 10 to the power of 1/2 as . (c) Is there a difference between the square root of 10 squared and (the square root of 10) squared? Yes No
failedAnswered
Simplify -√400 (Enter ∅ if your answer is not a real number.)
failedAnswered
Considered late if submitted after 11 pm on the due date. Question 6 5 pts is the force of gravity that is exerted on an object. While is the quantity of matter in a given sample of a substance.