Key Concepts and Formulas
- De Morgan's Laws:
- ∼(p∨q)≡(∼p)∧(∼q)
- ∼(p∧q)≡(∼p)∨(∼q)
- Distributive Law:
- p∧(q∨r)≡(p∧q)∨(p∧r)
- p∨(q∧r)≡(p∨q)∧(p∨r)
- Definition of Biconditional (Equivalence):
- p↔q≡(p→q)∧(q→p)≡(p∧q)∨(∼p∧∼q)
- Complement Law: p∧∼p≡F (False) and p∨∼p≡T (True)
Step-by-Step Solution
Step 1: Negate the given expression using De Morgan's Law.
We are given the expression p∨(∼p∧q). We want to find the negation of this expression, which is ∼[p∨(∼p∧q)].
Using De Morgan's Law, we have:
∼[p∨(∼p∧q)]≡∼p∧∼(∼p∧q)
Step 2: Simplify the second negated term using De Morgan's Law again.
Now we simplify ∼(∼p∧q). Using De Morgan's Law, we get:
∼(∼p∧q)≡∼(∼p)∨∼q≡p∨∼q
Substituting this back into the expression from Step 1, we have:
∼p∧(p∨∼q)
Step 3: Apply the Distributive Law.
We distribute ∼p over (p∨∼q):
∼p∧(p∨∼q)≡(∼p∧p)∨(∼p∧∼q)
Step 4: Simplify using the Complement Law.
We know that ∼p∧p is always false, i.e., ∼p∧p≡F. Thus, the expression becomes:
F∨(∼p∧∼q)≡∼p∧∼q
Step 5: Connect to the definition of Biconditional.
We want to show that ∼p∧∼q is equivalent to p↔q. Recall that p↔q≡(p∧q)∨(∼p∧∼q). Therefore, ∼(p↔q)≡∼[(p∧q)∨(∼p∧∼q)]≡∼(p∧q)∧∼(∼p∧∼q)≡(∼p∨∼q)∧(p∨q)≡(∼p∧q)∨(∼q∧p) This is not equal to ∼p∧∼q.
Let's consider the expression p↔q≡(p→q)∧(q→p)≡(∼p∨q)∧(∼q∨p).
The negation of this is:
∼(p↔q)≡∼[(∼p∨q)∧(∼q∨p)]≡∼(∼p∨q)∨∼(∼q∨p)≡(p∧∼q)∨(q∧∼p)
This is not equal to ∼p∧∼q.
We know that p↔q≡(p∧q)∨(∼p∧∼q). So, ∼(p↔q)≡∼[(p∧q)∨(∼p∧∼q)]≡[∼(p∧q)]∧[∼(∼p∧∼q)]≡(∼p∨∼q)∧(p∨q)≡(∼p∧p)∨(∼p∧q)∨(∼q∧p)∨(∼q∧q)≡(∼p∧q)∨(∼q∧p).
Let's reconsider our initial simplification. We have ∼p∧∼q.
If we are given that the answer is p↔q, then ∼[p∨(∼p∧q)]=p↔q is incorrect. Instead, we should have that ∼[p∨(∼p∧q)]=∼p∧∼q which is equal to ∼p∧∼q. However, ∼p∧∼q is not equivalent to p↔q. But, if we negate the biconditional, we get ∼(p↔q)≡(p∧∼q)∨(q∧∼p). Which is not what we have.
Let us use a truth table.
| p | q | ~p | ~q | ~p ∧ q | p ∨ (~p ∧ q) | ~[p ∨ (~p ∧ q)] | p↔q |
|---|
| T | T | F | F | F | T | F | T |
| T | F | F | T | F | T | F | F |
| F | T | T | F | T | T | F | F |
| F | F | T | T | F | F | T | T |
Now, let's evaluate ∼p∧∼q
| p | q | ~p | ~q | ~p ∧ ~q |
|---|
| T | T | F | F | F |
| T | F | F | T | F |
| F | T | T | F | F |
| F | F | T | T | T |
We see that ∼[p∨(∼p∧q)]≡∼p∧∼q. But the given answer is p↔q. This means that the given answer is incorrect. Let's re-examine the original expression and our derivation.
∼[p∨(∼p∧q)]≡∼p∧∼(∼p∧q)≡∼p∧(p∨∼q)≡(∼p∧p)∨(∼p∧∼q)≡F∨(∼p∧∼q)≡∼p∧∼q
The options are:
(A) p↔q
(B) ∼p∧∼q
(C) p∧q
(D) ∼p∨∼q
Since we derived ∼p∧∼q, option (B) is the correct answer. However, the provided "Correct Answer" is (A).
Let's analyze option (A) and force our solution to reach it. We have ∼[p∨(∼p∧q)]. We want to show this is equal to p↔q. Let's write out the truth table for p↔q as shown above. We need to get the negation of the original expression to be: F, T, T, F.
But our truth table for ∼[p∨(∼p∧q)] is F, F, F, T. Thus, it is impossible for ∼[p∨(∼p∧q)] to equal to p↔q.
Common Mistakes & Tips
- Carefully apply De Morgan's laws, paying close attention to the change in operators.
- Remember the distributive property and how to apply it correctly.
- Utilize truth tables to verify your simplifications, especially when unsure.
Summary
We started by negating the given expression using De Morgan's Laws and the distributive property. After simplification using these logical equivalences, we arrived at the expression ∼p∧∼q. The given correct answer is p↔q, but we have proven through simplification and truth tables that this is incorrect. The correct answer should be ∼p∧∼q.
Final Answer
The final answer is \boxed{~p \wedge ~q}, which corresponds to option (B).