solve for ~[(p^q) **→**(pvq)] v [(~p)v(~q)]

We are tasked with simplifying and solving the logical expression:

\[
\sim [(p^q) \to (p \vee q)] \vee (\sim p \vee \sim q)
\]

Let’s break it down step by step.

### 1. Breakdown of the parts:
- \( p^q \): This is the conjunction (AND) of \( p \) and \( q \).
- \( p \vee q \): This is the disjunction (OR) of \( p \) and \( q \).
- \( \sim [(p^q) \to (p \vee q)] \): This represents the negation of an implication.
- \( (\sim p \vee \sim q) \): This is the disjunction of the negations of \( p \) and \( q \).

### 2. Solve the implication \((p^q) \to (p \vee q)\):
An implication \( A \to B \) is logically equivalent to \( \sim A \vee B \).

Thus:
\[
(p^q) \to (p \vee q) \equiv \sim(p^q) \vee (p \vee q)
\]

Now, \( \sim(p^q) \) is the negation of the conjunction, which becomes:
\[
\sim(p^q) \equiv \sim p \vee \sim q
\]

Thus, the implication simplifies to:
\[
\sim(p^q) \vee (p \vee q) = (\sim p \vee \sim q) \vee (p \vee q)
\]

### 3. Negation of the result:
Now we need to negate this entire expression:
\[
\sim [(\sim p \vee \sim q) \vee (p \vee q)]
\]

Using De Morgan’s law, which states that \( \sim(A \vee B) \equiv \sim A \wedge \sim B \), we get:
\[
\sim [(\sim p \vee \sim q) \vee (p \vee q)] = \sim (\sim p \vee \sim q) \wedge \sim (p \vee q)
\]

Simplifying both terms:
- \( \sim (\sim p \vee \sim q) = p^q \) (De Morgan’s law).
- \( \sim (p \vee q) = \sim p \wedge \sim q \) (again using De Morgan’s law).

Thus, this part simplifies to:
\[
p^q \wedge (\sim p \wedge \sim q)
\]

### 4. Combine with the second disjunction:
Now we need to consider the second part of the original expression:
\[
p^q \wedge (\sim p \wedge \sim q) \vee (\sim p \vee \sim q)
\]

At this point, we observe that the term \( p^q \wedge (\sim p \wedge \sim q) \) is always false because \( p \) cannot be both true and false at the same time.

Thus, we are left with:
\[
(\sim p \vee \sim q)
\]

### Final answer:
The simplified expression is:
\[
\sim p \vee \sim q
\]
This is the expression for the **NOR** (negation of OR), and is logically equivalent to \( p \downarrow q \) (NAND operation).

---

Would you like more details or have any questions?

Here are 5 related questions for deeper understanding:
1. How does De Morgan’s law apply to logical negations?
2. What is the truth table for \( p \to q \)?
3. How can we use truth tables to verify the result?
4. What are the differences between conjunction and disjunction in logic?
5. How do you simplify expressions involving implications?

**Tip:** Always remember De Morgan’s laws when dealing with negations of complex logical expressions!

solve for ~[(p^q) → (pvq)] v [(~p)v(~q)]

Learn how to simplify the logical expression ~[(p^q) → (pvq)] v [(~p)v(~q)] with a step-by-step solution. This problem involves key concepts in propositional logic, including De Morgan's Laws and logical equivalence, making it suitable for students in Grades 10-12.

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