To analyze the logical equivalence of the expression `~(P ∧ Q) ↔ (~P ∨ ~Q)`, let's break it down step by step. 

### Step 1: Understand the symbols
- `~` means "NOT."
- `∧` means "AND."
- `∨` means "OR."
- `↔` means "if and only if" (bi-conditional).

The expression states that the negation of `(P ∧ Q)` is logically equivalent to `(¬P ∨ ¬Q)`.

### Step 2: Decompose `~(P ∧ Q)`
Using **De Morgan's Laws**, the negation of a conjunction can be expressed as:
\[
~(P ∧ Q) = (~P ∨ ~Q)
\]

### Step 3: Substitute back
The original statement now becomes:
\[
(~P ∨ ~Q) ↔ (~P ∨ ~Q)
\]

### Step 4: Simplify
The biconditional (`↔`) states that both sides are logically equivalent. Clearly:
\[
(~P ∨ ~Q) ↔ (~P ∨ ~Q)
\]
is a **tautology** (always true), as any statement is equivalent to itself.

### Conclusion:
The statement `~(P ∧ Q) ↔ (~P ∨ ~Q)` is **always true** due to De Morgan's laws.

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Would you like a truth table to illustrate this? Let me know!

### Follow-up Questions
1. What are De Morgan's laws, and how do they simplify expressions?
2. Can you provide examples of biconditional logical equivalences?
3. How can truth tables help verify logical equivalences?
4. What is the difference between tautology and contingency in logic?
5. How does the distributive property apply to logical statements?

### Tip:
When working with logical equivalences, always consider De Morgan’s laws to simplify negations of compound statements.

Explore the logical equivalence of the statement ~(P ∧ Q) ↔ (~P ∨ ~Q). This detailed guide uses De Morgan's laws and tautology principles to demonstrate the equivalence with step-by-step reasoning. Ideal for students in Grades 9-12 studying propositional logic.

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