The expression you've provided, `~p → (p → q)`, is a logical implication involving propositions \( p \), \( q \), and a negation of \( p \), written as \( \sim p \) (read as "not p"). Let’s break down and simplify this logical expression step by step:

### Step-by-Step Analysis:

1. **Expression:** \( \sim p \rightarrow (p \rightarrow q) \)
   - \( \sim p \): This means "not \( p \)", so the truth value of \( p \) is false.
   - \( p \rightarrow q \): This is a conditional statement that means "if \( p \), then \( q \)".

2. **Truth Table Construction:**
   To fully understand this expression, we need to construct a truth table for all possible truth values of \( p \) and \( q \).

| \( p \) | \( q \) | \( \sim p \) | \( p \rightarrow q \) | \( \sim p \rightarrow (p \rightarrow q) \) |
|--------|--------|--------------|----------------------|--------------------------------------------|
| T      | T      | F            | T                    | T                                          |
| T      | F      | F            | F                    | T                                          |
| F      | T      | T            | T                    | T                                          |
| F      | F      | T            | T                    | T                                          |

### Interpretation:
- **Case 1: \( p = T \), \( q = T \):**
   - \( \sim p = F \), and \( p \rightarrow q = T \).
   - So, \( \sim p \rightarrow (p \rightarrow q) = T \), because \( F \rightarrow T = T \).

- **Case 2: \( p = T \), \( q = F \):**
   - \( \sim p = F \), and \( p \rightarrow q = F \).
   - So, \( \sim p \rightarrow (p \rightarrow q) = T \), because \( F \rightarrow F = T \).

- **Case 3: \( p = F \), \( q = T \):**
   - \( \sim p = T \), and \( p \rightarrow q = T \).
   - So, \( \sim p \rightarrow (p \rightarrow q) = T \), because \( T \rightarrow T = T \).

- **Case 4: \( p = F \), \( q = F \):**
   - \( \sim p = T \), and \( p \rightarrow q = T \).
   - So, \( \sim p \rightarrow (p \rightarrow q) = T \), because \( T \rightarrow T = T \).

### Conclusion:
The statement \( \sim p \rightarrow (p \rightarrow q) \) is **always true**, which means this is a **tautology**. No matter the truth values of \( p \) and \( q \), the whole expression will evaluate to true.

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Would you like more details on how the tautology is derived or how conditionals work? Here are some related questions:

1. What is the definition of a tautology in propositional logic?
2. How do truth tables help verify logical statements?
3. Can you explain the properties of the conditional (→) operator in more depth?
4. How does the negation (\( \sim \)) of a proposition affect logical expressions?
5. What are other examples of tautologies in propositional logic?

**Tip:** A conditional statement is false only when the antecedent is true and the consequent is false.

Explore the logical expression ~p → (p → q) and learn how it simplifies into a tautology using a step-by-step truth table. This problem covers core concepts in propositional logic such as negation, implication, and tautology verification. Suitable for Grades 10-12 students.

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