If p always speaks against q, then ` p Rightarrow p vv ~ q ` is

Options

A. a tautology
B. contradiction
C. contingency
D. none of these

Answer

To solve the problem, we need to analyze the statement "If p always speaks against q, then \( p \Rightarrow (p \lor \neg q) \)".

### Step-by-Step Solution:

1. **Understanding the Statement**:
- The statement "p always speaks against q" implies that whenever p is true, q must be false. This can be represented as \( p \Rightarrow \neg q \).

2. **Rewriting the Implication**:
- The expression we need to analyze is \( p \Rightarrow (p \lor \neg q) \).
- Recall that \( p \Rightarrow r \) can be rewritten as \( \neg p \lor r \). Therefore, we can rewrite our expression:
\[ p \Rightarrow (p \lor \neg q) \equiv \neg p \lor (p \lor \neg q) \]

3. **Simplifying the Expression**:
- The expression \( \neg p \lor (p \lor \neg q) \) can be simplified using the associative property of disjunction:
\[ \neg p \lor p \lor \neg q \]
- The term \( \neg p \lor p \) is always true (this is known as the Law of Excluded Middle). Thus, we have:
\[ \text{True} \lor \neg q \]
- Since anything ORed with true is true, we conclude:
\[ \neg p \lor (p \lor \neg q) \equiv \text{True} \]

4. **Conclusion**:
- The expression \( p \Rightarrow (p \lor \neg q) \) is always true, which means it is a tautology.

### Final Answer:
The expression \( p \Rightarrow (p \lor \neg q) \) is a **tautology**.