Let p: Maths is intersting and q : Maths is easy, then `p rArr (~p vv q)` is equivalent to

Options

A. It Maths is easy then it is interesting
B. Either Maths is interesting or it easy
C. If Maths is interseting then it is easy
D. Maths is neither interesting nor easy

Answer

To solve the problem, we need to analyze the logical expression given: \( p \Rightarrow (\neg p \lor q) \).

### Step-by-Step Solution:

1. **Understand the Implication**:
The expression \( p \Rightarrow (\neg p \lor q) \) can be rewritten using the implication equivalence:
\[ p \Rightarrow r \equiv \neg p \lor r \]
Here, \( r \) is \( \neg p \lor q \). Thus, we can rewrite the expression as:
\[ \neg p \lor (\neg p \lor q) \]

2. **Apply Associative Law**:
The disjunction (or) operation is associative, which allows us to regroup the terms:
\[ \neg p \lor (\neg p \lor q) \equiv \neg p \lor \neg p \lor q \]

3. **Simplify Using Idempotent Law**:
According to the idempotent law in logic, \( X \lor X \equiv X \). Therefore:
\[ \neg p \lor \neg p \equiv \neg p \]
Thus, we can simplify the expression to:
\[ \neg p \lor q \]

4. **Reinterpret the Expression**:
Now, we can interpret \( \neg p \lor q \) in terms of implications. The expression \( \neg p \lor q \) can be rewritten as:
\[ p \Rightarrow q \]
This means "if \( p \) (Math is interesting), then \( q \) (Math is easy)".

### Final Result:
The expression \( p \Rightarrow (\neg p \lor q) \) is equivalent to:
\[ p \Rightarrow q \]