Which of the following relations is valid for Boolean algebra :-

Options

A. `A(B+bar(B))=A`
B. `A+AB =A`
C. `A+0=A`
D. all

Answer

To determine which of the given relations is valid for Boolean algebra, we will analyze each option step by step.

### Step 1: Analyze Option A
**Relation:** \( A \cdot (B + \overline{B}) = A \)

**Left-Hand Side (LHS):**
- We know that \( B + \overline{B} = 1 \) (This is a fundamental property of Boolean algebra, where a variable ORed with its complement equals 1).
- Therefore, LHS becomes:
\[ A \cdot (B + \overline{B}) = A \cdot 1 = A \]

**Right-Hand Side (RHS):**
- The RHS is simply \( A \).

**Conclusion for Option A:**
- Since LHS = RHS, the relation is valid.

### Step 2: Analyze Option B
**Relation:** \( A + AB = A \)

**Left-Hand Side (LHS):**
- We can factor out \( A \) from the expression:
\[ A + AB = A(1 + B) \]
- We know that \( 1 + B = 1 \) (Another fundamental property of Boolean algebra).
- Thus, LHS simplifies to:
\[ A(1 + B) = A \cdot 1 = A \]

**Right-Hand Side (RHS):**
- The RHS is simply \( A \).

**Conclusion for Option B:**
- Since LHS = RHS, the relation is valid.

### Step 3: Analyze Option C
**Relation:** \( A + 0 = A \)

**Left-Hand Side (LHS):**
- The LHS is simply \( A + 0 \).
- According to Boolean algebra, adding 0 to any variable does not change its value:
\[ A + 0 = A \]

**Right-Hand Side (RHS):**
- The RHS is simply \( A \).

**Conclusion for Option C:**
- Since LHS = RHS, the relation is valid.

### Final Conclusion
All options A, B, and C are valid relations in Boolean algebra.