Identifying Correct Boolean Algebra Rule

Boolean algebra is a fundamental branch of mathematics used extensively in digital logic design and computer science. It deals with variables that can have only two values: TRUE (1) or FALSE (0), and operations like AND (.), OR (+), and NOT ('). Understanding the basic rules and laws of Boolean algebra is crucial for simplifying complex logical expressions and designing efficient digital circuits.

The question asks us to identify the correct rule among the given options. Let's analyze each option:

Option 1: A . A’=1

This statement claims that the AND operation between a variable 'A' and its complement 'A’' results in 1.

According to the Complement Law in Boolean algebra, the AND operation between a variable and its complement is always 0.

The correct rule is: $$ A \cdot A' = 0 $$ Therefore, Option 1 is incorrect.

Option 2: A+AB=A+B

This statement suggests a simplification rule where 'A OR (A AND B)' equals 'A OR B'.

Let's simplify the left side, A + AB, using the Absorption Law of Boolean algebra. The Absorption Law states that A + AB = A.

So, the correct simplification is: $$ A + AB = A $$ The expression given in Option 2, A+B, is different from the correct simplification A. Therefore, Option 2 is incorrect.

Option 3: A+A’B = A+B

This statement proposes a specific Boolean algebra identity.

Let's verify this rule using basic Boolean algebra laws:

• Start with the expression: A + A’B
• We can use the distributive law, but first, we need to make the expression suitable for it. We know that A = A + A. So, we can rewrite the expression as: A + A + A’B.
• Rearranging using the commutative property: (A + A’B) + A. This doesn't seem helpful directly. Let's try another approach.
• We can use the consensus theorem or a variation of the distributive law. A useful identity is derived as follows: Start with A + A’B. Add A . A' to the expression (which equals 0, so it doesn't change the value): A + A’B + (A . A’). This doesn't simplify easily. Let's use the standard identity derivation: We know that A = 1 . A and 1 = A + A’ (Complement Law). Substitute A: $$ A + A'B = (A + A')(A + B) $$ This step uses the distributive property of OR over AND, applied in reverse: X + YZ = (X+Y)(X+Z). Here, X = A, Y = A', and Z = B. Now, apply the Complement Law (A + A’) = 1: $$ = 1 \cdot (A + B) $$ Finally, apply the Identity Law 1 . X = X: $$ = A + B $$ Thus, the rule A + A’B = A + B is correct.

Therefore, Option 3 is correct.

Option 4: A(A+B)+B

This option presents an expression rather than a rule equating two expressions. Let's simplify the expression:

• Expression: A(A+B)+B
• Apply the distributive law: (A . A) + (A . B) + B
• Simplify A . A using the Idempotent Law (X . X = X): A + AB + B
• Apply the Absorption Law (A + AB = A): A + B
• The expression simplifies to A + B. While the simplified result is A + B, the option itself is just the expression A(A+B)+B and not presented as an equality rule like Option 3. Given the context of the question asking for a correct *rule*, Option 3 is the explicitly stated and verifiable rule.

Conclusion

Based on the analysis, the rule A + A’B = A + B is a valid identity in Boolean algebra.