Core Boolean Simplification Laws Quiz

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1) Simplify A ∧ 1

According to the identity law in Boolean algebra, the conjunction of any variable with 1 is the variable itself. Therefore, A ∧ 1 simplifies to A.

Explanation

According to the identity law in Boolean algebra, the conjunction of any variable with 1 is the variable itself. Therefore, A ∧ 1 simplifies to A.

About This Quiz

Core Boolean Simplification Laws Quiz - Quiz

Want to make long logic expressions look short and clean? In this quiz, you’ll practice the core Boolean algebra simplification rules like identity, annihilator, complement, idempotent, and absorption. You’ll work through expressions such as A ∧ 1, A ∨ 0, A ∧ A', and mixed forms like A ∨ (A... see more∧ B) or A ∧ (A' ∨ B). Step by step, you’ll see how these laws help you reduce expressions to their simplest form without changing their truth. By the end, you’ll feel much more confident spotting patterns and quickly deciding which law to use to simplify a given Boolean expression.
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2) Simplify A ∧ 0

According to the annihilator law in Boolean algebra, the conjunction of any variable with 0 is 0. Therefore, A ∧ 0 simplifies to 0.

Explanation

According to the annihilator law in Boolean algebra, the conjunction of any variable with 0 is 0. Therefore, A ∧ 0 simplifies to 0.

3) Simplify A ∨ 1

According to the annihilator law in Boolean algebra, the disjunction of any variable with 1 is 1. Therefore, A ∨ 1 simplifies to 1.

Explanation

According to the annihilator law in Boolean algebra, the disjunction of any variable with 1 is 1. Therefore, A ∨ 1 simplifies to 1.

4) Simplify A ∨ (A ∧ B)

A ∧ B

A ∨ B

According to the absorption law in Boolean algebra, the disjunction of A and the conjunction of A and B simplifies to A. This is because if A is true, the entire expression is true, and if A is false, both A and A ∧ B are false, so the expression is false. Thus, A ∨ (A ∧ B) = A.

Explanation

5) Simplify A ∧ (A' ∨ B)

A ∧ B

A ∨ B

Using the distributive law, A ∧ (A' ∨ B) = (A ∧ A') ∨ (A ∧ B). Then, by the complement law, A ∧ A' = 0. So, the expression becomes 0 ∨ (A ∧ B). Finally, by the identity law, 0 ∨ (A ∧ B) = A ∧ B.

Explanation

6) Simplify (A ∧ 1) V (0 ∧ B)

A V B

A ∧ B

First, by the identity law, A ∧ 1 simplifies to A. Second, by the annihilator law, 0 ∧ B simplifies to 0. Then, by the identity law, A V 0 simplifies to A. Therefore, (A ∧ 1) V (0 ∧ B) = A V 0 = A.

Explanation

7) Simplify A ∨ 0

According to the identity law in Boolean algebra, the disjunction of any variable with 0 is the variable itself. Therefore, A ∨ 0 simplifies to A.

Explanation

According to the identity law in Boolean algebra, the disjunction of any variable with 0 is the variable itself. Therefore, A ∨ 0 simplifies to A.

8) Simplify A ∧ A'

According to the complement law in Boolean algebra, the conjunction of a variable and its complement is always 0. Therefore, A ∧ A' simplifies to 0.

Explanation

According to the complement law in Boolean algebra, the conjunction of a variable and its complement is always 0. Therefore, A ∧ A' simplifies to 0.

9) Simplify A ∧ (B ∨ B')

A ∧ B

First, according to the complement law, B ∨ B' simplifies to 1. Then, according to the identity law, A ∧ 1 simplifies to A. Therefore, A ∧ (B ∨ B') = A ∧ 1 = A.

Explanation

First, according to the complement law, B ∨ B' simplifies to 1. Then, according to the identity law, A ∧ 1 simplifies to A. Therefore, A ∧ (B ∨ B') = A ∧ 1 = A.

10) Simplify (A ∨ B) ∧ A

A ∨ B

A ∧ B

According to the absorption law, (A ∨ B) ∧ A simplifies to A. This can be verified by distribution: (A ∨ B) ∧ A = A ∧ A ∨ B ∧ A = A ∨ (B ∧ A). Since A ∨ (A ∧ B) = A by absorption, the expression simplifies to A.

Explanation

11) Simplify (A V 0) ∧ 1

First, by the identity law, A V 0 simplifies to A. Then, by the identity law, A ∧ 1 simplifies to A. Therefore, (A V 0) ∧ 1 = A ∧ 1 = A.

Explanation

First, by the identity law, A V 0 simplifies to A. Then, by the identity law, A ∧ 1 simplifies to A. Therefore, (A V 0) ∧ 1 = A ∧ 1 = A.

12) Simplify A ∨ A'

According to the complement law in Boolean algebra, the disjunction of a variable and its complement is always 1. Therefore, A ∨ A' simplifies to 1.

Explanation

According to the complement law in Boolean algebra, the disjunction of a variable and its complement is always 1. Therefore, A ∨ A' simplifies to 1.

13) Simplify A ∧ (A ∨ B)

A ∧ B

A ∨ B

According to the absorption law in Boolean algebra, the conjunction of A and the disjunction of A and B simplifies to A. This is because if A is true, the conjunction is true since A is true, and if A is false, the conjunction is false. Thus, A ∧ (A ∨ B) = A.

Explanation

14) Simplify A ∨ (B ∧ B')

A ∨ B

First, according to the complement law, B ∧ B' simplifies to 0. Then, according to the identity law, A ∨ 0 simplifies to A. Therefore, A ∨ (B ∧ B') = A ∨ 0 = A.

Explanation

First, according to the complement law, B ∧ B' simplifies to 0. Then, according to the identity law, A ∨ 0 simplifies to A. Therefore, A ∨ (B ∧ B') = A ∨ 0 = A.

15) Simplify p(A ∧ A)

First, by the idempotent law, A ∧ A simplifies to A. Then, ¬(A) is A'. Therefore, ¬(A ∧ A) = ¬A = A'.

Explanation

First, by the idempotent law, A ∧ A simplifies to A. Then, ¬(A) is A'. Therefore, ¬(A ∧ A) = ¬A = A'.

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Alva Benedict B. |PhD

College Expert

Alva Benedict B. is an experienced mathematician and math content developer with over 15 years of teaching and tutoring experience across high school, undergraduate, and test prep levels. He specializes in Algebra, Calculus, and Statistics, and holds advanced academic training in Mathematics with extensive expertise in LaTeX-based math content development.