De Morgan’s And Distributive Simplification Quiz

1) Simplify ¬(A ∧ B)

¬A ∧ ¬B

A ∨ B

¬A ∨ ¬B

A ∧ B

According to De Morgan's law, the negation of a conjunction is the disjunction of the negations. Therefore, ¬(A ∧ B) is equivalent to ¬A ∨ ¬B.

Explanation

According to De Morgan's law, the negation of a conjunction is the disjunction of the negations. Therefore, ¬(A ∧ B) is equivalent to ¬A ∨ ¬B.

De Morgans and Distributive Simplification Quiz - Quiz

Think you can recognize deeper patterns inside brackets and negations? This quiz gives you plenty of practice applying De Morgan’s laws, distributive laws, and factoring tricks to simplify more involved Boolean expressions. You’ll rewrite negations like ¬(A ∧ B) and ¬(A ∨ B), distribute terms across ∧ and ∨, and... see morefactor common parts out of expressions such as (A ∧ B) ∨ (A ∧ C) or (A ∨ B) ∧ (A ∨ C). Along the way, you’ll see how complement identities like X' ∨ X = 1 and B ∨ B' = 1 help collapse big formulas into small, elegant ones. By the end, turning a cluttered expression into a neat, equivalent version will start to feel natural.
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2) Simplify ¬(A ∨ B)

¬A ∨ ¬B

A ∧ B

A ∨ B

¬A ∧ ¬B

According to De Morgan's law, the negation of a disjunction is the conjunction of the negations. Therefore, ¬(A ∨ B) is equivalent to ¬A ∧ ¬B.

Explanation

According to De Morgan's law, the negation of a disjunction is the conjunction of the negations. Therefore, ¬(A ∨ B) is equivalent to ¬A ∧ ¬B.

3) Simplify ¬(A' ∧ B')

A ∨ B

A ∧ B

A' ∨ B'

¬A ∨ ¬B

Apply De Morgan's law: ¬(A' ∧ B') = ¬A' ∨ ¬B'. Since ¬A' is A and ¬B' is B, this simplifies to A ∨ B.

Explanation

Apply De Morgan's law: ¬(A' ∧ B') = ¬A' ∨ ¬B'. Since ¬A' is A and ¬B' is B, this simplifies to A ∨ B.

4) Simplify ¬(¬A ∨ ¬B)

A ∨ B

A ∧ B

¬A ∧ ¬B

¬A ∨ ¬B

Apply De Morgan's law: ¬(¬A ∨ ¬B) = ¬¬A ∧ ¬¬B. Since ¬¬A is A and ¬¬B is B, this simplifies to A ∧ B.

Explanation

Apply De Morgan's law: ¬(¬A ∨ ¬B) = ¬¬A ∧ ¬¬B. Since ¬¬A is A and ¬¬B is B, this simplifies to A ∧ B.

5) Simplify ¬(A' ∨ B)

A V ¬B

A ∧ ¬B

¬A ∧ ¬B

A' ∧ B'

Apply De Morgan's law: ¬(A' ∨ B) = ¬A' ∧ ¬B. Since ¬A' is A, this simplifies to A ∧ ¬B.

Explanation

Apply De Morgan's law: ¬(A' ∨ B) = ¬A' ∧ ¬B. Since ¬A' is A, this simplifies to A ∧ ¬B.

6) (A ∧ B)' V (A ∧ B)

This expression is of the form X' ∨ X, where X is (A ∧ B). According to the complement law, X' ∨ X = 1 for any X. Therefore, (A ∧ B)' ∨ (A ∧ B) = 1.

Explanation

This expression is of the form X' ∨ X, where X is (A ∧ B). According to the complement law, X' ∨ X = 1 for any X. Therefore, (A ∧ B)' ∨ (A ∧ B) = 1.

7) Simplify A ∧ (B V C)

(A ∧ B) V (A ∧ C)

(A V B) ∧ (A V C)

A ∧ B ∧ C

Cannot simplify further

According to the distributive law of conjunction over disjunction, A ∧ (B V C) is equivalent to (A ∧ B) V (A ∧ C).

Explanation

According to the distributive law of conjunction over disjunction, A ∧ (B V C) is equivalent to (A ∧ B) V (A ∧ C).

8) Simplify A V (B ∧ C)

(A ∧ B) V (A ∧ C)

(A V B) ∧ (A V C)

A V B V C

Cannot simplify further

According to the distributive law of disjunction over conjunction, A V (B ∧ C) is equivalent to (A V B) ∧ (A V C).

Explanation

According to the distributive law of disjunction over conjunction, A V (B ∧ C) is equivalent to (A V B) ∧ (A V C).

9) Simplify (A ∧ B) V (A ∧ C)

A ∧ (B V C)

A V (B ∧ C)

(A V B) ∧ C

A ∧ B ∧ C

According to the distributive law, (A ∧ B) V (A ∧ C) can be factored as A ∧ (B V C).

Explanation

According to the distributive law, (A ∧ B) V (A ∧ C) can be factored as A ∧ (B V C).

10) Simplify (A V B) ∧ (A V C)

A ∧ (B V C)

A V (B ∧ C)

A V B V C

A ∧ B ∧ C

According to the distributive law, (A V B) ∧ (A V C) can be simplified to A V (B ∧ C).

Explanation

According to the distributive law, (A V B) ∧ (A V C) can be simplified to A V (B ∧ C).

11) Simplify (A ∧ B) ∨ (A ∧ B')

A ∧ (B ∨ B')

A ∨ B

Factor out A: (A ∧ B) ∨ (A ∧ B') = A ∧ (B ∨ B'). Since B ∨ B' = 1 by complement law, then A ∧ 1 = A by identity law.

Explanation

Factor out A: (A ∧ B) ∨ (A ∧ B') = A ∧ (B ∨ B'). Since B ∨ B' = 1 by complement law, then A ∧ 1 = A by identity law.

12) Simplify A ∧ B ∧ (A ∨ C)

A ∧ B

A ∧ C

B ∧ C

A ∧ B ∧ C

According to the absorption law, A ∧ B ∧ (A ∨ C) simplifies to A ∧ B. This is because if A and B are true, the expression is true regardless of C, and if A or B is false, the expression is false.

Explanation

13) Simplify (A ∨ B) ∧ A

A ∧ B

A ∨ B

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

Explanation

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