Core Boolean Simplification Laws Quiz

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.

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.

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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

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.

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.