Boolean Simplification and De Morgan’s Theorems Quiz

The expression P AND Q OR R contains three distinct Boolean variables: P, Q, and R. The AND operation is between P and Q, and the result is then ORed with R, but the number of variables is three regardless of the order of operations.

The absorption law states that A OR (A AND B) equals A. This is because if A is true, the OR is true, and if A is false, both A and A AND B are false, so the result is A. Similarly, A AND (A OR B) = A is also an absorption law.

De Morgan’s Theorem for OR states that the negation of A OR B is equivalent to NOT A AND NOT B. This means that for A OR B to be false, both A and B must be false.

Using the absorption law, A OR (A AND B) simplifies to A. This can be derived by considering that if A is true, the expression is true, and if A is false, both terms are false, so the result is A.

Using the absorption law, A AND (A OR B) simplifies to A. This is because if A is true, the AND is true only if A is true, and if A is false, the AND is false, so the result is A.

Using the distributive law of OR over AND, A OR (NOT A AND B) equals (A OR NOT A) AND (A OR B). Since A OR NOT A is always true (1), and 1 AND (A OR B) is A OR B, the simplification results in A OR B.

Using the distributive law of AND over OR, A AND (NOT A OR B) equals (A AND NOT A) OR (A AND B). Since A AND NOT A is always false (0), and 0 OR (A AND B) is A AND B, the simplification results in A AND B.

Using the distributive law of AND over OR, (A OR B) AND (A OR NOT B) equals A OR (B AND NOT B). Since B AND NOT B = 0, we get A OR 0 = A by Identity Law. This is a common simplification pattern.

Using the distributive law of AND over OR, (A AND B) OR (A AND C) equals A AND (B OR C). This is factorizing A out of the two terms.

Using the distributive law, (A AND B) OR (A AND NOT B) equals A AND (B OR NOT B). Since B OR NOT B = 1 (by the complement law), and A AND 1 = A (by the identity law), the expression simplifies to A.

Using De Morgan’s theorem, NOT (A OR NOT B) equals NOT A AND NOT NOT B. Since NOT NOT B is B, the simplification results in NOT A AND B.

The distributive law of OR over AND states that OR can be distributed over AND, so A OR (B AND C) equals (A OR B) AND (A OR C). This is analogous to the distributive property of union over intersection in set theory.

De Morgan’s Theorem for AND states that the negation of A AND B is equivalent to NOT A OR NOT B. This means that for A AND B to be false, either A is false or B is false.

This expression represents the biconditional or equivalence of A and B, denoted as A IFF B. It is true when both A and B are true or both are false, which is the definition of logical equivalence.

To find the dual of a Boolean expression, we swap AND operations with OR operations, and swap 0s with 1s (while leaving variables unchanged). For A + BC + 0, we replace the OR operators (+) with AND operators (·), replace the AND operator (implicit between B and C) with an OR operator, and replace 0 with 1. This gives us A·(B + C)·1 as the dual expression. Note that the final ·1 term is important in the formal dual, though A·(B + C)·1 simplifies to A·(B + C) in practice.