Advanced Boolean Expression Simplification Quiz

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

Factor out B: (A ∨ B) ∧ (A' ∨ B) = B ∨ (A ∧ A'). Since A ∧ A' = 0 by complement law, then B ∨ 0 = B by identity law.

First, apply De Morgan's law: ¬(A ∧ B) = ¬A ∨ ¬B. Then, (¬A ∨ ¬B) ∧ A = (¬A ∧ A) ∨ (¬B ∧ A) = 0 ∨ (A ∧ ¬B) = A ∧ ¬B, which is A ∧ B'.

By idempotent law, (A ∧ B) ∨ (A ∧ B) is equivalent to (A ∧ B). Therefore, the expression simplifies to (A ∧ B) ∨ C.

First, apply distributive law: B ∧ (A ∨ C) = (B ∧ A) ∨ (B ∧ C). Then, A ∨ [(B ∧ A) ∨ (B ∧ C)] = A ∨ (B ∧ A) ∨ (B ∧ C). By absorption, A ∨ (B ∧ A) = A, so the expression simplifies to A ∨ (B ∧ C).

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

First, apply De Morgan's law: ¬(A ∨ B) = ¬A ∧ ¬B. Then, (¬A ∧ ¬B) ∨ A. Using distributive law, this is (¬A ∨ A) ∧ (¬B ∨ A) = 1 ∧ (A ∨ ¬B) = A ∨ ¬B.

By absorption law, A ∧ (A ∨ B) = A. Then, A ∧ (A ∨ C) = A. Therefore, the entire expression simplifies to A.

First, combine the first two terms: (A ∧ B) ∨ (A' ∧ B) = B ∧ (A ∨ A') = B ∧ 1 = B. Then, B ∨ (A ∧ B') = (B ∨ A) ∧ (B ∨ B') = (A ∨ B) ∧ 1 = A ∨ B. Thus, the expression simplifies to A ∨ B.

Factor out C: [ (A ∧ B) ∨ (A ∧ B') ∨ (A' ∧ B) ] ∧ C. Now, (A ∧ B) ∨ (A ∧ B') = A ∧ (B ∨ B') = A ∧ 1 = A. Then, A ∨ (A' ∧ B) = (A ∨ A') ∧ (A ∨ B) = 1 ∧ (A ∨ B) = A ∨ B. So overall, the expression is (A ∨ B) ∧ C.

To simplify ¬((A ∧ B) ∨ (A ∧ C)), we apply De Morgan's law to the outer negation, which gives ¬(A ∧ B) ∧ ¬(A ∧ C). Then, we apply De Morgan's law to each inner negation: ¬(A ∧ B) becomes ¬A ∨ ¬B, and ¬(A ∧ C) becomes ¬A ∨ ¬C. Therefore, the expression simplifies to (¬A ∨ ¬B) ∧ (¬A ∨ ¬C).

First, simplify A ∧ (A' ∨ B) using the distributive law: A ∧ (A' ∨ B) = (A ∧ A') ∨ (A ∧ B) = 0 ∨ (A ∧ B) = A ∧ B. Then, take this result and combine it with (A' ∨ C): (A ∧ B) ∧ (A' ∨ C) = (A ∧ B ∧ A') ∨ (A ∧ B ∧ C) = 0 ∨ (A ∧ B ∧ C) = A ∧ B ∧ C. Thus, the entire expression simplifies to A ∧ B ∧ C.

We can factor out A ∨ B from both terms: (A ∨ B) ∨ C ∧ (A ∨ B) ∨ C' = (A ∨ B) ∨ (C ∧ C'). Since C ∧ C' = 0 by the complement law, the expression becomes (A ∨ B) ∨ 0 = A ∨ B by the identity law.

First, apply De Morgan's law to ¬(¬A ∧ B): ¬(¬A ∧ B) = ¬¬A ∨ ¬B = A ∨ ¬B. Then, the expression becomes (A ∨ ¬B) ∨ (A ∧ B). Now, simplify (A ∨ ¬B) ∨ (A ∧ B). Since A ∨ ¬B is true whenever A is true or B is false, and A ∧ B is true only when both A and B are true, but if A is true, A ∨ ¬B is already true, so the addition of A ∧ B does not change the outcome. Therefore, the expression simplifies to A ∨ ¬B.

Factor out (A ∨ B) from both terms: (A ∨ B) ∧ (C ∨ C'). Since C ∨ C' = 1 by the complement law, the expression becomes (A ∨ B) ∧ 1 = A ∨ B by the identity law.