Boolean logic questions and answers

Answers for the Boolean expressions simplicity:

1. A+AB=A(1+B)
due to 1+B=1 ( Null Law)
So, A(1+B)=1A=A (Identity Law)

2. 𝐴𝐵 + 𝐴𝐵̅ = 𝐴(𝐵 + 𝐵̅)
Due to 𝐵 + 𝐵̅ = 1 (Inverse Law)
So, 𝐴(𝐵 + 𝐵̅) = 1 ∙ 𝐴 = 𝐴

3. 𝐴𝐵 + 𝐴(𝐶𝐷 + 𝐶𝐷 ̅ )= 𝐴𝐵 + 𝐴𝐶(𝐷 + 𝐷
̅)
̅ = 1 (Inverse Law)
Due to 𝐷 + 𝐷
So, 𝐴𝐵 + 𝐴𝐶(𝐷 + 𝐷 ̅ ) = 𝐴𝐵 + 𝐴𝐶

4. (𝐵𝐶̅ + 𝐴̅𝐷 )(𝐴𝐵̅ + 𝐶𝐷 ̅ ) = (𝐵𝐶̅ ) ∙ (𝐴𝐵̅) + (𝐵𝐶̅ ) ∙ (𝐶𝐷
̅ ) + (𝐴̅𝐷 ) ∙ (𝐴𝐵̅) + (𝐴̅𝐷 ) ∙ (𝐶𝐷̅)
(distributive law)
(𝐵𝐶̅ ) ∙ (𝐴𝐵̅)=(𝐴𝐶̅ ) ∙ (𝐵𝐵̅) = (𝐴𝐶̅ ) ∙ 0 = 0 (commutative law and null law)
(𝐵𝐶̅ ) ∙ (𝐶𝐷̅ )=(𝐶𝐶̅ ) ∙ (𝐵𝐷̅ )=0 ∙ (𝐵𝐷 ̅ )=0 (commutative law and null law)
(𝐴̅𝐷 ) ∙ (𝐴𝐵̅)=(𝐴̅𝐴) ∙ (𝐷𝐵̅)=0 ∙ (𝐷𝐵̅)=0 (commutative law and null law)
(𝐴̅𝐷 ) ∙ (𝐶𝐷̅ ) = (𝐴̅𝐶 ) ∙ (𝐷𝐷 ̅ )=(𝐴̅𝐶 ) ∙ 0=0 (commutative law and null law)
So, (𝐵𝐶̅ + 𝐴̅𝐷 )(𝐴𝐵̅ + 𝐶𝐷 ̅ ) = (𝐵𝐶̅ ) ∙ (𝐴𝐵̅) + (𝐵𝐶̅ ) ∙ (𝐶𝐷̅ ) + (𝐴̅𝐷 ) ∙ (𝐴𝐵̅) + (𝐴̅𝐷 ) ∙ (𝐶𝐷
̅)
=0+0+0+0=0

5. ̅̅̅̅̅̅̅̅̅̅̅̅
𝐴(𝐴 + 𝐶)=𝐴̅ + ̅̅̅̅̅̅̅̅
𝐴 + 𝐶 (De Morgan’s law)
Due to ̅̅̅̅̅̅̅̅
𝐴 + 𝐶 = 𝐴̅𝐶̅ (De Morgan’s law)
̅̅̅̅̅̅̅̅̅̅̅̅
𝐴(𝐴 + 𝐶)=𝐴̅ + ̅̅̅̅̅̅̅̅
𝐴 + 𝐶 = 𝐴̅ + 𝐴̅𝐶̅ =𝐴̅(1 + 𝐶̅ )= 𝐴̅ ∙ 1=𝐴̅

6. ̅̅̅̅̅̅̅̅̅
𝐴 + ̅̅̅̅
𝐵𝐶 = 𝐴̅ ∙ ̅̅̅̅
𝐵𝐶 = 𝐴̅𝐵𝐶 (De Morgan’s law)

7. 𝐴̅𝐵𝐶 + 𝐴𝐶 = (𝐴̅𝐵 + 𝐴)𝐶 = (𝐴̅ + 𝐴)(𝐴 + 𝐵)𝐶 (distributive law)
𝐷𝑢𝑒 𝑡𝑜 (𝐴̅ + 𝐴) = 1 (Inverse law)
So, (𝐴̅ + 𝐴)(𝐴 + 𝐵)𝐶 = 1 ∙ (𝐴 + 𝐵)𝐶 = (𝐴 + 𝐵)𝐶

8. 𝐴(𝐵 + 𝐴)(𝐶 + 𝐵)𝐵̅ = (𝐴𝐵 + 𝐴𝐴)(𝐶𝐵̅ + 𝐵𝐵̅) (distributive law)
Due to 𝐴𝐴 = 𝐴 𝐵𝐵̅ = 0 (Idempotent law and inverse law)
(𝐴𝐵 + 𝐴𝐴)(𝐶𝐵̅ + 𝐵𝐵̅) = (𝐴𝐵 + 𝐴)(𝐶𝐵̅ + 0) = (𝐴𝐵 + 𝐴)𝐶𝐵̅
𝐴𝐵 + 𝐴 = 𝐴 (please refer to question 1)
So, (𝐴𝐵 + 𝐴)𝐶𝐵̅ = 𝐴𝐶𝐵̅

9. 𝐴𝐵 + 𝐵𝐶̅ + 𝐴𝐶= 𝐵𝐶̅ + 𝐴𝐶 (directly applying the formula on slide 9)

10. 𝐴(𝐵 + 𝐴(𝐵 + 𝐶𝐵̅))
Due to 𝐵 + 𝐶𝐵̅ = (𝐵 + 𝐵̅)(𝐵 + 𝐶) = 𝐵 + 𝐶 (distributive law and inverse law)
So, 𝐴(𝐵 + 𝐴(𝐵 + 𝐶𝐵̅)) = 𝐴(𝐵 + 𝐴(𝐵 + 𝐶))
And 𝐴(𝐵 + 𝐴(𝐵 + 𝐶)) = 𝐴𝐵 + 𝐴𝐴(𝐵 + 𝐶) (distributive law)
= 𝐴𝐵 + 𝐴(𝐵 + 𝐶) (Idempotent law)
= 𝐴𝐵 + 𝐴𝐵 + 𝐴𝐶 (distributive law)