prove the following rules if True or False using Boolean algebra
(a)A+AB = A+B
(b)(A+B)(A+C) = A+BC
To prove or disprove the given rules using Boolean algebra, we need to simplify both sides of the equation and check if they are equivalent.
(a) A + AB = A + B
To simplify the left-hand side (LHS):
A + AB
= A(1 + B) (Distributive Law)
= A (Identity Law)
To simplify the right-hand side (RHS):
A + B
Since A = A + 0 (Identity Law), then:
A + B
= A + 0 + B (Identity Law)
= A(1 + 0) + B (Distributive Law)
= A + B (Identity Law)
Therefore, we have simplified the LHS and RHS to A, which means that the equation A + AB = A + B is true in Boolean algebra.
(b) (A + B)(A + C) = A + BC
To simplify the left-hand side (LHS):
(A + B)(A + C)
= A(A + C) + B(A + C) (Distributive Law)
= AA + AC + BA + BC (Distributive Law)
= A + AC + BA + BC (Idempotent Law)
= A(1 + C) + B(1 + C) (Distributive Law)
= A + AC + B + BC (Distributive Law)
To simplify the right-hand side (RHS):
A + BC
Now, we need to convert A + BC to a multiplication expression:
A + BC
= A(1 + BC) (Distributive Law)
= A (Identity Law)
We have simplified the RHS to A, and the earlier step simplified the LHS to A + AC + B + BC. Since these two expressions are not equivalent, the equation (A + B)(A + C) = A + BC is false in Boolean algebra.
Therefore, we have proven (a) to be true and (b) to be false using Boolean algebra.