prove the following rules if true or false using Boolean algebra
(a)A+AB = A+B
(b)(A+B)(A+C) = A+BC
A+AB = A(1+B) = A(1) = A
(A+B)(A+C) = A(A+C)+B(A+C) = A+AB+BC = A+BC
To prove or disprove the rules using Boolean algebra, we will begin by simplifying both sides of the equation and then comparing them.
(a) A + AB = A + B
To simplify both sides of the equation, we can use the distributive law of Boolean algebra which states that A + AB = A.
Let's simplify the left-hand side (LHS):
A + AB = A(1 + B) [Using the distributive law]
= A(1) [Since (1 + B) is always 1]
= A [Simplifying A(1) to A]
Now, let's simplify the right-hand side (RHS):
A + B = A [Since A + B simplifies to A]
Since the LHS (A) is equal to the RHS (A), we can conclude that the rule (a) is true using Boolean algebra.
(b) (A + B)(A + C) = A + BC
To simplify both sides of the equation, we can again use the distributive law of Boolean algebra.
Let's simplify the left-hand side (LHS):
(A + B)(A + C) = A(A + C) + B(A + C) [Using the distributive law]
= A + AC + AB + BC [Applying distribution]
= A(1 + C) + B(1 + C) [Factoring out A and B respectively]
= A + AC + B + BC [Simplifying (1 + C) to 1]
= A(1 + C) + BC + B [Rearranging terms]
= A(1) + B(C + 1) [Applying distribution]
= A + B [Simplifying A(1) to A and B(C + 1) to B]
Now, let's simplify the right-hand side (RHS):
A + BC = A + BC [No further simplification possible]
Since the LHS (A + B) is equal to the RHS (A + BC), we can conclude that the rule (b) is true using Boolean algebra.
Hence, both rules (a) and (b) are true according to Boolean algebra.