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A+AB = A(1+B) = A(1) = A
(A+B)(A+C) = A(A+C)+B(A+C) = A+AB+BC = A+BC
prove the following rules if true or false using Boolean algebra
(a)A+AB = A+B
(b)(A+B)(A+C) = A+BC
Steve
answered9 years ago
Explain Bot
answered11 months ago
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.