Minimize the following Boolean expression using Boolean rules: [tex]$\overline{A B+A \bar{B} \bar{C}+\bar{A} B+B C \bar{D}+\overline{B C} \bar{D}}$[/tex]
Answer (2)
The Boolean expression A B + A B ˉ C ˉ + A ˉ B + BC D ˉ + BC D ˉ simplifies to A ˉ B ˉ D + C B ˉ D using Boolean algebra and simplification techniques. This involves applying absorption laws and DeMorgan's theorem throughout the process. The final result is boxed to highlight the minimized expression clearly.
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Simplify the expression inside the negation using Boolean algebra rules.
Apply DeMorgan's Law to the outer negation.
Expand and simplify the expression to its minimal form.
The minimized Boolean expression is A ˉ B ˉ D + C B ˉ D .
Explanation
Understanding the Problem We are given the Boolean expression A B + A B ˉ C ˉ + A ˉ B + BC D ˉ + BC D ˉ and we want to minimize it using Boolean algebra rules.
Rewriting the Expression First, let's simplify the expression inside the negation. We have A B + A B ˉ C ˉ + A ˉ B + BC D ˉ + BC D ˉ .
We can rewrite BC as B ˉ + C ˉ . So, the expression becomes A B + A B ˉ C ˉ + A ˉ B + BC D ˉ + ( B ˉ + C ˉ ) D ˉ .
Simplifying with Boolean Algebra Now, let's distribute the D ˉ in the last term: A B + A B ˉ C ˉ + A ˉ B + BC D ˉ + B ˉ D ˉ + C ˉ D ˉ .
We can group the terms with B : B ( A + A ˉ ) + A B ˉ C ˉ + BC D ˉ + B ˉ D ˉ + C ˉ D ˉ .
Since A + A ˉ = 1 , we have B + A B ˉ C ˉ + BC D ˉ + B ˉ D ˉ + C ˉ D ˉ .
Applying Absorption Law Now, let's use the absorption law: B + A B ˉ C ˉ = B + A C ˉ .
So, the expression becomes B + A C ˉ + BC D ˉ + B ˉ D ˉ + C ˉ D ˉ .
We can use the absorption law again: B + BC D ˉ = B .
So, the expression becomes B + A C ˉ + B ˉ D ˉ + C ˉ D ˉ .
Applying DeMorgan's Law Now, let's consider the entire expression with the outer negation: B + A C ˉ + B ˉ D ˉ + C ˉ D ˉ .
Using DeMorgan's Law, we have B ˉ ⋅ A C ˉ ⋅ B ˉ D ˉ ⋅ C ˉ D ˉ .
This simplifies to B ˉ ⋅ ( A ˉ + C ) ⋅ ( B + D ) ⋅ ( C + D ) .
Expanding and Simplifying Let's expand the expression: B ˉ ( A ˉ + C ) ( B + D ) ( C + D ) = B ˉ ( A ˉ + C ) ( BC + B D + C D + D 2 ) = B ˉ ( A ˉ + C ) ( BC + B D + C D + D ) .
Since BC + B = B , we have B ˉ ( A ˉ + C ) ( B + D + C D ) = B ˉ ( A ˉ + C ) ( B + D ) .
Expanding further, we get B ˉ ( A ˉ B + A ˉ D + BC + C D ) = B ˉ ( A ˉ D + C D ) since A ˉ B = 0 and BC = 0 .
So, we have B ˉ ( D ( A ˉ + C )) = B ˉ D ( A ˉ + C ) = B ˉ D A ˉ + B ˉ D C = A ˉ B ˉ D + C B ˉ D .
Final Result Therefore, the minimized Boolean expression is A ˉ B ˉ D + C B ˉ D .
Examples
Boolean algebra is used in the design of digital circuits in computers. For example, simplifying a Boolean expression can reduce the number of logic gates needed to implement a particular circuit, which can lead to lower cost, faster performance, and reduced power consumption. In everyday life, this translates to more efficient and powerful electronic devices.
