Find a simplified expression for bar (F) (the inverse of F) using DeMorgan’s Theorem. You will likely need to apply the theorem more than once. F(A,B,C,D) = bar ((bar (((bar (A)) + B)) * C)) + bar (D) bar (F)(A,B,C,D) =
The correct answer and explanation is :
We are tasked with simplifying the inverse (complement) of the function ( F(A,B,C,D) ), which is given as:
[
F(A,B,C,D) = \overline{(\overline{(\overline{A} + B)} \cdot C)} + \overline{D}
]
Where the overline denotes the NOT operation, and the multiplication represents the AND operation, while the addition represents the OR operation.
To find ( \overline{F(A,B,C,D)} ), we need to apply DeMorgan’s Theorem. This theorem helps us simplify Boolean expressions involving AND and OR operations. It states that:
[
\overline{(A \cdot B)} = \overline{A} + \overline{B}
]
[
\overline{(A + B)} = \overline{A} \cdot \overline{B}
]
We will simplify ( F(A,B,C,D) ) by first addressing the expression inside the overline (complement) and then applying DeMorgan’s law multiple times.
Step 1: Break down the expression of ( F(A,B,C,D) )
We begin with:
[
F(A,B,C,D) = \overline{(\overline{(\overline{A} + B)} \cdot C)} + \overline{D}
]
We can first simplify the inner term, ( \overline{(\overline{A} + B)} ), by applying DeMorgan’s Theorem:
[
\overline{(\overline{A} + B)} = \overline{\overline{A}} \cdot \overline{B} = A \cdot \overline{B}
]
So, now we have:
[
F(A,B,C,D) = \overline{(A \cdot \overline{B} \cdot C)} + \overline{D}
]
Step 2: Apply DeMorgan’s Theorem to ( \overline{(A \cdot \overline{B} \cdot C)} )
Now, we simplify the expression inside the first overline by applying DeMorgan’s Theorem to ( A \cdot \overline{B} \cdot C ):
[
\overline{(A \cdot \overline{B} \cdot C)} = \overline{A} + B + \overline{C}
]
Thus, the expression for ( F(A,B,C,D) ) becomes:
[
F(A,B,C,D) = (\overline{A} + B + \overline{C}) + \overline{D}
]
Step 3: Now, find ( \overline{F(A,B,C,D)} )
Finally, to find ( \overline{F(A,B,C,D)} ), we apply the complement to the entire expression:
[
\overline{F(A,B,C,D)} = \overline{(\overline{A} + B + \overline{C}) + \overline{D}}
]
Now, apply DeMorgan’s Theorem to this expression. First, we rewrite it as:
[
\overline{F(A,B,C,D)} = \overline{(\overline{A} + B + \overline{C})} \cdot \overline{(\overline{D})}
]
Next, simplify each part using DeMorgan’s Theorem. For the first part:
[
\overline{(\overline{A} + B + \overline{C})} = A \cdot \overline{B} \cdot C
]
And for the second part:
[
\overline{(\overline{D})} = D
]
Thus, we have:
[
\overline{F(A,B,C,D)} = (A \cdot \overline{B} \cdot C) \cdot D
]
Final Expression
The simplified expression for ( \overline{F(A,B,C,D)} ) is:
[
\overline{F(A,B,C,D)} = A \cdot \overline{B} \cdot C \cdot D
]
Explanation
DeMorgan’s Theorem is fundamental in simplifying Boolean expressions by transforming ANDs into ORs and vice versa. In this case, we started with a relatively complex Boolean function and systematically applied the theorem to break it down step by step. By doing so, we simplified the expression for ( \overline{F(A,B,C,D)} ) into a much more straightforward form.
This process of applying DeMorgan’s theorem repeatedly is crucial when dealing with complex Boolean logic in digital circuit design or in simplifying expressions for Boolean algebra problems. The final result shows how the original logic can be represented more compactly, which is often useful in circuit optimization or further algebraic manipulations.