Certainly! Let's simplify the given Boolean expression using algebraic simplification step-by-step.
The original expression is:
[tex]\[ F(A, B, C) = \overline{A} B + B \overline{C} + B C + A \overline{B C} \][/tex]
### Step-by-Step Simplification:
1. Combine like terms:
Notice that the term [tex]\(B\)[/tex] is common in the first three parts of the expression:
[tex]\[ \overline{A} B + B \overline{C} + B C + A \overline{B C} \][/tex]
2. Factor out [tex]\(B\)[/tex]:
[tex]\[ B(\overline{A} + \overline{C} + C) + A \overline{B C} \][/tex]
3. Simplify inside the parentheses:
We know that [tex]\(\overline{C} + C = 1\)[/tex] (because a variable OR'ed with its complement is always 1):
[tex]\[ B(\overline{A} + 1) + A \overline{B C} \][/tex]
Since [tex]\(\overline{A} + 1\)[/tex] is always 1 regardless of the state of [tex]\(A\)[/tex]:
[tex]\[ B \cdot 1 + A \overline{B C} \][/tex]
Simplifying further:
[tex]\[ B + A \overline{B C} \][/tex]
4. Simplify [tex]\(A \overline{B C}\)[/tex]:
Recall from De Morgan's laws and the complement rule that:
[tex]\[ \overline{B C} = \overline{B} + \overline{C} \][/tex]
So:
[tex]\[ A \overline{B C} = A (\overline{B} + \overline{C}) \][/tex]
5. Distribute [tex]\(A\)[/tex]:
[tex]\[ B + A \overline{B} + A \overline{C} \][/tex]
6. Observe redundancy:
Notice that [tex]\(B\)[/tex] already covers some cases where [tex]\(A \overline{B}\)[/tex] and [tex]\(A \overline{C}\)[/tex] might be present.
Let's consider the terms [tex]\(A \overline{B}\)[/tex] and [tex]\(A \overline{C}\)[/tex]:
- If [tex]\(B = 1\)[/tex], [tex]\(B\)[/tex] itself is true, making the whole expression true.
- If [tex]\(B = 0\)[/tex], we need to check [tex]\(A\)[/tex]'s impact on [tex]\(\overline{C}\)[/tex].
7. Test for full simplification:
Consider the simplified Boolean expression. Combining all elements, the simplified form covers all possible truth value permutations beyond the individual terms. We don't actually need the complex terms once [tex]\(B\)[/tex] handles most scenarios.
The final simplified form is therefore:
[tex]\[ F(A, B, C) = A + B \][/tex]
Thus, the minimized Boolean expression [tex]\(F(A, B, C)\)[/tex] is:
[tex]\[ F(A, B, C) = A + B \][/tex]
This concludes our step-by-step algebraic simplification of the given Boolean expression.
