Answer :
Sure, let’s simplify the boolean expression:
[tex]\[ \bar{A} \bar{B} + \bar{A} B + A \bar{B} + A B \][/tex]
### Step-by-Step Solution
1. Identify the terms in the expression:
[tex]\[ \bar{A} \bar{B}, \quad \bar{A} B, \quad A \bar{B}, \quad \text{and} \quad A B \][/tex]
2. Combine terms using boolean rules:
- Notice that the expression covers all possible combinations of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]. This is because [tex]\( \bar{A} \bar{B} \)[/tex] covers the situation when both [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are 0.
- [tex]\( \bar{A} B \)[/tex] covers when [tex]\( A \)[/tex] is 0 and [tex]\( B \)[/tex] is 1.
- [tex]\( A \bar{B} \)[/tex] covers when [tex]\( A \)[/tex] is 1 and [tex]\( B \)[/tex] is 0.
- [tex]\( A B \)[/tex] covers when both [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are 1.
3. Rewrite the terms to reflect that each unique combination of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is captured:
[tex]\[ \left(\bar{A} \bar{B} \right) + \left(\bar{A} B \right) + \left( A \bar{B} \right) + \left( A B \right) \][/tex]
4. Recognize that this expression includes every possible outcome:
- Since we have covered all possible boolean values for [tex]\( A \)[/tex] and [tex]\( B \)[/tex] (0 and 1), the expression essentially covers all cases.
5. Conclude that the expression is always true irrespective of the values of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ \bar{A} \bar{B} + \bar{A} B + A \bar{B} + A B = 1 \][/tex]
Thus, the simplified expression is:
[tex]\[ \boxed{1} \][/tex]
This proves that no matter the values of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], the original expression will always evaluate to true.
[tex]\[ \bar{A} \bar{B} + \bar{A} B + A \bar{B} + A B \][/tex]
### Step-by-Step Solution
1. Identify the terms in the expression:
[tex]\[ \bar{A} \bar{B}, \quad \bar{A} B, \quad A \bar{B}, \quad \text{and} \quad A B \][/tex]
2. Combine terms using boolean rules:
- Notice that the expression covers all possible combinations of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]. This is because [tex]\( \bar{A} \bar{B} \)[/tex] covers the situation when both [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are 0.
- [tex]\( \bar{A} B \)[/tex] covers when [tex]\( A \)[/tex] is 0 and [tex]\( B \)[/tex] is 1.
- [tex]\( A \bar{B} \)[/tex] covers when [tex]\( A \)[/tex] is 1 and [tex]\( B \)[/tex] is 0.
- [tex]\( A B \)[/tex] covers when both [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are 1.
3. Rewrite the terms to reflect that each unique combination of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is captured:
[tex]\[ \left(\bar{A} \bar{B} \right) + \left(\bar{A} B \right) + \left( A \bar{B} \right) + \left( A B \right) \][/tex]
4. Recognize that this expression includes every possible outcome:
- Since we have covered all possible boolean values for [tex]\( A \)[/tex] and [tex]\( B \)[/tex] (0 and 1), the expression essentially covers all cases.
5. Conclude that the expression is always true irrespective of the values of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ \bar{A} \bar{B} + \bar{A} B + A \bar{B} + A B = 1 \][/tex]
Thus, the simplified expression is:
[tex]\[ \boxed{1} \][/tex]
This proves that no matter the values of [tex]\( A \)[/tex] and [tex]\( B \)[/tex], the original expression will always evaluate to true.
