(A+B) B+ A (B+C)

Решение:

1. Исходное выражение: (A + B) ⋅ B + A̅ ⋅ (B + C)
2. Раскроем скобки: A ⋅ B + B ⋅ B + A̅ ⋅ B + A̅ ⋅ C
3. По закону идемпотентности: B ⋅ B = B
4. Тогда: A ⋅ B + B + A̅ ⋅ B + A̅ ⋅ C
5. Сгруппируем: (A ⋅ B + A̅ ⋅ B) + B + A̅ ⋅ C
6. Вынесем B: B ⋅ (A + A̅) + B + A̅ ⋅ C
7. По закону исключенного третьего: A + A̅ = 1
8. Тогда: B ⋅ 1 + B + A̅ ⋅ C = B + B + A̅ ⋅ C = B + A̅ ⋅ C

Ответ: B + A̅ ⋅ C