explain this B + A’C + AB’ = B (1) + A’C + AB’

The equation you provided:

\[
B + A'C + AB' = B(1) + A'C + AB'
\]

is an expression of Boolean algebra, and we're going to break it down.

### Step-by-step explanation:

1. **Left-hand side:**
   - \( B \) is one term.
   - \( A'C \) is the second term, where \( A' \) represents the negation of \( A \), and \( C \) is a variable.
   - \( AB' \) is the third term, where \( B' \) represents the negation of \( B \).

   So the expression is the sum (logical OR) of these three terms: \( B + A'C + AB' \).

2. **Right-hand side:**
   - \( B(1) \) simplifies to \( B \), since anything ANDed with 1 in Boolean algebra remains unchanged: \( B \cdot 1 = B \).
   - The other terms, \( A'C \) and \( AB' \), remain the same.

Thus, the equation simplifies to:

\[
B + A'C + AB' = B + A'C + AB'
\]

This shows that the equation is **true** as both sides are equivalent. The step where \( B(1) = B \) simply clarifies that multiplying (ANDing) by 1 does not change the term.

### Conclusion:

This equation is valid in Boolean algebra, and the right-hand side is just a more explicit form of the left-hand side after applying the identity \( B \cdot 1 = B \).

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Would you like more details or have any questions?

Here are 5 related questions:
1. How does negation work in Boolean algebra?
2. What are the common identities used in Boolean algebra simplifications?
3. How can we simplify other complex Boolean expressions?
4. How does the distributive property work in Boolean algebra?
5. Can this equation be simplified further using other Boolean theorems?

**Tip:** In Boolean algebra, multiplying (ANDing) a variable with 1 always leaves the variable unchanged, and ANDing with 0 always results in 0.

explain this B + A'C + AB' = B(1) + A'C + AB'

Explore the step-by-step solution to simplifying the Boolean algebra expression B + A'C + AB' = B(1) + A'C + AB'. This problem demonstrates key concepts such as the identity law and logical operations used in digital logic design.

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