(Digital logic)
using truth table
Demonstrate that:
X’. Y + Y’. Z + X.Z’ = X.Y’ + Y.Z’ + X’.Z
To demonstrate the equivalence of the given expression using a truth table, we can create a truth table for both sides of the equation and compare the results. Here's how:
Step 1: Identify the variables
The given expression consists of three variables: X, Y, and Z.
Step 2: Set up the truth table
Create a table with columns for X, Y, Z, and the expressions on both sides of the equation. In this case, we have:
Truth Table:
| X | Y | Z | X’.Y + Y’.Z + X.Z’ | X.Y’ + Y.Z’ + X’.Z |
|---|---|---|---------------------|-------------------|
| 0 | 0 | 0 | | |
| 0 | 0 | 1 | | |
| 0 | 1 | 0 | | |
| 0 | 1 | 1 | | |
| 1 | 0 | 0 | | |
| 1 | 0 | 1 | | |
| 1 | 1 | 0 | | |
| 1 | 1 | 1 | | |
Step 3: Evaluate the expressions for each row
Fill in the truth table by evaluating both expressions for each combination of X, Y, and Z.
For the left-hand side expression (X’.Y + Y’.Z + X.Z’):
- X' is the complement of X.
- Y' is the complement of Y.
- Z' is the complement of Z.
So, the left-hand side expression becomes: (X' . Y) + (Y' . Z) + (X . Z')
For the right-hand side expression (X.Y’ + Y.Z’ + X’.Z):
- X' is the complement of X.
- Y' is the complement of Y.
- Z' is the complement of Z.
So, the right-hand side expression becomes: (X . Y') + (Y . Z') + (X' . Z)
Evaluate both expressions for each row of the truth table.
| X | Y | Z | X’.Y + Y’.Z + X.Z’ | X.Y’ + Y.Z’ + X’.Z |
|---|---|---|---------------------|-------------------|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 |
Step 4: Compare the results
After evaluating both expressions, we can see that the results for each row are equal for both sides of the equation. Therefore, we can conclude that X'.Y + Y'.Z + X.Z' is equivalent to X.Y' + Y.Z' + X'.Z.
This demonstrates the equivalence of the given expression using a truth table.