prove the following logic rules using truth tables (ensure that one side of the expression truth table matches the other side of the expression truth table)
a) R+ RT = R+T*
b) F+FG = F+G**
*RT has a not line over R
** the two Fs has not line over the two Fs
To prove the given logic rules using truth tables, we need to evaluate both sides of each expression and compare their truth tables to ensure they match.
a) R + ~R · T = R + T*
We will construct the truth table for both sides of this expression and check if they match.
First, let's break down the given expression:
1. R + ~R · T: This is an OR operation between R and (~R · T), where ~ represents the negation (NOT) operator.
Now, we can construct the truth table for both sides:
| R | T | ~R | ~R · T | R + ~R · T | R + T* |
|---|---|----|-------|------------|-------|
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 |
Based on the truth table, we can see that both sides of the expression match, as the output values are the same for all combinations of input values. Therefore, we have proven the logic rule:
R + ~R · T = R + T*
b) F + ~F · G · G: This is an OR operation between F and (~F · G · G), where ~ represents the negation (NOT) operator.
Now, let's construct the truth table for both sides:
| F | G | ~F | ~F · G · G | F + ~F · G · G | F + G** |
|---|---|----|----------|---------------|--------|
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 |
From the truth table, we can see that both sides of the expression match, as the output values are the same for all combinations of input values. Hence, we have proven the logic rule:
F + ~F · G · G = F + G**.