Prove that \( p \Rightarrow q = \bar{p} + q \).
Prove that \( p \Rightarrow q = \bar{p} + q \).
To prove this equivalence, a truth table is constructed to compare the truth values of the conditional expression \( p \Rightarrow q \) and the expression \( \bar{p} + q \) for all possible combinations of \( p \) and \( q \).
Truth Table for Identity Proof:
| \( p \) | \( q \) | \( \bar{p} \) | \( p \Rightarrow q \) | \( \bar{p} + q \) |
|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 |
Conclusion:
From the truth table above, it is observed that the columns for \( p \Rightarrow q \) and \( \bar{p} + q \) possess the same truth set: (1, 1, 0, 1).
Therefore, it is proved that: \( p \Rightarrow q = \bar{p} + q \).
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