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Verify if the following proposition is valid using the truth table:
Verify if the following proposition is valid using the truth table:
$(X \wedge Y) \Rightarrow Z = (Y \Rightarrow Z) \wedge (X \Rightarrow Y)$
Answer
Answer
Official answer keyTruth table:
The LHS column $(X \wedge Y) \Rightarrow Z$ and the RHS column $(Y \Rightarrow Z) \wedge (X \Rightarrow Y)$ differ in three rows (e.g. X=0,Y=1,Z=0 gives LHS=1 but RHS=0; also X=1,Y=0,Z=0 and X=1,Y=0,Z=1). Since the two sides are not identical for every row, the given proposition is INVALID (it is not a tautology).
| X | Y | Z | X.Y | (X.Y)=>Z | Y=>Z | X=>Y | (Y=>Z).(X=>Y) |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
From ISC 2025 Specimen Computer Science Paper 1, question 5(ii).