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Mathematical Reasoning
normal
The statement $[(p \wedge q) \rightarrow p] \rightarrow (q \wedge \sim q)$ is
A
tautology
B
contradiction
C
open statement
D
neither tautology nor contradiction
Solution
| $p$ | $q$ | $p \wedge q$ | $\left( {p \wedge q} \right) \to p$ | $ \sim q$ | $q \wedge \sim q$ | $\left[ {\left( {p \wedge q} \right) \to p} \right] \to \left( {q \wedge \sim q} \right)$ |
| $T$ | $T$ | $T$ | $T$ | $F$ | $F$ | $F$ |
| $T$ | $F$ | $F$ | $T$ | $T$ | $F$ | $F$ |
| $F$ | $T$ | $F$ | $T$ | $F$ | $F$ | $F$ |
| $F$ | $F$ | $F$ | $T$ | $T$ | $F$ | $F$ |
Given compound statement is always false. So it is a contradiction.
Standard 11
Mathematics
