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Mathematical Reasoning
normal
$(p \wedge \, \sim q)\, \wedge \,( \sim p \vee q)$ is :-
A
A contradiction
B
A tautology
C
Either $(A)$ or $(B)$
D
Neither $(A)$ nor $(B)$
Solution
| $p$ | $q$ | $ \sim p$ | ${ \sim q}$ | $\left( {p \wedge \sim q} \right)$ | ${ \sim p \vee q}$ | $\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \vee q} \right)$ |
| $T$ | $T$ | $F$ | $F$ | $F$ | $T$ | $F$ |
| $T$ | $F$ | $F$ | $T$ | $T$ | $F$ | $F$ |
| $F$ | $T$ | $T$ | $F$ | $F$ | $T$ | $F$ |
| $F$ | $F$ | $T$ | $T$ | $F$ | $T$ | $F$ |
Clearly $ – \left( {p \wedge \sim q} \right) \wedge \left( {p \vee \sim q} \right)$ is a contradiction.
Standard 11
Mathematics
