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Mathematical Reasoning
normal
$(p\rightarrow q) \leftrightarrow (q \vee ~ p)$ is
A
equivalent to $p \wedge q$
B
Tautology
C
Fallacy
D
Neither tautology nor fallacy
Solution
| $p$ | $ \sim p$ | $q$ | $p \to q$ | $q \vee \sim p$ | $\left( {p \to q} \right) \leftrightarrow \left( {q \vee \sim p} \right)$ |
| $T$ | $F$ | $T$ | $T$ | $T$ | $T$ |
| $T$ | $F$ | $F$ | $F$ | $F$ | $T$ |
| $F$ | $T$ | $T$ | $T$ | $T$ | $T$ |
| $F$ | $T$ | $F$ | $T$ | $T$ | $T$ |
Standard 11
Mathematics
