The statement $[(\mathrm{p} \rightarrow \mathrm{q}) \wedge \sim \mathrm{q}] \rightarrow \mathrm{r}$ is a…

The statement $[(\mathrm{p} \rightarrow \mathrm{q}) \wedge \sim \mathrm{q}] \rightarrow \mathrm{r}$ is a tautology, when $\mathrm{r}$ is equivalent to
  1. $\mathrm{p} \wedge \sim \mathrm{q}$
  2. $q \vee p$
  3. $\mathrm{p} \wedge \mathrm{q}$
  4. $\sim q$

Solution

\begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{p} & \mathrm{q} & \mathrm{r} & \mathrm{p} \rightarrow \mathrm{q} & \sim \mathrm{q} & (\mathrm{p} \rightarrow \mathrm{q}) \wedge \sim \mathrm{q} & {[(\mathrm{p} \rightarrow \mathrm{q}) \wedge \sim \mathrm{q}] \rightarrow \mathrm{r}} \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\ \hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\ \hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} \\ \hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\ \hline \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} \\ \hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{T} \\ \hline \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} \\ \hline \end{array} $\therefore \quad[(p \rightarrow q) \wedge \sim q] \rightarrow r$ is a tautology when all the entries in the last column are $T$, which is only possible when $r \equiv \sim q$

Asked in: MHT CET 2023 (14 May Shift 1)

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