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If \mathrm{p}, \mathrm{q}$ and $\mathrm{r} three propositions, then which of the following combination of truth values of \mathrm{p}, \mathrm{q}$ and $\mathrm{r} makes the logical expression\{(p \vee q) \wedge((\sim p) \vee r)\} \rightarrow((\sim q) \vee r) false?


 

Option: 1

\mathrm{p}=\mathrm{T}, \mathrm{q}=\mathrm{T}, \mathrm{r}=\mathrm{F}
 


Option: 2

\mathrm{p}=\mathrm{T}, \mathrm{q}=\mathrm{F}, \mathrm{r}=\mathrm{T}
 


Option: 3

\mathrm{p}=\mathrm{F}, \mathrm{q}=\mathrm{T}, \mathrm{r}=\mathrm{F}
 


Option: 4

\mathrm{p}=\mathrm{T}, \mathrm{q}=\mathrm{F}, \mathrm{r}=\mathrm{F}


Answers (1)

best_answer

\begin{aligned} & \quad(\mathrm{p} \vee \mathrm{q}) \vee(\sim \mathrm{p}) \vee \mathrm{r}) \rightarrow((\sim \mathrm{q}) \vee \mathrm{r}) \\ \end{aligned}

\begin{aligned} & \mathrm{T} \rightarrow \mathrm{F} \equiv \mathrm{F} \\ \end{aligned}

\begin{aligned} & \therefore(\mathrm{p} \vee \mathrm{q}) \wedge((\sim \mathrm{p}) \vee \mathrm{r}) \equiv \mathrm{T} \\ \end{aligned}            .................(1)

\begin{aligned} & (\sim \mathrm{q}) \vee \mathrm{r} \equiv \mathrm{F} \\ \end{aligned}                                       ..................(2)

\begin{aligned} & \Rightarrow \sim \mathrm{q}=\mathrm{F}, \mathrm{r}=\mathrm{F} \\ \end{aligned}

\begin{aligned} & \Rightarrow \mathrm{q}=\mathrm{T} \\ \end{aligned}

\begin{aligned} & \text { From }(1) \mathrm{p} \vee \mathrm{q} \equiv \mathrm{T} \\ \end{aligned}

\begin{aligned} & \sim \mathrm{p} \vee \mathrm{r} \equiv \mathrm{T} \\ & \therefore \mathrm{r}=\mathrm{F} \\ & \Rightarrow \sim \mathrm{p}=\mathrm{T} \\ & \Rightarrow \mathrm{p}=\mathrm{F} \end{aligned}

\mathrm{\therefore p=F,q=T,r=F}

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