0
19views
Find $\operatorname{dnf}$ of $(p \rightarrow(q \wedge r)) \wedge(\sim p \rightarrow(\sim p \wedge \sim r))$ by truth table method.
0
0views

Solution:

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \mathbf{p} & \mathbf{q} & \mathbf{r} & \sim \mathbf{p} & \sim \mathbf{r} & \mathbf{q} \wedge \mathbf{r} & \mathbf{p} \rightarrow \mathbf{q} \wedge \mathbf{r} & \sim \mathbf{p} \wedge \sim \mathbf{r} & \sim \mathbf{p} \rightarrow(\sim \mathbf{p} \wedge \sim \mathbf{r}) & (\mathbf{p} \rightarrow \mathbf{q} \wedge \mathbf{r}) ( \wedge( \sim \mathbf{p} \rightarrow(\sim \mathbf{p} \wedge \sim \mathbf{r})) \\ \hline \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \text { T} \\\\ \hline \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\\\ \hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\\\ \hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} \\\\ \hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\\\ \hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{F} \\\\ \hline \text { F } & \text { T } & \text { F } & \text { T } & \text { T } & \text { F } & \text { T } & \text { T } & \text { T } & \text { T}\\\\ \hline \text { F } & \text { F } & \text { F } & \text { T } & \text { T } & \text { F } & \text { T } & \text { T } & \text { T} & \text { T} \\\\ \hline \end{array}$$ Hence the logically equivalent form is $$(p \wedge q \wedge r) \vee\left(p^{\prime} \wedge q \wedge r^{\prime}\right) \vee\left(p^{\prime} \wedge q^{\prime} \wedge r^{\prime}\right)$$