If the Boolean expression

Question:

If the Boolean expression $(\mathrm{p} \wedge \mathrm{q}) \circledast(\mathrm{p} \otimes \mathrm{q})$ is a tautology, then $\circledast$ and $\otimes$ are respectively given by

  1. $\rightarrow, \rightarrow$

  2. $\wedge, \vee$

  3. $\vee, \rightarrow$

  4. $\wedge, \rightarrow$


Correct Option: 1

Solution:

Option (1)

$(\mathrm{p} \wedge \mathrm{q}) \longrightarrow(\mathrm{p} \rightarrow \mathrm{q})$

$=\sim(\mathrm{p} \wedge \mathrm{q}) \vee(\sim \mathrm{p} \vee \mathrm{q})$

$=(\sim \mathrm{p} \vee \sim \mathrm{q}) \vee(\sim \mathrm{p} \vee \mathrm{q})$

$=\sim \mathrm{p} \vee(\sim \mathrm{q} \vee \mathrm{q})$

$=\sim \mathrm{p} \vee \mathrm{t}$

$=\mathrm{t}$

Option (2)

$(\mathrm{p} \wedge \mathrm{q}) \wedge(\mathrm{p} \vee \mathrm{q})=(\mathrm{p} \wedge \mathrm{q}) \quad($ Not a tautology $)$

Option (3)

$(\mathrm{p} \wedge \mathrm{q}) \vee(\mathrm{p} \rightarrow \mathrm{q})$

$=(\mathrm{p} \wedge \mathrm{q}) \vee(\sim \mathrm{p} \vee \mathrm{q})$

$=\sim \mathrm{p} \vee \mathrm{q}$

(Not a tautology)

Option (4)

$=(\mathrm{p} \wedge \mathrm{q}) \wedge(\sim \mathrm{p} \vee \mathrm{q})$

$=\mathrm{p} \wedge \mathrm{q}$

(Not a tautology)

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