Question

If $S^{\ast }(p,q,r)$ is the dual of the compound statement $S(p,q,r)$ and $S(p,q,r)=\sim p\wedge [\sim (q\vee r)]$ then $S^{\ast }(\sim p,\sim q,\sim r)$ is equivalent to -

• A. $S(p,q,r)$
• B. $\sim S(\sim p,\sim q,\sim r)$
• C. $\sim S(p,q,r)$
• D. $S^{\ast }(p,q,r)$

Answer 1

Answer: (C)

$S^{\ast }(p,q,r)=\sim p\vee [\sim (q\wedge r)]$
$S^{\ast }(\sim p,\sim q,\sim r)=\sim (\sim p)\vee [\sim (\sim q\wedge \sim r)]$
$=p\vee [q\vee r]$
$\sim S(p,q,r)=\sim [\sim p\wedge [\sim (q\vee r)]]$
$=\sim (\sim p)\vee \sim [\sim (q\vee r)]=p\vee (q\vee r)$
$=S^{\ast }(\sim p,\sim q,\sim r)$