Question

The proposition $\sim (p\vee \sim q)\vee \sim (p\vee q)$ is logically equivalent to

• A. $p$
• B. $q$
• C. $\sim p$
• D. $\sim q$

Answer 1

The correct option is C $\sim p$

$\sim (p\vee \sim q)\vee \sim (p\vee q)$
$=(\sim p\wedge q)\vee (\sim p\wedge \sim q)$
$=\sim p\wedge (q\vee \sim q)$

$=\sim p$

Alternate :
Let $p$ and $q$ substatements represent $A$ and $B$ respectively.
Using sets analogy
$\sim (p\vee \sim q)\vee \sim (p\vee q)$ is same as
$(A\cup B^{\prime }{)}^{{}^{\prime }}\cup (A\cup B{)}^{{}^{\prime }}$

$\left(i\right)(A\cup B^{\prime }{)}^{{}^{\prime }}=$


$\left(ii\right)(A\cup B{)}^{{}^{\prime }}=$


$\left(iii\right)(A\cup B^{\prime }{)}^{{}^{\prime }}\cup (A\cup B{)}^{{}^{\prime }}=$


Clearly, $(A\cup B^{\prime }{)}^{{}^{\prime }}\cup (A\cup B{)}^{{}^{\prime }}=A^{\prime }=\sim p$