Discussion Forum

1)

Let  $a: \sim (p \wedge \sim r)\vee (\sim q \vee s)$   and 

$b: \ (p \vee s)\leftrightarrow ( q \wedge r)$ .If the truth  values of p and q are true and that of r and s are false, then the truth values of a and b  be respectively...........

Answer:

Option A

Explanation:

 Key Idea  Use   $p\rightarrow q= \sim p \vee q$

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

 Given,  $p,q \rightarrow T$ and $r,s \rightarrow F$

 $\therefore$    $a:\sim (p \wedge \sim r)\vee (\sim q \vee s)$

 $\equiv \sim(T \wedge T)\vee(F\vee F)$

$\equiv \sim(T) \vee(F)$

$\equiv F \vee F= F$

 and  $b: \ (p \vee s)\leftrightarrow ( q \wedge r)$

$\equiv  (\sim (p \vee s)\vee(q\wedge r))\wedge ((p \vee s)\vee\sim (q \wedge r))$

$\because p\leftrightarrow q \equiv(\sim p \vee q)\wedge(p \vee \sim q)$

$\equiv(\sim (T \vee F)\vee (T \wedge F))\wedge(( T \vee F)\vee \sim(T \wedge F))$

$\equiv( F \vee F)\wedge(T \vee T)$

 $\equiv F \wedge T\equiv F$